Conditional Probability and Bayes' Theorem for CUET Maths

Q: What is conditional probability?

The probability of A given that B has happened is P(A B) = (P(A B))/(P(B)), P(B) > 0. Knowing B shrinks the sample space to B. Within it, the usual rules hold: P(A' B) = 1 - P(A B).

Q: How do you find the probability that two events both happen?

P(A B) = P(A) P(B A). In draws without replacement, the second probability is worked out after the first draw has changed what is left.

Q: When are two events independent?

When knowing one does not change the chance of the other: P(A B) = P(A) P(B), or P(A B) = P(A). For independent events, the probability that at least one happens is 1 - P(A')P(B').

Q: How do you find a probability through several cases?

If E₁, E₂,, Eₙ are mutually exclusive and together cover the sample space, then P(A) = P(E₁)P(A E₁) + P(E₂)P(A E₂) + + P(Eₙ)P(A Eₙ).

Q: What does Bayes' theorem find?

The probability of a case given the outcome: P(Eᵢ A) = (P(Eᵢ)P(A Eᵢ))/(Σⱼ P(Eⱼ)P(A Eⱼ)). The denominator is the total probability of A.

About 5 min to learn · 125 questions

Section A of CUET Maths asks simple probability; Section B1 asks what changes when you already know something. Conditional probability, independent events, total probability and Bayes' theorem bring three or four questions a paper, and each follows a short formula. Learn each rule here, from conditional probability and the multiplication theorem to independent events, the theorem of total probability and Bayes' theorem, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section B1: Mathematics

On this page · 11 sections

01Probability with something known
02Conditional probability
03The multiplication theorem
04Independent events
05The theorem of total probability
06Bayes' theorem
07How CUET asks this chapter
08Exam shortcuts
09Common mistakes
10Quick revision
11Check your understanding

Probability with something known

In a college, 60% of students come by bus and 40% walk. A tenth of the bus riders and a twentieth of the walkers arrive late. A student walks in late. How likely is it that they came by bus? Knowing the student is late changes the odds, and this chapter is about exactly that: probability once something is known.

Conditional probability

What is conditional probability?

The probability of $A$ given that $B$ has happened is

P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0.

Knowing $B$ shrinks the sample space to $B$ . Within it, the usual rules hold: $P(A' \mid B) = 1 - P(A \mid B)$ .

Example 1 · Counting inside the condition

A die is thrown. Given that the number is odd, what is the probability that it is prime?

The odd outcomes are $1, 3, 5$ .
Of these, 3 and 5 are prime.

Answer

\frac{2}{3}

Watch out

$P(A \mid B)$ and $P(B \mid A)$ divide the same $P(A \cap B)$ by different numbers. Read which event is given.

The multiplication theorem

How do you find the probability that two events both happen?

$P(A \cap B) = P(A)\,P(B \mid A)$ . In draws without replacement, the second probability is worked out after the first draw has changed what is left.

Example 2 · Two kings

Two cards are drawn without replacement from a pack of 52. What is the probability that both are kings?

$P(\text{first a king}) = \frac{4}{52}$ ; then 3 kings remain in 51 cards.
$\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}$ .

Answer

\frac{1}{221}

Independent events

When are two events independent?

When knowing one does not change the chance of the other: $P(A \cap B) = P(A)\,P(B)$ , or $P(A \mid B) = P(A)$ . For independent events, the probability that at least one happens is $1 - P(A')P(B')$ .

Key idea

Mutually exclusive and independent are opposites, not partners. If $A$ and $B$ cannot happen together and both have positive probability, knowing one rules the other out, so they are not independent.

The theorem of total probability

How do you find a probability through several cases?

If $E_1, E_2, \ldots, E_n$ are mutually exclusive and together cover the sample space, then

P(A) = P(E_1)P(A \mid E_1) + P(E_2)P(A \mid E_2) + \cdots + P(E_n)P(A \mid E_n).

Example 3 · Late students

With 60% by bus (10% late) and 40% walking (5% late), what fraction of students arrive late?

$0.6 \times 0.1 + 0.4 \times 0.05$ .
$0.06 + 0.02 = 0.08$ .

Answer

Bayes' theorem

What does Bayes' theorem find?

The probability of a case given the outcome:

P(E_i \mid A) = \frac{P(E_i)P(A \mid E_i)}{\sum_j P(E_j)P(A \mid E_j)}.

The denominator is the total probability of $A$ .

Example 4 · The late student

A student arrives late. What is the probability that they came by bus?

$P(\text{bus} \cap \text{late}) = 0.06$ and $P(\text{late}) = 0.08$ .
$\frac{0.06}{0.08} = \frac{3}{4}$ .

Answer

\frac{3}{4}

How CUET asks this chapter

Conditional probability: from given probabilities, or by counting dice, coins and cards inside a condition.
Multiplication theorem: draws without replacement, one after another.
Independence: testing it, or using it for "both", "neither" and "at least one".
Total probability and Bayes: bags of balls, machines and their defects, tests and their errors.

Exam shortcuts

Given $B$ : count only inside $B$ .
At least one of independent events: $1 - \prod P(\text{fail})$ .
Bayes: the case's share of the total probability of the outcome.
A rare condition with an imperfect test: most positives are false alarms.

Common mistakes

Swapping $P(A \mid B)$ and $P(B \mid A)$ .
Replacing a card or ball when the question draws without replacement.
Treating mutually exclusive events as independent.
Averaging the cases instead of weighting them by their probabilities.

Quick revision

Conditional

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
$P(A \cap B) = P(A)P(B \mid A)$

Independent

$P(A \cap B) = P(A)P(B)$
At least one: $1 - P(A')P(B')$

Bayes

$P(A) = \sum P(E_i)P(A \mid E_i)$
$P(E_i \mid A) = \frac{P(E_i)P(A \mid E_i)}{P(A)}$

All of Section B1's formulas are on one page: the Mathematics (B1) formula sheet.

Next · 10Quick revision

Check your understanding

125 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Conditional probabilityQ1 of 125Q1

Conditional probability

29 questions

Question 1

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$P(A) = \frac{3}{5}$ , $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{3}{10}$ . What is $P(A \mid B)$ ?

\frac{3}{10}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{5}{6}

Correct answerYour answer

Show the explanationHide the explanation

AnswerC · $\frac{3}{5}$

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , and $P(A' \mid B) = 1 - P(A \mid B)$ .
$P(A \mid B) = \frac{\frac{3}{10}}{\frac{1}{2}} = \frac{3}{5}$ .

Worth noting

$\frac{3}{10}$ is $P(A \cap B)$ itself, before dividing by $P(B)$ .

Question 2

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Two dice are thrown. Given that the sum is 8, what is the probability that both show the same number?

\frac{1}{36}

Correct answerYour answer

\frac{5}{36}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

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AnswerD · $\frac{1}{5}$

Sum 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ , 5 outcomes.
Only $(4, 4)$ is a double: $\frac{1}{5}$ .

Worth noting

$\frac{1}{6}$ is the chance of a double with no condition; knowing the sum is 8 narrows the outcomes to 5.

Question 3

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$P(A) = \frac{7}{13}$ , $P(B) = \frac{9}{13}$ and $P(A \cap B) = \frac{4}{13}$ . What is $P(B \mid A)$ ?

\frac{4}{13}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

\frac{4}{7}

Correct answerYour answer

\frac{7}{9}

Correct answerYour answer

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AnswerC · $\frac{4}{7}$

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , and $P(A' \mid B) = 1 - P(A \mid B)$ .
$P(B \mid A) = \frac{\frac{4}{13}}{\frac{7}{13}} = \frac{4}{7}$ .

Worth noting

$\frac{4}{9}$ is $P(A \mid B)$ ; given $A$ , divide by $P(A)$ .

Question 4

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Two fair dice are thrown. $E$ : the sum is 9. $F$ : at least one die shows a 5.

What is $P(E)$ ?

\frac{1}{12}

Correct answerYour answer

\frac{1}{9}

Correct answerYour answer

\frac{5}{36}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

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AnswerB · $\frac{1}{9}$

Sum 9: $(3, 6), (4, 5), (5, 4), (6, 3)$ .
$\frac{4}{36} = \frac{1}{9}$ .

Worth noting

$\frac{1}{12}$ counts $(4, 5)$ and $(5, 4)$ as one outcome, and $(3, 6)$ , $(6, 3)$ as one.

Question 5

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A family has two children. Given that at least one is a boy, what is the probability that both are boys?

\frac{1}{4}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

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AnswerB · $\frac{1}{3}$

At least one boy: $BB, BG, GB$ , 3 equally likely outcomes.
Only $BB$ has two boys: $\frac{1}{3}$ .

Worth noting

$\frac{1}{2}$ treats it as the chance that the other child is a boy, but "at least one" does not say which child.

Question 6

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A fair die is thrown. Given that the number is odd, what is the probability that it is prime?

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

1

Correct answerYour answer

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AnswerC · $\frac{2}{3}$

Odd: 1, 3, 5.
Of these, 3 and 5 are prime (1 is not): $\frac{2}{3}$ .

Worth noting

1 counts 1 as a prime; a prime has exactly two factors.

Question 7

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$P(A \mid C) = 0.5$ , $P(B \mid C) = 0.4$ and $P(A \cap B \mid C) = 0.2$ . What is $P(A \cup B \mid C)$ ?

0.3

Correct answerYour answer

0.7

Correct answerYour answer

0.8

Correct answerYour answer

0.9

Correct answerYour answer

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AnswerB · $0.7$

Given $C$ , the addition rule holds as usual: $P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C)$ .
$0.5 + 0.4 - 0.2 = 0.7$ .

Worth noting

0.9 adds the two conditional probabilities and so counts the overlap $A \cap B$ twice.

Question 8

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Two dice are thrown. Given that they show different numbers, what is the probability that the sum is 6?

\frac{1}{9}

Correct answerYour answer

\frac{2}{15}

Correct answerYour answer

\frac{5}{36}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

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AnswerB · $\frac{2}{15}$

Different numbers: $36 - 6 = 30$ outcomes.
Sum 6 with different numbers: $(1, 5), (5, 1), (2, 4), (4, 2)$ , so $\frac{4}{30} = \frac{2}{15}$ .

Worth noting

$\frac{5}{36}$ ignores the condition: it keeps $(3, 3)$ and divides by all 36 outcomes.

Question 9

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A group of 60 students has 36 girls and 24 boys. Of the girls, 20 study Physics; of the boys, 8 study Physics. One student is picked at random from the group.

Given that the student is a girl, what is the probability that she studies Physics?

\frac{1}{3}

Correct answerYour answer

\frac{7}{15}

Correct answerYour answer

\frac{5}{9}

Correct answerYour answer

\frac{5}{7}

Correct answerYour answer

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AnswerC · $\frac{5}{9}$

Given a girl, only the 36 girls count.
20 of them study Physics: $\frac{20}{36} = \frac{5}{9}$ .

Worth noting

$\frac{1}{3} = \frac{20}{60}$ divides by all 60 students; it is the chance of picking a girl who studies Physics.

Question 10

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A group of 60 students has 36 girls and 24 boys. Of the girls, 20 study Physics; of the boys, 8 study Physics. One student is picked at random from the group.

Given that the student studies Physics, what is the probability that the student is a girl?

\frac{1}{3}

Correct answerYour answer

\frac{5}{9}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{5}{7}

Correct answerYour answer

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AnswerD · $\frac{5}{7}$

Physics students: $20 + 8 = 28$ .
20 of them are girls: $\frac{20}{28} = \frac{5}{7}$ .

Worth noting

$\frac{5}{9}$ is the chance that a girl studies Physics, the condition the other way round.

Question 11

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Read the two statements and choose the correct option.

I. $P(A \mid A) = 1$ whenever $P(A) > 0$ .

II. $P(A \mid B) = P(B \mid A)$ for any two events.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $\frac{P(A \cap A)}{P(A)} = 1$ . True.
II is false: the two divide the same $P(A \cap B)$ by different numbers, $P(B)$ and $P(A)$ .

Worth noting

Swapping the condition changes the denominator.

Question 12

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$P(A) = \frac{2}{5}$ , $P(B) = \frac{1}{4}$ and $P(A \cap B) = \frac{1}{10}$ . What is $P(A' \mid B)$ ?

\frac{1}{10}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

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AnswerC · $\frac{3}{5}$

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , and $P(A' \mid B) = 1 - P(A \mid B)$ .
$P(A' \mid B) = 1 - \frac{\frac{1}{10}}{\frac{1}{4}} = \frac{3}{5}$ .

Worth noting

$\frac{2}{5}$ is $P(A \mid B)$ ; the complement within $B$ is $1 - \frac{2}{5}$ .

Question 13

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Three fair coins are tossed. Given that at least one shows a head, what is the probability that exactly two show heads?

\frac{2}{7}

Correct answerYour answer

\frac{3}{8}

Correct answerYour answer

\frac{3}{7}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{3}{7}$

At least one head: all 8 outcomes but $TTT$ , so 7.
Exactly two heads: $HHT, HTH, THH$ : $\frac{3}{7}$ .

Worth noting

$\frac{3}{8}$ ignores the condition and divides by all 8 outcomes.

Question 14

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Two dice are thrown. Given that the first die shows a 4, what is the probability that the sum is more than 8?

\frac{1}{6}

Correct answerYour answer

\frac{5}{18}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{1}{3}$

With the first die at 4, the second must be 5 or 6 for a sum above 8.
2 of the 6 outcomes: $\frac{1}{3}$ .

Worth noting

$\frac{5}{18}$ is the chance of a sum above 8 without knowing the first die.

Question 15

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Assertion (A): $P(A' \mid B) = 1 - P(A \mid B)$ .

Reason (R): $P(A \mid B) + P(A' \mid B) = P(S \mid B) = 1$ , since $A$ and $A'$ are disjoint and make up $S$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerA · Both A and R are true, and R is the correct explanation of A.

A is true for any $B$ with $P(B) > 0$ .
R is true and is exactly the reason: rearranging it gives A.

Question 16

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$P(A) = 0.6$ , $P(B) = 0.4$ and $P(A \cup B) = 0.7$ . What is $P(A \mid B)$ ?

0.3

Correct answerYour answer

0.5

Correct answerYour answer

0.6

Correct answerYour answer

0.75

Correct answerYour answer

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AnswerD · $0.75$

$P(A \cap B) = 0.6 + 0.4 - 0.7 = 0.3$ .
$P(A \mid B) = \frac{0.3}{0.4} = 0.75$ .

Worth noting

0.5 is $P(B \mid A) = \frac{0.3}{0.6}$ , dividing by the wrong event.

Question 17

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A number is picked at random from 1 to 20. Given that it is even, what is the probability that it is divisible by 4?

\frac{1}{10}

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerD · $\frac{1}{2}$

Even numbers: 10 of them.
Divisible by 4: 4, 8, 12, 16, 20, so $\frac{5}{10} = \frac{1}{2}$ .

Worth noting

$\frac{1}{4}$ divides by all 20 numbers; the condition leaves only the 10 even ones.

Question 18

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In a class of 40, 25 play cricket, 20 play football and 10 play both. A football player is picked at random. What is the probability that they also play cricket?

\frac{1}{4}

Correct answerYour answer

\frac{3}{8}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerD · $\frac{1}{2}$

Given football, only its 20 players count.
10 of them play cricket: $\frac{10}{20} = \frac{1}{2}$ .

Worth noting

$\frac{2}{5} = \frac{10}{25}$ is the chance that a cricketer plays football, the condition the other way round.

Question 19

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Two fair dice are thrown. $E$ : the sum is 9. $F$ : at least one die shows a 5.

What is $P(E \mid F)$ ?

\frac{1}{18}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

\frac{2}{11}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{2}{11}$

$F$ has 11 outcomes: 6 with the first die 5, 6 with the second die 5, less $(5, 5)$ counted twice.
Of these, $(4, 5)$ and $(5, 4)$ sum to 9: $\frac{2}{11}$ .

Worth noting

$\frac{1}{6} = \frac{2}{12}$ counts $(5, 5)$ twice when listing $F$ .

Question 20

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Two fair dice are thrown. $E$ : the sum is 9. $F$ : at least one die shows a 5.

What is $P(F \mid E)$ ?

\frac{2}{11}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{11}{36}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerD · $\frac{1}{2}$

Given $E$ , only its 4 outcomes remain.
Two of them, $(4, 5)$ and $(5, 4)$ , show a 5: $\frac{2}{4} = \frac{1}{2}$ .

Worth noting

$\frac{2}{11}$ is $P(E \mid F)$ , the condition the other way round.

Question 21

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Read the two statements and choose the correct option.

I. $P(A \mid B) \ge P(A \cap B)$ whenever $P(B) > 0$ .

II. If $P(A) > 0$ , $P(B) > 0$ and $P(A \mid B) = P(A)$ , then $P(B \mid A) = P(B)$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerC · Both I and II

I: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , and dividing by $P(B) \le 1$ cannot make a number smaller. True.
II: $P(A \mid B) = P(A)$ gives $P(A \cap B) = P(A)\,P(B)$ ; dividing by $P(A)$ gives $P(B \mid A) = P(B)$ . True.

Worth noting

If knowing $B$ leaves the chance of $A$ unchanged, then knowing $A$ leaves the chance of $B$ unchanged too.

Question 22

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For events $A$ and $B$ with $P(A) > 0$ and $P(B) > 0$ , which of the following statements are true?

I. If $A \subset B$ , then $P(B \mid A) = 1$ .

II. If $A \subset B$ , then $P(A \mid B) = \frac{P(A)}{P(B)}$ .

III. If $A$ and $B$ are mutually exclusive, then $P(A \mid B) = 0$ .

IV. $P(A \mid B) \ge P(A)$ for any two such events.

I, II and III onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerI and III onlyCorrect answerYour answer

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AnswerA · I, II and III only

I: $A \cap B = A$ , so $P(B \mid A) = \frac{P(A)}{P(A)} = 1$ . II: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$ . III: $P(A \cap B) = 0$ , so $P(A \mid B) = 0$ .
IV is false: for mutually exclusive events $P(A \mid B) = 0$ , which is less than $P(A)$ .

Worth noting

A condition can lower a probability as well as raise it.

Question 23

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A card is drawn at random from a well-shuffled pack of 52. Given that it is a king or a heart, what is the probability that it is the king of hearts?

\frac{1}{17}

Correct answerYour answer

\frac{1}{16}

Correct answerYour answer

\frac{1}{13}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

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AnswerB · $\frac{1}{16}$

Kings or hearts: $4 + 13 - 1 = 16$ cards, with the king of hearts counted once.
Only one of the 16 is the king of hearts: $\frac{1}{16}$ .

Worth noting

$\frac{1}{17}$ counts the king of hearts twice, once among the kings and once among the hearts.

Question 24

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Two dice are thrown. Given that the product of the two numbers is even, what is the probability that their sum is even?

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

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AnswerA · $\frac{1}{3}$

The product is odd only when both numbers are odd, in 9 outcomes, so it is even in $36 - 9 = 27$ .
An even product with an even sum needs both numbers even: 9 outcomes, so $\frac{9}{27} = \frac{1}{3}$ .

Worth noting

$\frac{1}{2}$ is the chance of an even sum with no condition; an even product rules out the 9 odd–odd outcomes, and all of them have even sums.

Question 25

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A group of 60 students has 36 girls and 24 boys. Of the girls, 20 study Physics; of the boys, 8 study Physics. One student is picked at random from the group.

Given that the student does not study Physics, what is the probability that the student is a girl?

\frac{4}{15}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerC · $\frac{1}{2}$

Not Physics: $36 - 20 = 16$ girls and $24 - 8 = 16$ boys, 32 students in all.
16 of the 32 are girls: $\frac{16}{32} = \frac{1}{2}$ .

Worth noting

$\frac{4}{9} = \frac{16}{36}$ is the chance that a girl does not study Physics; the condition here is not studying Physics, so divide by 32.

Question 26

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Two fair dice are thrown. $E$ : the sum is 9. $F$ : at least one die shows a 5.

Are $E$ and $F$ independent?

No:

P(E \cap F) = \frac{1}{18}

but

P(E)\,P(F) = \frac{11}{324}

Correct answerYour answerYes: the two dice are thrown independently of each otherCorrect answerYour answerNo: they are mutually exclusive, so they cannot happen togetherCorrect answerYour answerYes:

P(E \mid F) = P(E) = \frac{1}{9}

, so knowing

F

changes nothingCorrect answerYour answer

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AnswerA · No: $P(E \cap F) = \frac{1}{18}$ but $P(E)\,P(F) = \frac{11}{324}$

$P(E \cap F) = \frac{2}{36} = \frac{1}{18}$ .
$P(E)\,P(F) = \frac{1}{9} \times \frac{11}{36} = \frac{11}{324} \ne \frac{1}{18}$ : not independent.

Worth noting

They share $(4, 5)$ and $(5, 4)$ , so they are not exclusive; and $P(E \mid F) = \frac{2}{11} \ne \frac{1}{9}$ .

Question 27

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$P(A) = 0.5$ , $P(B) = 0.3$ and $P(A \cap B) = 0.2$ . What is $P(A' \mid B')$ ?

\frac{1}{3}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{3}{7}

Correct answerYour answer

\frac{4}{7}

Correct answerYour answer

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AnswerD · $\frac{4}{7}$

$P(A \cup B) = 0.5 + 0.3 - 0.2 = 0.6$ , so $P(A' \cap B') = 1 - 0.6 = 0.4$ .
$P(B') = 1 - 0.3 = 0.7$ , and $P(A' \mid B') = \frac{0.4}{0.7} = \frac{4}{7}$ .

Worth noting

$\frac{1}{3} = 1 - \frac{0.2}{0.3}$ is $P(A' \mid B)$ : it takes the complement of $A$ but keeps $B$ as the condition.

Question 28

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$P(A \mid B) = \frac{1}{2}$ , $P(B \mid A) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{6}$ . What is $P(A \cup B)$ ?

\frac{5}{12}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{7}{12}

Correct answerYour answer

1

Correct answerYour answer

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AnswerA · $\frac{5}{12}$

$P(B) = \frac{P(A \cap B)}{P(A \mid B)} = \frac{1/6}{1/2} = \frac{1}{3}$ and $P(A) = \frac{P(A \cap B)}{P(B \mid A)} = \frac{1/6}{2/3} = \frac{1}{4}$ .
$P(A \cup B) = \frac{1}{4} + \frac{1}{3} - \frac{1}{6} = \frac{5}{12}$ .

Worth noting

$\frac{7}{12} = \frac{1}{4} + \frac{1}{3}$ leaves the overlap in twice; $P(A \cap B)$ must be taken away once.

Question 29

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A group of 60 students has 36 girls and 24 boys. Of the girls, 20 study Physics; of the boys, 8 study Physics. One student is picked at random from the group.

Given that the student studies Physics or is a boy, what is the probability that the student is a girl?

\frac{5}{13}

Correct answerYour answer

\frac{5}{11}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{5}{7}

Correct answerYour answer

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AnswerB · $\frac{5}{11}$

Physics or a boy: $28 + 24 - 8 = 44$ students, the 8 boys who study Physics counted once.
The girls among them are the 20 who study Physics: $\frac{20}{44} = \frac{5}{11}$ .

Worth noting

$\frac{5}{13} = \frac{20}{52}$ counts the 8 boys who study Physics twice.

The multiplication theorem

23 questions

Question 30

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A bag has 7 red and 5 black balls. Two balls are drawn one after the other without replacement. What is the probability that both are red?

\frac{7}{22}

Correct answerYour answer

\frac{49}{144}

Correct answerYour answer

\frac{7}{12}

Correct answerYour answer

\frac{13}{22}

Correct answerYour answer

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AnswerA · $\frac{7}{22}$

Let $R_1$ , $R_2$ be red on the first and second draws: $P(R_1) = \frac{7}{12}$ , and with 6 red of 11 left, $P(R_2 \mid R_1) = \frac{6}{11}$ .
$\frac{7}{12} \times \frac{6}{11} = \frac{7}{22}$ .

Worth noting

$\frac{49}{144}$ puts the first ball back; without replacement the second draw has 11 balls.

Question 31

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In a class, 40% of the students take Mathematics, and 25% of those also take Biology. What is the probability that a student picked at random takes both?

0.1

Correct answerYour answer

0.15

Correct answerYour answer

0.625

Correct answerYour answer

0.65

Correct answerYour answer

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AnswerA · $0.1$

$P(M) = 0.4$ and $P(B \mid M) = 0.25$ .
$P(M \cap B) = 0.4 \times 0.25 = 0.1$ .

Worth noting

0.15 subtracts, $0.4 - 0.25$ ; "25% of those" means a fraction of the 40%.

Question 32

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A box has 6 white and 4 black marbles. Two are drawn one after the other without replacement. What is the probability that both are white?

\frac{1}{3}

Correct answerYour answer

\frac{9}{25}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerA · $\frac{1}{3}$

First white: $\frac{6}{10}$ ; then 5 white of 9 left: $\frac{5}{9}$ .
$\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$ .

Worth noting

$\frac{9}{25} = \left(\frac{3}{5}\right)^{2}$ puts the first marble back.

Question 33

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$P(A \cap B) = 0.12$ and $P(B \mid A) = 0.3$ . What is $P(A)$ ?

0.036

Correct answerYour answer

0.18

Correct answerYour answer

0.4

Correct answerYour answer

0.42

Correct answerYour answer

Show the explanationHide the explanation

AnswerC · $0.4$

$P(A \cap B) = P(A)\,P(B \mid A)$ , so $P(A) = \frac{P(A \cap B)}{P(B \mid A)}$ .
$\frac{0.12}{0.3} = 0.4$ .

Worth noting

0.036 multiplies, $0.12 \times 0.3$ ; to undo the theorem's product, divide.

Question 34

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Only one of 5 keys opens a lock. The keys are tried one at a time in a random order, and a key that fails is set aside. What is the probability that the lock opens within the first three tries?

\frac{1}{5}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{61}{125}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerD · $\frac{3}{5}$

The lock stays shut through three tries with probability $\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{2}{5}$ .
So it opens within three tries with probability $1 - \frac{2}{5} = \frac{3}{5}$ .

Worth noting

$\frac{61}{125} = 1 - \left(\frac{4}{5}\right)^{3}$ puts each failed key back, so the same key could be tried again.

Question 35

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$P(A) = 0.7$ and $P(B \mid A) = 0.4$ . What is $P(A \cap B')$ ?

0.28

Correct answerYour answer

0.3

Correct answerYour answer

0.42

Correct answerYour answer

0.6

Correct answerYour answer

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AnswerC · $0.42$

Given $A$ , $P(B' \mid A) = 1 - 0.4 = 0.6$ .
$P(A \cap B') = P(A)\,P(B' \mid A) = 0.7 \times 0.6 = 0.42$ .

Worth noting

0.28 is $P(A \cap B) = 0.7 \times 0.4$ , with $B$ where $B'$ was asked for.

Question 36

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Two cards are drawn one after the other, without replacement, from a well-shuffled pack of 52. What is the probability that they are of the same suit?

\frac{1}{17}

Correct answerYour answer

\frac{4}{17}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{13}{51}

Correct answerYour answer

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AnswerB · $\frac{4}{17}$

The first card can be any card; 12 of the 51 cards left share its suit.
$1 \times \frac{12}{51} = \frac{4}{17}$ .

Worth noting

$\frac{1}{4}$ puts the first card back; without replacement only 12 cards of its suit remain among 51.

Question 37

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$P(A) = 0.8$ and $P(B \mid A) = 0.5$ . What is $P(A \cap B)$ ?

0.3

Correct answerYour answer

0.4

Correct answerYour answer

0.5

Correct answerYour answer

0.625

Correct answerYour answer

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AnswerB · $0.4$

Multiplication theorem: $P(A \cap B) = P(A)\,P(B \mid A)$ .
$0.8 \times 0.5 = 0.4$ .

Worth noting

0.625 divides, $\frac{0.5}{0.8}$ ; the theorem multiplies.

Question 38

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Three cards are drawn one after the other, without replacement, from a pack of 52. What is the probability that all three are aces?

\frac{1}{5525}

Correct answerYour answer

\frac{1}{2197}

Correct answerYour answer

\frac{3}{52}

Correct answerYour answer

\frac{1}{13}

Correct answerYour answer

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AnswerA · $\frac{1}{5525}$

$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50}$ , each draw leaving one ace and one card fewer.
$= \frac{24}{132600} = \frac{1}{5525}$ .

Worth noting

$\frac{1}{2197} = \left(\frac{1}{13}\right)^{3}$ replaces each card before the next draw.

Question 39

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Two cards are drawn one after the other, without replacement, from a pack of 52. What is the probability that the first is a king and the second a queen?

\frac{1}{2652}

Correct answerYour answer

\frac{1}{663}

Correct answerYour answer

\frac{1}{169}

Correct answerYour answer

\frac{4}{663}

Correct answerYour answer

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AnswerD · $\frac{4}{663}$

$P(K_1) = \frac{4}{52}$ ; all 4 queens are still among the 51 cards left: $P(Q_2 \mid K_1) = \frac{4}{51}$ .
$\frac{4}{52} \times \frac{4}{51} = \frac{4}{663}$ .

Worth noting

$\frac{1}{169} = \frac{4}{52} \times \frac{4}{52}$ puts the king back before the second draw.

Question 40

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A candidate clears the first round of a selection with probability 0.8. Having cleared it, the candidate clears the second round with probability 0.6, and having cleared both, the third round with probability 0.75. What is the probability that the candidate clears the first two rounds but not the third?

0.12

Correct answerYour answer

0.25

Correct answerYour answer

0.36

Correct answerYour answer

0.48

Correct answerYour answer

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AnswerA · $0.12$

Once the first two rounds are cleared, the third is missed with probability $1 - 0.75 = 0.25$ .
$0.8 \times 0.6 \times 0.25 = 0.12$ .

Worth noting

0.25 is the chance of missing the third round once the first two are cleared; the candidate must first clear them, with probability $0.8 \times 0.6$ .

Question 41

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A basket has 10 mangoes, of which 3 are spoilt. Three mangoes are taken out one after the other without replacement. What is the probability that at least one of them is spoilt?

\frac{1}{120}

Correct answerYour answer

\frac{7}{24}

Correct answerYour answer

\frac{657}{1000}

Correct answerYour answer

\frac{17}{24}

Correct answerYour answer

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AnswerD · $\frac{17}{24}$

None spoilt: $\frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} = \frac{7}{24}$ .
At least one spoilt: $1 - \frac{7}{24} = \frac{17}{24}$ .

Worth noting

$\frac{657}{1000} = 1 - \left(\frac{7}{10}\right)^{3}$ puts each mango back before the next one is taken.

Question 42

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In a quiz, 50% of the contestants answer the first question correctly, and 60% of those also answer the second correctly. In all, 12% of the contestants answer all three questions correctly. What is the probability that a contestant who answers the first two correctly also answers the third correctly?

0.2

Correct answerYour answer

0.24

Correct answerYour answer

0.3

Correct answerYour answer

0.4

Correct answerYour answer

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AnswerD · $0.4$

Let $Q_1, Q_2, Q_3$ be correct answers to the three questions: $P(Q_1 \cap Q_2) = 0.5 \times 0.6 = 0.3$ , and $P(Q_1 \cap Q_2 \cap Q_3) = 0.12$ .
$P(Q_3 \mid Q_1 \cap Q_2) = \frac{0.12}{0.3} = 0.4$ .

Worth noting

0.24 divides 0.12 by 0.5 alone, the chance of the first answer; the condition is both of the first two answers, with probability 0.3.

Question 43

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A box of 12 items has 4 defective ones. The items are tested one by one in a random order, without replacement. What is the probability that the first defective item is found on the third test?

\frac{4}{27}

Correct answerYour answer

\frac{28}{165}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{14}{33}

Correct answerYour answer

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AnswerB · $\frac{28}{165}$

The first two items must be good and the third defective: $\frac{8}{12} \times \frac{7}{11} \times \frac{4}{10}$ .
$= \frac{224}{1320} = \frac{28}{165}$ .

Worth noting

$\frac{4}{27} = \left(\frac{2}{3}\right)^{2} \times \frac{1}{3}$ keeps the box unchanged between tests, but a tested item is not put back.

Question 44

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Assertion (A): Two balls are drawn one after the other, without replacement, from a bag of 4 red and 4 blue balls. The probability that both are red is $\frac{1}{4}$ .

Reason (R): $P(E \cap F) = P(E)\,P(F \mid E)$ for any events $E$ and $F$ with $P(E) > 0$ .

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AnswerD · A is false, but R is true.

R is the multiplication theorem: true.
By it, both are red with probability $\frac{4}{8} \times \frac{3}{7} = \frac{3}{14}$ , not $\frac{1}{4}$ : A is false.

Worth noting

$\frac{1}{4} = \frac{1}{2} \times \frac{1}{2}$ puts the first ball back before the second draw.

Question 45

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A bag has 3 red and 2 black balls. Two balls are drawn one after the other without replacement. Which of the following statements are true?

I. The probability that both are red is $\frac{3}{10}$ .

II. Given that the first ball is red, the probability that the second is red is $\frac{3}{5}$ .

III. The probability that the second ball is red is $\frac{3}{5}$ .

IV. Red then black and black then red are equally likely.

I, III and IV onlyCorrect answerYour answerI and IV onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI, II and IV onlyCorrect answerYour answer

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AnswerA · I, III and IV only

I: $\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ . III: $\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{5}$ . IV: $\frac{3}{5} \times \frac{2}{4} = \frac{2}{5} \times \frac{3}{4} = \frac{3}{10}$ .
II is false: after a red ball, 2 red balls of 4 remain, so the chance is $\frac{1}{2}$ .

Worth noting

Unseen, the second ball is red with the same chance as the first; once the first is known to be red, the chance falls.

Question 46

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A box holds 5 good and 3 faulty bulbs. Two bulbs are taken out one after the other without replacement. What is the probability that exactly one is faulty?

\frac{15}{56}

Correct answerYour answer

\frac{3}{8}

Correct answerYour answer

\frac{15}{32}

Correct answerYour answer

\frac{15}{28}

Correct answerYour answer

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AnswerD · $\frac{15}{28}$

Good then faulty: $\frac{5}{8} \times \frac{3}{7}$ ; faulty then good: $\frac{3}{8} \times \frac{5}{7}$ .
$\frac{15}{56} + \frac{15}{56} = \frac{15}{28}$ .

Worth noting

$\frac{15}{56}$ counts only one order; the faulty bulb can come first or second.

Question 47

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A bag has 3 red, 2 blue and 5 green beads. Three are drawn one after the other without replacement. What is the probability of red, then blue, then green, in that order?

\frac{3}{100}

Correct answerYour answer

\frac{1}{24}

Correct answerYour answer

\frac{1}{15}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

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AnswerB · $\frac{1}{24}$

$\frac{3}{10}$ for red, then $\frac{2}{9}$ for blue, then $\frac{5}{8}$ for green.
$\frac{3 \times 2 \times 5}{10 \times 9 \times 8} = \frac{30}{720} = \frac{1}{24}$ .

Worth noting

$\frac{1}{4}$ allows the three colours in any of the 6 orders; the question fixes one order.

Question 48

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A bag has 4 red and 6 black balls. Two are drawn one after the other without replacement. What is the probability that the second ball is red?

\frac{2}{15}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

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AnswerC · $\frac{2}{5}$

Split on the first ball: $\frac{4}{10} \times \frac{3}{9} + \frac{6}{10} \times \frac{4}{9}$ .
$\frac{12 + 24}{90} = \frac{2}{5}$ , the same as for the first ball.

Worth noting

$\frac{1}{3}$ assumes the first ball was red; it may have been black.

Question 49

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Three cards are drawn one after the other, without replacement, from a well-shuffled pack of 52. What is the probability that the three cards are of three different suits?

\frac{169}{2550}

Correct answerYour answer

\frac{3}{8}

Correct answerYour answer

\frac{169}{425}

Correct answerYour answer

\frac{13}{17}

Correct answerYour answer

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AnswerC · $\frac{169}{425}$

The first card can be any card; the second must avoid its suit, $\frac{39}{51}$ ; the third must avoid both suits drawn, $\frac{26}{50}$ .
$\frac{39}{51} \times \frac{26}{50} = \frac{13}{17} \times \frac{13}{25} = \frac{169}{425}$ .

Worth noting

$\frac{3}{8} = \frac{3}{4} \times \frac{2}{4}$ puts each card back, as if 13 cards of every suit stayed in the pack.

Question 50

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A bag has 2 red and 3 white balls. A and B take turns to draw one ball each, without replacement, with A going first; whoever first draws a red ball wins. What is the probability that A wins?

\frac{2}{5}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{5}{8}

Correct answerYour answer

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AnswerC · $\frac{3}{5}$

A wins on the first draw with probability $\frac{2}{5}$ , or on the third, after two white balls, with probability $\frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} = \frac{1}{5}$ .
With only 3 white balls, a red one appears by the fourth draw, so A gets no fifth draw: $\frac{2}{5} + \frac{1}{5} = \frac{3}{5}$ .

Worth noting

$\frac{5}{8}$ puts each ball back, as if the game could go on for ever; here the white balls run out.

Question 51

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A crate has 5 red and 3 green apples. Three apples are taken out one after the other without replacement. What is the probability that their colours alternate?

\frac{5}{56}

Correct answerYour answer

\frac{5}{28}

Correct answerYour answer

\frac{15}{64}

Correct answerYour answer

\frac{15}{56}

Correct answerYour answer

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AnswerD · $\frac{15}{56}$

Red, green, red: $\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} = \frac{60}{336}$ ; green, red, green: $\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} = \frac{30}{336}$ .
$\frac{60 + 30}{336} = \frac{90}{336} = \frac{15}{56}$ .

Worth noting

$\frac{5}{28} = \frac{60}{336}$ counts only red, green, red; green, red, green alternates too.

Question 52

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Two cards are drawn one after the other, without replacement, from a well-shuffled pack of 52. What is the probability that the first card is a heart and the second an ace?

\frac{1}{68}

Correct answerYour answer

\frac{4}{221}

Correct answerYour answer

\frac{1}{52}

Correct answerYour answer

\frac{1}{51}

Correct answerYour answer

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AnswerC · $\frac{1}{52}$

Split on the first card. The ace of hearts first: $\frac{1}{52} \times \frac{3}{51}$ ; another heart first: $\frac{12}{52} \times \frac{4}{51}$ .
$\frac{3 + 48}{2652} = \frac{51}{2652} = \frac{1}{52}$ .

Worth noting

$\frac{1}{51} = \frac{13}{52} \times \frac{4}{51}$ leaves all 4 aces in the pack, but the first heart may have been the ace of hearts.

Independent events

25 questions

Question 53

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A problem is given to two students, who solve it independently with probabilities $\frac{1}{2}$ and $\frac{1}{3}$ . What is the probability that the problem is solved?

\frac{1}{6}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{5}{6}

Correct answerYour answer

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AnswerC · $\frac{2}{3}$

The problem stays unsolved only if both fail: $\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$ .
$1 - \frac{1}{3} = \frac{2}{3}$ .

Worth noting

$\frac{5}{6}$ adds the two chances, counting twice the case where both solve it.

Question 54

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$A$ and $B$ are independent with $P(A) = \frac{1}{4}$ and $P(B) = \frac{2}{5}$ . What is the probability that neither happens?

\frac{9}{20}

Correct answerYour answer

\frac{11}{20}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

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AnswerA · $\frac{9}{20}$

$A'$ and $B'$ are independent too: $P(A' \cap B') = \frac{3}{4} \times \frac{3}{5}$ .
$= \frac{9}{20}$ .

Worth noting

$\frac{11}{20}$ is the chance that at least one happens, $1 - \frac{9}{20}$ .

Question 55

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A coin is tossed and a die is thrown. What is the probability of a head on the coin and a number greater than 4 on the die?

\frac{1}{12}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{5}{6}

Correct answerYour answer

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AnswerB · $\frac{1}{6}$

The coin and the die do not affect each other: a head has probability $\frac{1}{2}$ , and a number greater than 4 is 5 or 6, with probability $\frac{2}{6} = \frac{1}{3}$ .
$\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ .

Worth noting

$\frac{5}{6} = \frac{1}{2} + \frac{1}{3}$ adds; for both of two independent events to happen, multiply.

Question 56

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$A$ and $B$ are independent with $P(A) = 0.35$ and $P(B) = 0.6$ . What is $P(A' \mid B)$ ?

0.35

Correct answerYour answer

0.39

Correct answerYour answer

0.4

Correct answerYour answer

0.65

Correct answerYour answer

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AnswerD · $0.65$

$A'$ and $B$ are independent too, so knowing $B$ changes nothing: $P(A' \mid B) = P(A')$ .
$P(A') = 1 - 0.35 = 0.65$ .

Worth noting

0.35 is $P(A \mid B) = P(A)$ ; the question asks for the complement.

Question 57

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A die is thrown three times. What is the probability that a six turns up at least once?

\frac{1}{216}

Correct answerYour answer

\frac{91}{216}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{125}{216}

Correct answerYour answer

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AnswerB · $\frac{91}{216}$

The throws are independent, so no six in all three has probability $\left(\frac{5}{6}\right)^{3} = \frac{125}{216}$ .
$1 - \frac{125}{216} = \frac{91}{216}$ .

Worth noting

$\frac{1}{2} = 3 \times \frac{1}{6}$ adds the three chances, counting the throws with two or three sixes more than once.

Question 58

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Which of the following statements are true?

I. If $A$ and $B$ are independent, so are $A'$ and $B'$ .

II. Two mutually exclusive events with positive probabilities are independent.

III. If $A$ and $B$ are independent, $P(A \mid B) = P(A)$ .

IV. If $A$ and $B$ are independent, $P(A \cap B) = P(A)\,P(B)$ .

All four: I, II, III and IVCorrect answerYour answerI, III and IV onlyCorrect answerYour answerIII and IV onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerB · I, III and IV only

I, III and IV hold for independent events.
II is false: mutually exclusive means $P(A \cap B) = 0$ , but $P(A)\,P(B) > 0$ .

Worth noting

Exclusive events are as dependent as can be: if one happens, the other cannot.

Question 59

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$A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.4$ . What is $P(A \cup B)$ ?

0.12

Correct answerYour answer

0.42

Correct answerYour answer

0.58

Correct answerYour answer

0.7

Correct answerYour answer

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AnswerC · $0.58$

Independent: $P(A \cap B) = 0.3 \times 0.4 = 0.12$ .
$P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$ .

Worth noting

0.7 treats the events as mutually exclusive; independent events can happen together.

Question 60

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Read the two statements and choose the correct option.

I. If $A$ and $B$ are independent, then $P(A \cup B) = P(A) + P(B)$ .

II. Independent events with positive probabilities cannot happen together.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerD · Neither I nor II

I: that formula needs $P(A \cap B) = 0$ ; independence gives $P(A \cap B) = P(A)P(B) > 0$ . False.
II: $P(A \cap B) = P(A)P(B) > 0$ , so they can happen together. False.

Question 61

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Three shooters hit a target independently with probabilities $\frac{4}{5}$ , $\frac{5}{6}$ and $\frac{1}{2}$ . What is the probability that all three hit it?

\frac{1}{60}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{32}{45}

Correct answerYour answer

\frac{59}{60}

Correct answerYour answer

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AnswerB · $\frac{1}{3}$

For independent events, multiply.
$\frac{4}{5} \times \frac{5}{6} \times \frac{1}{2} = \frac{1}{3}$ .

Worth noting

$\frac{1}{60}$ is the chance that all three miss.

Question 62

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$A$ and $B$ are independent with $P(A) = 0.6$ and $P(B) = 0.25$ . What is $P(A \cap B')$ ?

0.15

Correct answerYour answer

0.35

Correct answerYour answer

0.45

Correct answerYour answer

0.75

Correct answerYour answer

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AnswerC · $0.45$

$A$ and $B'$ are independent too, and $P(B') = 0.75$ .
$0.6 \times 0.75 = 0.45$ .

Worth noting

0.15 is $P(A \cap B)$ , with $B$ in place of its complement.

Question 63

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$A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.6$ . What is the probability that exactly one of them happens?

0.18

Correct answerYour answer

0.54

Correct answerYour answer

0.72

Correct answerYour answer

0.9

Correct answerYour answer

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AnswerB · $0.54$

$A$ only: $0.3 \times 0.4 = 0.12$ ; $B$ only: $0.7 \times 0.6 = 0.42$ .
$0.12 + 0.42 = 0.54$ .

Worth noting

0.72 is at least one, $1 - 0.7 \times 0.4$ , which also counts both happening.

Question 64

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Read the two statements and choose the correct option.

I. A number is picked at random from 1 to 10. The events $E$ : the number is even, and $M$ : the number is a multiple of 3, are independent.

II. A number is picked at random from 1 to 12. The events $E$ : the number is even, and $M$ : the number is a multiple of 3, are independent.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerB · Only II

I: $P(E) = \frac{1}{2}$ , $P(M) = \frac{3}{10}$ and $E \cap M = \{6\}$ : $\frac{1}{10} \ne \frac{1}{2} \times \frac{3}{10}$ . False.
II: $P(E) = \frac{1}{2}$ , $P(M) = \frac{1}{3}$ and $E \cap M = \{6, 12\}$ : $\frac{2}{12} = \frac{1}{6} = \frac{1}{2} \times \frac{1}{3}$ . True.

Worth noting

Independence is a fact about the numbers, not about how the events are described: the same two events are independent on 1 to 12 but not on 1 to 10.

Question 65

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$A$ and $B$ are independent, $P(A \cap B) = 0.12$ and $P(A' \cap B) = 0.28$ . What is $P(A)$ ?

0.12

Correct answerYour answer

0.3

Correct answerYour answer

0.4

Correct answerYour answer

0.7

Correct answerYour answer

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AnswerB · $0.3$

$B$ splits into $A \cap B$ and $A' \cap B$ : $P(B) = 0.12 + 0.28 = 0.4$ .
By independence, $P(A) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.3$ .

Worth noting

0.4 is $P(B)$ , the sum of the two pieces, not $P(A)$ .

Question 66

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$A$ and $B$ are independent with $P(A) = 0.4$ and $P(B) = 0.7$ . What is $P(A' \cup B')$ ?

0.18

Correct answerYour answer

0.28

Correct answerYour answer

0.72

Correct answerYour answer

0.82

Correct answerYour answer

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AnswerC · $0.72$

$A' \cup B'$ is the complement of $A \cap B$ : the two do not both happen.
$1 - 0.4 \times 0.7 = 1 - 0.28 = 0.72$ .

Worth noting

0.18 is $P(A' \cap B') = 0.6 \times 0.3$ , neither happening; $A' \cup B'$ means at most one happens.

Question 67

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$P(A) = 0.4$ , $P(B) = 0.5$ and $P(A \cup B) = 0.7$ . Which of the following statements are true?

I. $A$ and $B$ are independent.

II. $A$ and $B$ are mutually exclusive.

III. $P(A' \cap B) = 0.3$ .

IV. $P(A \mid B) = 0.5$ .

I and III onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answerII and III onlyCorrect answerYour answerIII and IV onlyCorrect answerYour answer

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AnswerA · I and III only

$P(A \cap B) = 0.4 + 0.5 - 0.7 = 0.2 = 0.4 \times 0.5$ : I is true, and II is false since $P(A \cap B) \ne 0$ .
III: $0.5 - 0.2 = 0.3$ , true. IV: $\frac{0.2}{0.5} = 0.4$ , false.

Worth noting

IV takes $P(B)$ for $P(A \mid B)$ ; for independent events $P(A \mid B) = P(A) = 0.4$ .

Question 68

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A card is drawn at random from a well-shuffled pack of 52. Which of the following pairs of events is independent?

The card is a spade; the card is an aceCorrect answerYour answerThe card is a spade; the card is blackCorrect answerYour answerThe card is a king; the card is an aceCorrect answerYour answerThe card is a heart; the card is a red aceCorrect answerYour answer

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AnswerA · The card is a spade; the card is an ace

Spade and ace: only the ace of spades is both, $\frac{1}{52}$ , and $\frac{13}{52} \times \frac{4}{52} = \frac{1}{52}$ : independent.
The rest fail: spade and black give $\frac{13}{52} \ne \frac{13}{52} \times \frac{26}{52}$ ; a king and an ace cannot happen together; heart and red ace give $\frac{1}{52} \ne \frac{13}{52} \times \frac{2}{52}$ .

Worth noting

Every spade is black, so knowing the card is a spade makes it certainly black: the chance moves from $\frac{1}{2}$ to 1.

Question 69

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$E$ and $F$ are independent events with $P(E) = \frac{1}{3}$ and $P(F) = \frac{3}{5}$ . Match each probability in List I with its value in List II.

List I	List II
A. $P(E \cap F')$	I. $\frac{11}{15}$
B. $P(E' \cap F)$	II. $\frac{2}{15}$
C. $P(E \cup F)$	III. $\frac{1}{3}$
D. $P(E \mid F')$	IV. $\frac{2}{5}$

A-II, B-IV, C-I, D-IIICorrect answerYour answerA-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answer

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AnswerA · A-II, B-IV, C-I, D-III

$P(E \cap F') = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}$ and $P(E' \cap F) = \frac{2}{3} \times \frac{3}{5} = \frac{2}{5}$ .
$P(E \cup F) = \frac{1}{3} + \frac{3}{5} - \frac{1}{5} = \frac{11}{15}$ ; $E$ and $F'$ are independent, so $P(E \mid F') = P(E) = \frac{1}{3}$ .

Worth noting

Complements of independent events are independent, so each piece is a product of $P(E)$ or $P(E')$ with $P(F)$ or $P(F')$ .

Question 70

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Assertion (A): If $P(E) = 0.3$ , $P(F) = 0.4$ and $P(E \cup F) = 0.7$ , then $E$ and $F$ are not independent.

Reason (R): If $P(E \cup F) = P(E) + P(F)$ , then $E$ and $F$ are independent.

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AnswerC · A is true, but R is false.

$P(E \cap F) = 0.3 + 0.4 - 0.7 = 0$ , while $P(E)\,P(F) = 0.12$ : A is true.
R is false: $P(E \cup F) = P(E) + P(F)$ means $P(E \cap F) = 0$ , so the events are mutually exclusive, and with positive probabilities that rules out independence.

Worth noting

Exclusive events with positive probabilities are never independent: if one happens, the other cannot.

Question 71

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Two dice are thrown. $A$ : the first die shows an even number. $B$ : the sum is 7. Are $A$ and $B$ independent?

No: they can happen togetherCorrect answerYour answerYes:

P(A \cap B) = \frac{1}{12} = P(A)\,P(B)

Correct answerYour answerNo: the sum depends on the first dieCorrect answerYour answerYes: they are mutually exclusiveCorrect answerYour answer

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AnswerB · Yes: $P(A \cap B) = \frac{1}{12} = P(A)\,P(B)$

$P(A) = \frac{1}{2}$ , $P(B) = \frac{6}{36} = \frac{1}{6}$ , and $A \cap B = \{(2, 5), (4, 3), (6, 1)\}$ , so $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$ .
$\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ : independent.

Worth noting

Whatever the first die shows, exactly one value of the second makes 7, so knowing $A$ does not change the chance of $B$ .

Question 72

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$A$ and $B$ are independent, $P(A) = \frac{1}{2}$ and $P(A \cup B) = \frac{3}{5}$ . What is $P(B)$ ?

\frac{1}{10}

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

\frac{3}{10}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerB · $\frac{1}{5}$

$P(A \cup B) = P(A) + P(B) - P(A)P(B)$ : $\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}$ .
$\frac{p}{2} = \frac{1}{10}$ , so $p = \frac{1}{5}$ .

Worth noting

$\frac{1}{10} = \frac{3}{5} - \frac{1}{2}$ treats the events as mutually exclusive.

Question 73

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Three fair coins are tossed. $E$ : the first coin shows a head. $F$ : exactly two coins show heads. Are $E$ and $F$ independent?

Yes: the three coins are tossed independently of one anotherCorrect answerYour answerYes:

P(E \cap F) = P(E)\,P(F) = \frac{1}{2} \times \frac{3}{8} = \frac{3}{16}

Correct answerYour answerNo: they are mutually exclusive, so they cannot happen togetherCorrect answerYour answerNo:

P(E \cap F) = \frac{1}{4}

but

P(E)\,P(F) = \frac{3}{16}

Correct answerYour answer

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AnswerD · No: $P(E \cap F) = \frac{1}{4}$ but $P(E)\,P(F) = \frac{3}{16}$

$P(E) = \frac{1}{2}$ and $P(F) = \frac{3}{8}$ ( $HHT, HTH, THH$ ).
$E \cap F = \{HHT, HTH\}$ : $\frac{2}{8} = \frac{1}{4} \ne \frac{1}{2} \times \frac{3}{8}$ . Not independent.

Worth noting

Independent tosses do not make every pair of events independent: knowing the first is a head raises the chance of exactly two heads from $\frac{3}{8}$ to $\frac{1}{2}$ .

Question 74

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$A$ and $B$ are independent, with $P(A \cap B) = \frac{1}{6}$ , $P(A' \cap B') = \frac{1}{3}$ and $P(A) > P(B)$ . What is $P(A)$ ?

\frac{1}{6}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{5}{6}

Correct answerYour answer

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AnswerC · $\frac{1}{2}$

With $p = P(A)$ and $q = P(B)$ : $pq = \frac{1}{6}$ and $(1 - p)(1 - q) = 1 - (p + q) + pq = \frac{1}{3}$ , so $p + q = \frac{5}{6}$ .
$p$ and $q$ are the roots of $t^{2} - \frac{5}{6}t + \frac{1}{6} = 0$ , namely $\frac{1}{2}$ and $\frac{1}{3}$ ; the larger is $P(A) = \frac{1}{2}$ .

Worth noting

$\frac{1}{3}$ is the smaller root, $P(B)$ ; the question says $P(A) > P(B)$ .

Question 75

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In a quiz, three contestants answer the same question independently; the probabilities that they answer it correctly are $\frac{4}{5}$ , $\frac{1}{2}$ and $\frac{1}{4}$ . What is the probability that exactly one of them answers it correctly?

\frac{3}{40}

Correct answerYour answer

\frac{1}{10}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{37}{40}

Correct answerYour answer

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AnswerC · $\frac{2}{5}$

Only the first: $\frac{4}{5} \times \frac{1}{2} \times \frac{3}{4} = \frac{12}{40}$ ; only the second: $\frac{1}{5} \times \frac{1}{2} \times \frac{3}{4} = \frac{3}{40}$ ; only the third: $\frac{1}{5} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{40}$ .
$\frac{12 + 3 + 1}{40} = \frac{16}{40} = \frac{2}{5}$ .

Worth noting

$\frac{37}{40} = 1 - \frac{1}{5} \times \frac{1}{2} \times \frac{3}{4}$ is at least one correct answer, which also counts two or three.

Question 76

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Two archers shoot at a target independently, hitting it with probabilities 0.6 and 0.3. Each shoots once, and the target is hit exactly once. What is the probability that the first archer hit it?

\frac{21}{50}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{7}{9}

Correct answerYour answer

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AnswerD · $\frac{7}{9}$

Only the first hits: $0.6 \times 0.7 = 0.42$ ; only the second: $0.4 \times 0.3 = 0.12$ ; so exactly one hit has probability $0.54$ .
$\frac{0.42}{0.54} = \frac{7}{9}$ .

Worth noting

$\frac{3}{5} = 0.6$ is the first archer's chance before the result is known; a single hit makes the better archer likelier still.

Question 77

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A machine runs when its motor works and at least one of its two fans works. The motor and the two fans work independently, with probabilities 0.9, 0.8 and 0.7. What is the probability that the machine runs?

0.504

Correct answerYour answer

0.846

Correct answerYour answer

0.94

Correct answerYour answer

0.994

Correct answerYour answer

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AnswerB · $0.846$

Both fans fail with probability $0.2 \times 0.3 = 0.06$ , so at least one works with probability $0.94$ .
The motor must work as well: $0.9 \times 0.94 = 0.846$ .

Worth noting

0.504 is $0.9 \times 0.8 \times 0.7$ , which needs both fans, but one is enough.

The theorem of total probability

22 questions

Question 78

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A coin is tossed. On a head, a ball is drawn from urn A (2 red, 3 blue); on a tail, from urn B (4 red, 1 blue). What is the probability that the ball is red?

\frac{6}{25}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerD · $\frac{3}{5}$

$P(R) = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{4}{5}$ .
$\frac{1}{5} + \frac{2}{5} = \frac{3}{5}$ .

Worth noting

$\frac{2}{5}$ uses urn A only, as if the coin always showed a head.

Question 79

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Bag I has 3 red and 2 black balls; bag II has 2 red and 3 black balls. One ball is drawn at random from bag I and put into bag II without its colour being seen. Then a ball is drawn at random from bag II.

If the ball moved was red, what is the probability that the ball drawn from bag II is red?

\frac{3}{10}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerC · $\frac{1}{2}$

After a red ball is moved in, bag II has 3 red and 3 black balls.
$\frac{3}{6} = \frac{1}{2}$ .

Worth noting

$\frac{2}{5}$ uses bag II as it was, forgetting the ball moved into it.

Question 80

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A bag holds three coins: a fair coin, a coin with a head on both sides, and a coin that shows a head with probability $\frac{1}{4}$ . A coin is picked at random and tossed. What is the probability of a head?

\frac{1}{3}

Correct answerYour answer

\frac{5}{12}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{7}{12}

Correct answerYour answer

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AnswerD · $\frac{7}{12}$

Each coin is picked with probability $\frac{1}{3}$ , and the coins show a head with probabilities $\frac{1}{2}$ , 1 and $\frac{1}{4}$ .
$\frac{1}{3}\left(\frac{1}{2} + 1 + \frac{1}{4}\right) = \frac{1}{3} \times \frac{7}{4} = \frac{7}{12}$ .

Worth noting

$\frac{5}{12}$ gives the two-headed coin a head only half the time; it shows a head every time.

Question 81

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$E_1$ , $E_2$ , $E_3$ are mutually exclusive events, each with positive probability, that together make up the sample space, and $A$ is any event. Read the two statements and choose the correct option.

I. $P(A)$ cannot be larger than the largest of $P(A \mid E_1)$ , $P(A \mid E_2)$ and $P(A \mid E_3)$ .

II. The theorem of total probability holds only when $P(E_1) = P(E_2) = P(E_3)$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $P(A) = \sum P(E_i)P(A \mid E_i)$ is a weighted average of the $P(A \mid E_i)$ , with weights adding to 1, so it lies between the smallest and the largest of them. True.
II: the weights $P(E_i)$ can be any positive numbers adding to 1. False.

Worth noting

The theorem needs a partition, not equal weights; unequal weights are exactly what it handles.

Question 82

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Bag I has 3 red and 4 black balls; bag II has 5 red and 4 black. A bag is picked at random and a ball drawn from it. What is the probability that the ball is red?

\frac{8}{63}

Correct answerYour answer

\frac{3}{7}

Correct answerYour answer

\frac{31}{63}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{31}{63}$

$P(R) = P(\text{I})P(R \mid \text{I}) + P(\text{II})P(R \mid \text{II}) = \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{9}$ .
$\frac{3}{14} + \frac{5}{18} = \frac{27 + 35}{126} = \frac{31}{63}$ .

Worth noting

$\frac{1}{2}$ , 8 red balls of 16, pools the balls of both bags, but the bags hold different numbers of balls and are equally likely to be picked.

Question 83

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A die is thrown. If it shows 1 or 2, a ball is drawn from bag A (3 red, 2 white); otherwise from bag B (1 red, 4 white). What is the probability that the ball is red?

\frac{1}{5}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{4}{5}

Correct answerYour answer

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AnswerB · $\frac{1}{3}$

$P(A) = \frac{2}{6} = \frac{1}{3}$ , $P(B) = \frac{2}{3}$ .
$P(R) = \frac{1}{3} \times \frac{3}{5} + \frac{2}{3} \times \frac{1}{5} = \frac{3 + 2}{15} = \frac{1}{3}$ .

Worth noting

$\frac{2}{5}$ gives the two bags equal weight; the die picks bag B twice as often.

Question 84

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At a school, 35% of the students live in the hostel and 65% are day scholars. 40% of hostel students and 20% of day scholars score an A. What is the probability that a student picked at random scores an A?

0.27

Correct answerYour answer

0.3

Correct answerYour answer

0.4

Correct answerYour answer

0.6

Correct answerYour answer

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AnswerA · $0.27$

$0.35 \times 0.4 + 0.65 \times 0.2$ .
$0.14 + 0.13 = 0.27$ .

Worth noting

0.3 averages 40% and 20% as if the two groups were the same size.

Question 85

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Box 1 has 3 gold and 1 silver coin, box 2 has 1 gold and 3 silver, and box 3 has 2 gold and 2 silver. A box is picked at random and a coin drawn from it. What is the probability that the coin is gold?

\frac{1}{4}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

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AnswerC · $\frac{1}{2}$

Each box has probability $\frac{1}{3}$ : $\frac{1}{3}\left(\frac{3}{4} + \frac{1}{4} + \frac{2}{4}\right)$ .
$\frac{1}{3} \times \frac{6}{4} = \frac{1}{2}$ .

Worth noting

$\frac{1}{3}$ is the chance of picking any one box, not of drawing gold.

Question 86

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A factory's day shift makes 70% of its output with 1% defective; the night shift makes the rest with 4% defective. What is the probability that an item picked at random is defective?

0.012

Correct answerYour answer

0.019

Correct answerYour answer

0.025

Correct answerYour answer

0.05

Correct answerYour answer

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AnswerB · $0.019$

$0.7 \times 0.01 + 0.3 \times 0.04$ .
$0.007 + 0.012 = 0.019$ .

Worth noting

0.025 averages 1% and 4%; the day shift makes more than half the items.

Question 87

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$P(A) = 0.5$ , $P(B \mid A) = 0.4$ and $P(B \mid A') = 0.2$ . What is $P(B)$ ?

0.3

Correct answerYour answer

0.4

Correct answerYour answer

0.5

Correct answerYour answer

0.6

Correct answerYour answer

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AnswerA · $0.3$

$A$ and $A'$ split the sample space: $P(B) = P(A)P(B \mid A) + P(A')P(B \mid A')$ .
$0.5 \times 0.4 + 0.5 \times 0.2 = 0.3$ .

Worth noting

0.6 adds the two conditional probabilities without weighting them by $P(A)$ and $P(A')$ .

Question 88

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$P(A) = 0.4$ , $P(B \mid A) = 0.7$ and $P(B) = 0.46$ . What is $P(B \mid A')$ ?

0.18

Correct answerYour answer

0.3

Correct answerYour answer

0.45

Correct answerYour answer

0.54

Correct answerYour answer

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AnswerB · $0.3$

$P(B) = P(A)P(B \mid A) + P(A')P(B \mid A')$ , so $0.46 = 0.4 \times 0.7 + 0.6x$ .
$0.6x = 0.46 - 0.28 = 0.18$ , so $x = 0.3$ .

Worth noting

0.18 is $P(A' \cap B)$ ; it still has to be divided by $P(A') = 0.6$ .

Question 89

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What is the probability that the ball moved and the ball drawn are of the same colour?

\frac{4}{15}

Correct answerYour answer

\frac{3}{10}

Correct answerYour answer

\frac{12}{25}

Correct answerYour answer

\frac{17}{30}

Correct answerYour answer

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AnswerD · $\frac{17}{30}$

Both red: $\frac{3}{5} \times \frac{3}{6} = \frac{9}{30}$ ; both black: after a black ball moves in, bag II has 4 black balls of 6, so $\frac{2}{5} \times \frac{4}{6} = \frac{8}{30}$ .
$\frac{9}{30} + \frac{8}{30} = \frac{17}{30}$ .

Worth noting

$\frac{12}{25} = \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5}$ draws from bag II as it was, without the ball moved in.

Question 90

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A die is thrown. If it shows 1, a ball is drawn from urn I (2 white, 3 black); if it shows 2 or 3, from urn II (4 white, 1 black); otherwise from urn III (3 white, 2 black). What is the probability that the ball is white?

\frac{3}{10}

Correct answerYour answer

\frac{17}{30}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{19}{30}

Correct answerYour answer

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AnswerD · $\frac{19}{30}$

$P(\text{I}) = \frac{1}{6}$ , $P(\text{II}) = \frac{2}{6}$ and $P(\text{III}) = \frac{3}{6}$ .
$P(W) = \frac{1}{6} \times \frac{2}{5} + \frac{2}{6} \times \frac{4}{5} + \frac{3}{6} \times \frac{3}{5} = \frac{2 + 8 + 9}{30} = \frac{19}{30}$ .

Worth noting

$\frac{3}{5}$ gives the three urns equal weight; the die picks urn III three times as often as urn I.

Question 91

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What is the probability that the ball drawn from bag II is red?

\frac{3}{10}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{5}{12}

Correct answerYour answer

\frac{13}{30}

Correct answerYour answer

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AnswerD · $\frac{13}{30}$

Split on the ball moved: $P(R) = \frac{3}{5} \times \frac{3}{6} + \frac{2}{5} \times \frac{2}{6}$ .
$\frac{9}{30} + \frac{4}{30} = \frac{13}{30}$ .

Worth noting

$\frac{2}{5}$ is bag II's share of red before the move; the ball moved in changes it.

Question 92

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One card is lost from a well-shuffled pack of 52. A card is then drawn at random from the remaining 51. What is the probability that it is a heart?

\frac{1}{17}

Correct answerYour answer

\frac{4}{17}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{13}{51}

Correct answerYour answer

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AnswerC · $\frac{1}{4}$

Split on the lost card: a heart with probability $\frac{13}{52}$ , leaving 12 hearts; not a heart with probability $\frac{39}{52}$ , leaving 13.
$\frac{13}{52} \times \frac{12}{51} + \frac{39}{52} \times \frac{13}{51} = \frac{156 + 507}{2652} = \frac{1}{4}$ .

Worth noting

$\frac{4}{17} = \frac{12}{51}$ assumes the lost card was a heart, which is only one of the two cases.

Question 93

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In a town, 25% of the days are rainy. A bus runs late on 60% of rainy days, and it is on time on 80% of the other days. What is the probability that the bus runs late on a day picked at random?

0.15

Correct answerYour answer

0.3

Correct answerYour answer

0.4

Correct answerYour answer

0.75

Correct answerYour answer

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AnswerB · $0.3$

On the other days the bus is late with probability $1 - 0.8 = 0.2$ .
$0.25 \times 0.6 + 0.75 \times 0.2 = 0.15 + 0.15 = 0.3$ .

Worth noting

0.75 uses 80% as the chance of running late on the other days, but 80% is the chance of being on time.

Question 94

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Bag I has 3 red and 2 blue balls; bag II has 3 red and 1 blue ball. A bag is picked at random and two balls are drawn from it, one after the other, without replacement. What is the probability that both are red?

\frac{1}{4}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{369}{800}

Correct answerYour answer

\frac{27}{40}

Correct answerYour answer

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AnswerB · $\frac{2}{5}$

Both red from bag I: $\frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ ; from bag II: $\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ .
$\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{20} + \frac{5}{20} = \frac{2}{5}$ .

Worth noting

$\frac{369}{800} = \frac{1}{2}\left(\frac{9}{25} + \frac{9}{16}\right)$ puts the first ball back before the second draw.

Question 95

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Machines A, B and C make 50%, 30% and 20% of a factory's bolts, and 1%, 2% and 3% of their bolts are faulty. What is the probability that a bolt picked at random is faulty?

0.01

Correct answerYour answer

0.017

Correct answerYour answer

0.02

Correct answerYour answer

0.06

Correct answerYour answer

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AnswerB · $0.017$

$0.5 \times 0.01 + 0.3 \times 0.02 + 0.2 \times 0.03$ .
$0.005 + 0.006 + 0.006 = 0.017$ .

Worth noting

0.02 averages the three fault rates as if each machine made a third of the bolts.

Question 96

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The ball drawn from bag II is red. What is the probability that the ball moved was red?

\frac{4}{13}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{9}{13}

Correct answerYour answer

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AnswerD · $\frac{9}{13}$

Let $E_1$ be that the ball moved is red and $R$ that the ball drawn is red: $P(E_1)\,P(R \mid E_1) = \frac{3}{5} \times \frac{1}{2} = \frac{9}{30}$ , and $P(R) = \frac{13}{30}$ .
$P(E_1 \mid R) = \frac{9/30}{13/30} = \frac{9}{13}$ .

Worth noting

$\frac{3}{5}$ is the chance that the ball moved was red before the drawn ball is seen; a red draw makes it likelier.

Question 97

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A bag has 4 red and 3 black balls. A ball is drawn at random and put back into the bag together with 2 more balls of its colour. A second ball is then drawn. What is the probability that the second ball is red?

\frac{8}{21}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

\frac{4}{7}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

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AnswerC · $\frac{4}{7}$

First red, with probability $\frac{4}{7}$ : the bag then has 6 red balls of 9. First black, with probability $\frac{3}{7}$ : 4 red balls of 9.
$\frac{4}{7} \times \frac{6}{9} + \frac{3}{7} \times \frac{4}{9} = \frac{24 + 12}{63} = \frac{4}{7}$ .

Worth noting

$\frac{2}{3} = \frac{6}{9}$ assumes the first ball was red; it is black with probability $\frac{3}{7}$ .

Question 98

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A shop's seeds come from two growers, X and Y. 80% of X's seeds and 30% of Y's seeds sprout, and 50% of all the shop's seeds sprout. What is the probability that a seed picked at random came from X?

0.2

Correct answerYour answer

0.4

Correct answerYour answer

0.6

Correct answerYour answer

0.625

Correct answerYour answer

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AnswerB · $0.4$

With $p = P(X)$ , total probability gives $0.8p + 0.3(1 - p) = 0.5$ .
$0.5p = 0.2$ , so $p = 0.4$ .

Worth noting

0.625 is $\frac{0.5}{0.8}$ , which leaves out Y's seeds, though they sprout too.

Question 99

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Bag I has 4 white and 2 black balls; bag II has 2 white and 4 black balls. Two balls are drawn at random from bag I and put into bag II. A ball is then drawn at random from bag II. What is the probability that it is white?

\frac{1}{3}

Correct answerYour answer

\frac{3}{8}

Correct answerYour answer

\frac{5}{12}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{5}{12}$

The two balls moved are both white, one of each, or both black, with probabilities $\frac{6}{15}$ , $\frac{8}{15}$ and $\frac{1}{15}$ ; bag II then holds 4, 3 or 2 white balls of 8.
$\frac{6}{15} \times \frac{4}{8} + \frac{8}{15} \times \frac{3}{8} + \frac{1}{15} \times \frac{2}{8} = \frac{24 + 24 + 2}{120} = \frac{5}{12}$ .

Worth noting

$\frac{1}{3} = \frac{2}{6}$ uses bag II as it was, before the two balls were added.

Bayes' theorem

26 questions

Question 100

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A shop gets 70% of its eggs from farm P and 30% from farm Q. Of P's eggs, 5% are cracked; of Q's, 10% are cracked. An egg is picked at random.

What is the probability that the egg is from farm Q and cracked?

0.03

Correct answerYour answer

0.1

Correct answerYour answer

0.13

Correct answerYour answer

0.3

Correct answerYour answer

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AnswerA · $0.03$

$P(Q \cap C) = P(Q)\,P(C \mid Q)$ .
$0.3 \times 0.1 = 0.03$ .

Worth noting

0.1 is $P(C \mid Q)$ , the rate within Q's eggs, not the share of all eggs.

Question 101

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Box I has 3 red and 7 white balls; box II has 6 red and 4 white balls. A box is picked at random and a ball drawn from it is red. What is the probability that it came from box I?

\frac{3}{20}

Correct answerYour answer

\frac{3}{10}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{1}{3}$

$P(R) = \frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{6}{10} = \frac{9}{20}$ .
$P(\text{I} \mid R) = \frac{3/20}{9/20} = \frac{1}{3}$ .

Worth noting

$\frac{3}{10}$ is $P(R \mid \text{I})$ , the chance of red from box I, the condition the other way round.

Question 102

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$P(E) = 0.2$ , $P(A \mid E) = 0.5$ and $P(A) = 0.25$ . What is $P(E \mid A)$ ?

0.1

Correct answerYour answer

0.4

Correct answerYour answer

0.5

Correct answerYour answer

0.8

Correct answerYour answer

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AnswerB · $0.4$

$P(E \mid A) = \frac{P(E)\,P(A \mid E)}{P(A)}$ .
$\frac{0.2 \times 0.5}{0.25} = \frac{0.1}{0.25} = 0.4$ .

Worth noting

0.1 is $P(E \cap A)$ , the numerator, before dividing by $P(A)$ .

Question 103

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Coin A is fair; coin B has a head on both sides. One coin is picked at random and tossed, and it shows a head. What is the probability that it is coin B?

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

\frac{3}{4}

Correct answerYour answer

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AnswerC · $\frac{2}{3}$

$P(H) = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times 1 = \frac{3}{4}$ .
$P(B \mid H) = \frac{1/2}{3/4} = \frac{2}{3}$ .

Worth noting

$\frac{1}{2}$ ignores the head; coin B shows a head every time, so a head points to it.

Question 104

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Read the two statements and choose the correct option.

I. Bayes' theorem finds $P(A \mid E_i)$ when the $P(E_i \mid A)$ are known.

II. The probability $P(E_i \mid A)$ after $A$ is seen can be larger than $P(E_i)$ .

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerB · Only II

I: it is the other way round; Bayes' theorem turns the known $P(A \mid E_i)$ into $P(E_i \mid A)$ . False.
II: a head lifted coin B from $\frac{1}{2}$ to $\frac{2}{3}$ above. True.

Question 105

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A shop gets 70% of its eggs from farm P and 30% from farm Q. Of P's eggs, 5% are cracked; of Q's, 10% are cracked. An egg is picked at random.

What is the probability that the egg is cracked?

0.035

Correct answerYour answer

0.065

Correct answerYour answer

0.075

Correct answerYour answer

0.15

Correct answerYour answer

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AnswerB · $0.065$

$P(C) = 0.7 \times 0.05 + 0.3 \times 0.1$ .
$0.035 + 0.03 = 0.065$ .

Worth noting

0.075 averages 5% and 10%, as if the farms sent equal numbers.

Question 106

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Assertion (A): In Bayes' theorem the events $E_1, E_2, \ldots, E_n$ must be mutually exclusive and together cover the sample space.

Reason (R): The denominator of Bayes' theorem is $P(A) = \sum P(E_i)\,P(A \mid E_i)$ , and the theorem of total probability gives this sum only when the $E_i$ are mutually exclusive and cover the sample space.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

A is true: the $E_i$ form a partition.
R is true and explains A: the partition is what makes the denominator of Bayes' theorem the total probability of $A$ .

Question 107

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Three bags hold 5 balls each: bag I has 2 white, bag II has 3 white and bag III has 4 white, the rest being black. A bag is picked at random and a ball drawn from it is white. What is the probability that it came from bag III?

\frac{4}{15}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

\frac{4}{5}

Correct answerYour answer

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AnswerC · $\frac{4}{9}$

$P(W) = \frac{1}{3}\left(\frac{2}{5} + \frac{3}{5} + \frac{4}{5}\right) = \frac{3}{5}$ .
$P(\text{III} \mid W) = \frac{\frac{1}{3} \times \frac{4}{5}}{\frac{3}{5}} = \frac{4}{9}$ .

Worth noting

$\frac{1}{3}$ is bag III's chance before the ball is seen; a white ball points towards the bag with the most white balls.

Question 108

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$P(E) = 0.4$ , $P(A \mid E) = 0.5$ and $P(A \mid E') = 0.2$ . What is $P(E \mid A)$ ?

0.2

Correct answerYour answer

0.4

Correct answerYour answer

0.5

Correct answerYour answer

0.625

Correct answerYour answer

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AnswerD · $0.625$

$P(A) = 0.4 \times 0.5 + 0.6 \times 0.2 = 0.2 + 0.12 = 0.32$ .
$P(E \mid A) = \frac{0.2}{0.32} = 0.625$ .

Worth noting

0.5 is $P(A \mid E)$ , the conditional the other way round.

Question 109

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A courier takes the highway, the city road or the hill road with probabilities $\frac{1}{2}$ , $\frac{1}{3}$ and $\frac{1}{6}$ . A parcel is delayed with probability $\frac{1}{10}$ on the highway, $\frac{1}{5}$ on the city road and $\frac{1}{2}$ on the hill road. A parcel is delayed. What is the probability that it went by the hill road?

\frac{1}{6}

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

\frac{5}{12}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

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AnswerC · $\frac{5}{12}$

$P(D) = \frac{1}{2} \times \frac{1}{10} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{6} \times \frac{1}{2} = \frac{3 + 4 + 5}{60} = \frac{1}{5}$ .
The hill road's share: $\frac{1/12}{1/5} = \frac{5}{12}$ .

Worth noting

$\frac{1}{6}$ is the hill road's chance before the delay is known; delays are likeliest there.

Question 110

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Bag I has 3 red balls and 1 black ball; bag II has 2 red and 2 black balls. A bag is picked at random and a ball is drawn from it; $R$ is the event that the ball is red. Which of the following statements are true?

I. $P(\text{I} \mid R) = \frac{3}{5}$ .

II. $P(\text{I} \mid R) + P(\text{II} \mid R) = 1$ .

III. $P(\text{I} \mid R) + P(\text{I} \mid R') = 1$ .

IV. $P(\text{II} \mid R') = \frac{2}{3}$ .

I, II and IV onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerI and II onlyCorrect answerYour answerII, III and IV onlyCorrect answerYour answer

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AnswerA · I, II and IV only

$P(R) = \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} = \frac{5}{8}$ , so $P(\text{I} \mid R) = \frac{3/8}{5/8} = \frac{3}{5}$ ; $P(R') = \frac{3}{8}$ , so $P(\text{II} \mid R') = \frac{2/8}{3/8} = \frac{2}{3}$ . I and IV are true.
II is true: given $R$ , the ball came from exactly one bag. III is false: $P(\text{I} \mid R') = \frac{1}{3}$ , and $\frac{3}{5} + \frac{1}{3} = \frac{14}{15}$ .

Worth noting

The chances of the two bags add to 1 under one condition; the chances of one bag under two different conditions need not.

Question 111

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Printers $X$ , $Y$ and $Z$ print 30%, 45% and 25% of an office's pages, and 2%, 4% and 6% of their pages, respectively, are smudged. A page is picked at random, and $S$ is the event that it is smudged. Match each probability in List I with its value in List II.

List I	List II
A. $P(S)$	I. $\frac{2}{13}$
B. $P(X \mid S)$	II. $\frac{5}{13}$
C. $P(Y \mid S)$	III. $0.039$
D. $P(Z \mid S)$	IV. $\frac{6}{13}$

A-III, B-I, C-IV, D-IICorrect answerYour answerA-III, B-IV, C-I, D-IICorrect answerYour answerA-III, B-I, C-II, D-IVCorrect answerYour answerA-III, B-II, C-IV, D-ICorrect answerYour answer

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AnswerA · A-III, B-I, C-IV, D-II

$P(S) = 0.3 \times 0.02 + 0.45 \times 0.04 + 0.25 \times 0.06 = 0.006 + 0.018 + 0.015 = 0.039$ .
$P(X \mid S) = \frac{6}{39} = \frac{2}{13}$ , $P(Y \mid S) = \frac{18}{39} = \frac{6}{13}$ and $P(Z \mid S) = \frac{15}{39} = \frac{5}{13}$ .

Worth noting

The three printers' chances add to 1, since a smudged page came from exactly one of them.

Question 112

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Assertion (A): Bag I has 1 red and 4 black balls, and bag II has 3 red and 2 black balls. A bag is picked at random and a red ball is drawn from it. Then bag II is three times as likely as bag I to have been picked.

Reason (R): When $P(\text{I}) = P(\text{II})$ , $\frac{P(\text{II} \mid R)}{P(\text{I} \mid R)} = \frac{P(R \mid \text{II})}{P(R \mid \text{I})}$ .

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AnswerA · Both A and R are true, and R is the correct explanation of A.

R: by Bayes' theorem both posterior chances have the same denominator $P(R)$ , and the equal $P(\text{I})$ and $P(\text{II})$ cancel, leaving the ratio of $P(R \mid \text{II})$ to $P(R \mid \text{I})$ . True.
A: that ratio is $\frac{3/5}{1/5} = 3$ . True, and R is the reason.

Worth noting

When the bags are equally likely to be picked, each bag's chance after a red ball is in proportion to its chance of giving red.

Question 113

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Bag I has 3 red and 4 black balls; bag II has 5 red and 4 black. A bag is picked at random and the ball drawn from it is red. What is the probability that it came from bag II?

\frac{1}{2}

Correct answerYour answer

\frac{5}{9}

Correct answerYour answer

\frac{35}{62}

Correct answerYour answer

\frac{5}{8}

Correct answerYour answer

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AnswerC · $\frac{35}{62}$

$P(\text{II} \mid R) = \frac{P(\text{II})P(R \mid \text{II})}{P(R)} = \frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{31}{63}}$ .
$\frac{5}{18} \times \frac{63}{31} = \frac{35}{62}$ .

Worth noting

$\frac{5}{9}$ is $P(R \mid \text{II})$ , the chance the other way round.

Question 114

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Machines A, B and C make 50%, 30% and 20% of a factory's bolts, and 1%, 2% and 3% of their bolts are faulty. A bolt picked at random is faulty. What is the probability that machine C made it?

\frac{3}{100}

Correct answerYour answer

\frac{1}{5}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{6}{17}

Correct answerYour answer

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AnswerD · $\frac{6}{17}$

$P(C \mid F) = \frac{0.2 \times 0.03}{0.017} = \frac{0.006}{0.017}$ .
$= \frac{6}{17}$ .

Worth noting

$\frac{1}{5}$ is C's share of all bolts; among faulty bolts its share is larger, as its fault rate is highest.

Question 115

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A condition affects 1% of people. A test detects it in 90% of those who have it, and wrongly shows it in 10% of those who do not. A person tests positive. What is the probability that they have the condition?

\frac{9}{1000}

Correct answerYour answer

\frac{1}{12}

Correct answerYour answer

\frac{1}{10}

Correct answerYour answer

\frac{9}{10}

Correct answerYour answer

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AnswerB · $\frac{1}{12}$

Let $D$ be having the condition and $T$ a positive test: $P(T) = 0.01 \times 0.9 + 0.99 \times 0.1 = 0.009 + 0.099 = 0.108$ .
$P(D \mid T) = \frac{0.009}{0.108} = \frac{1}{12}$ .

Worth noting

0.9 is the chance of a positive test for someone who has it, not the chance of having it after a positive test.

Question 116

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A die is thrown. If it shows 1 or 2, a ball is drawn from bag A (3 red, 2 white); otherwise from bag B (1 red, 4 white). The ball drawn is red. What is the probability that it came from bag A?

\frac{1}{5}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

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AnswerD · $\frac{3}{5}$

$P(R) = \frac{1}{3} \times \frac{3}{5} + \frac{2}{3} \times \frac{1}{5} = \frac{1}{3}$ .
$P(A \mid R) = \frac{\frac{1}{3} \times \frac{3}{5}}{\frac{1}{3}} = \frac{3}{5}$ .

Worth noting

$\frac{1}{3}$ is the chance of bag A before the ball is seen; a red ball makes bag A likelier.

Question 117

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Machine $M_1$ makes 60% of a factory's pens and machine $M_2$ the rest; 2% of $M_1$ 's pens and 3% of $M_2$ 's are defective. A pen picked at random is defective. What is the probability that $M_2$ made it?

0.03

Correct answerYour answer

0.4

Correct answerYour answer

0.5

Correct answerYour answer

0.6

Correct answerYour answer

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AnswerC · $0.5$

$P(D) = 0.6 \times 0.02 + 0.4 \times 0.03 = 0.012 + 0.012 = 0.024$ .
$P(M_2 \mid D) = \frac{0.012}{0.024} = 0.5$ .

Worth noting

0.4 is $M_2$ 's share of all pens; its higher defect rate lifts its share of defective pens to a half.

Question 118

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In a test with 5 options per question, a student knows the answer with probability $\frac{3}{5}$ and otherwise guesses. The student answers a question correctly. What is the probability that they knew the answer?

\frac{1}{5}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{17}{25}

Correct answerYour answer

\frac{15}{17}

Correct answerYour answer

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AnswerD · $\frac{15}{17}$

Let $K$ be knowing the answer and $C$ answering correctly: $P(C) = \frac{3}{5} \times 1 + \frac{2}{5} \times \frac{1}{5} = \frac{17}{25}$ .
$P(K \mid C) = \frac{3/5}{17/25} = \frac{15}{17}$ .

Worth noting

$\frac{17}{25}$ is the chance of a correct answer, the denominator, not the answer itself.

Question 119

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Box I has 2 white and 3 black balls; box II has 4 white and 1 black. A box is picked at random and the ball drawn from it is white. What is the probability that it came from box II?

\frac{1}{3}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

\frac{3}{5}

Correct answerYour answer

\frac{2}{3}

Correct answerYour answer

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AnswerD · $\frac{2}{3}$

$P(W) = \frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{4}{5} = \frac{3}{5}$ .
$P(\text{II} \mid W) = \frac{\frac{1}{2} \times \frac{4}{5}}{\frac{3}{5}} = \frac{2}{3}$ .

Worth noting

$\frac{3}{5}$ is $P(W)$ , the chance of white at all; Bayes divides by it.

Question 120

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A shop gets 70% of its eggs from farm P and 30% from farm Q. Of P's eggs, 5% are cracked; of Q's, 10% are cracked. An egg is picked at random.

The egg is cracked. What is the probability that it came from farm Q?

\frac{1}{10}

Correct answerYour answer

\frac{3}{10}

Correct answerYour answer

\frac{6}{13}

Correct answerYour answer

\frac{7}{13}

Correct answerYour answer

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AnswerC · $\frac{6}{13}$

$P(Q \mid C) = \frac{P(Q \cap C)}{P(C)} = \frac{0.03}{0.065}$ .
$= \frac{30}{65} = \frac{6}{13}$ .

Worth noting

$\frac{7}{13}$ is the chance it came from farm P.

Question 121

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A shop gets 70% of its eggs from farm P and 30% from farm Q. Of P's eggs, 5% are cracked; of Q's, 10% are cracked. An egg is picked at random.

The egg is not cracked. What is the probability that it came from farm P?

\frac{54}{187}

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\frac{7}{10}

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\frac{133}{187}

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\frac{19}{20}

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AnswerC · $\frac{133}{187}$

$P(C') = 1 - 0.065 = 0.935$ , and $P(P \cap C') = 0.7 \times 0.95 = 0.665$ .
$P(P \mid C') = \frac{0.665}{0.935} = \frac{133}{187}$ .

Worth noting

0.95 is $P(C' \mid P)$ , the reverse conditional; the answer is only a little above P's share of 0.7.

Question 122

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A die is thrown. If it shows 1, a ball is drawn from urn I (2 white, 3 black); if it shows 2 or 3, from urn II (4 white, 1 black); otherwise from urn III (3 white, 2 black). The ball drawn is white. What is the probability that it came from urn II?

\frac{1}{3}

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\frac{8}{19}

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\frac{4}{9}

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\frac{4}{5}

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AnswerB · $\frac{8}{19}$

$P(W) = \frac{19}{30}$ , and $P(\text{II})\,P(W \mid \text{II}) = \frac{2}{6} \times \frac{4}{5} = \frac{8}{30}$ .
$P(\text{II} \mid W) = \frac{8/30}{19/30} = \frac{8}{19}$ .

Worth noting

$\frac{1}{3} = \frac{2}{6}$ is urn II's chance before the ball is seen; urn II gives white most often, so a white ball raises it.

Question 123

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20% of the items reaching an inspector are defective. The inspector flags 90% of the defective items, and also wrongly flags 5% of the good ones. An item is not flagged. What is the probability that it is defective?

\frac{1}{39}

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\frac{1}{10}

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\frac{1}{5}

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\frac{39}{50}

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AnswerA · $\frac{1}{39}$

Not flagged: defective $0.2 \times 0.1 = 0.02$ and good $0.8 \times 0.95 = 0.76$ , in all $0.78$ .
$\frac{0.02}{0.78} = \frac{1}{39}$ .

Worth noting

$\frac{1}{10}$ is the chance that a defective item escapes, not the chance that an unflagged item is defective.

Question 124

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Bag I has 3 red and 2 blue balls; bag II has 3 red and 1 blue ball. A bag is picked at random and two balls are drawn from it, one after the other, without replacement. Both are red. What is the probability that bag II was picked?

\frac{3}{8}

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\frac{1}{2}

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\frac{5}{9}

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\frac{5}{8}

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AnswerD · $\frac{5}{8}$

$P(RR \mid \text{I}) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ and $P(RR \mid \text{II}) = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ , so $P(RR) = \frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{1}{2} = \frac{2}{5}$ .
$P(\text{II} \mid RR) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{2}{5}} = \frac{5}{8}$ .

Worth noting

$\frac{5}{9}$ is the answer after one red ball; a second red ball, likelier from bag II, raises the chance further.

Question 125

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A bag holds 3 fair coins and 1 coin with a head on both sides. A coin is picked at random and tossed twice, and both tosses give a head. What is the probability that it is the two-headed coin?

\frac{1}{4}

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\frac{2}{5}

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\frac{4}{7}

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\frac{4}{5}

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AnswerC · $\frac{4}{7}$

Two heads come from a fair coin with probability $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ and from the two-headed coin with probability 1, so $P(HH) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times 1 = \frac{7}{16}$ .
$\frac{1/4}{7/16} = \frac{4}{7}$ .

Worth noting

$\frac{2}{5}$ is the answer after one head; the second head makes the two-headed coin likelier still.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.