Probability for CUET Maths

AnswerC · $216$

1. Multiply the numbers of outcomes of the separate parts; for a choice without order, use combinations.

2. $6 \times 6 \times 6 = 216$.

AnswerB · $12$

1. Multiply the numbers of outcomes of the separate parts; for a choice without order, use combinations.

2. $2 \times 6 = 12$.

AnswerD · A number greater than 6

1. An impossible event contains no outcome of the sample space $\{1, 2, 3, 4, 5, 6\}$.

2. No face is greater than 6, so that event is $\varnothing$. 'Less than 7' is every outcome, a sure event; 2 is an even prime; 3 and 6 are multiples of 3.

AnswerC · $3$

1. The sample space is $\{HH, HT, TH, TT\}$.

2. At least one head: $HH$, $HT$ and $TH$, 3 outcomes. It has more than one outcome, so it is a compound event.

AnswerD · $32$

1. Each toss has 2 outcomes, and the tosses multiply.

2. $2^5 = 32$.

AnswerB · $10$

1. A committee has no order, so count combinations: ${}^{5}C_2$.

2. ${}^{5}C_2 = \frac{5 \times 4}{2} = 10$.

AnswerD · Getting a 4

1. An elementary event has exactly one outcome.

2. 'Getting a 4' is $\{4\}$. The others are $\{2, 4, 6\}$, $\{1, 2\}$ and $\{2, 3, 5\}$, compound events.

AnswerB · The sum is at least 2

1. A sure event contains every outcome of the sample space.

2. The smallest sum is $1 + 1 = 2$, so the sum is always at least 2.

AnswerC · $5$

1. List the ordered pairs with sum 8.

2. $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$: 5 outcomes.

AnswerA · $1{,}326$

1. Multiply the numbers of outcomes of the separate parts; for a choice without order, use combinations.

2. ${}^{52}C_2 = \frac{52 \times 51}{2} = 1{,}326$.

AnswerC · $72$

1. Multiply the outcomes of the three parts: $6 \times 6 \times 2$.

2. $= 72$.

AnswerB · $2$

1. 'A but not B' is $A \cap B'$: the outcomes of $A$ that are not in $B$.

2. Removing 3 and 4 from $A$ leaves $\{1, 2\}$: 2 outcomes.

AnswerC · $4$

1. $A = \{2, 4, 6\}$ and $B = \{3, 6\}$.

2. $A \cup B = \{2, 3, 4, 6\}$: 4 outcomes, with 6 counted once.

AnswerB · A-II, B-IV, C-I, D-III

1. $A = \{2, 4, 6\}$ and $B = \{4, 5, 6\}$. 'A or B' joins them: $\{2, 4, 5, 6\}$; 'A and B' keeps the common part: $\{4, 6\}$.

2. 'A but not B' removes $B$'s outcomes from $A$: $\{2\}$; 'not A' is the odd numbers $\{1, 3, 5\}$.

AnswerD · At most one head

1. $E'$ is every outcome not in $E$: fewer than two heads, that is 0 or 1 head.

2. So $E'$ is 'at most one head': $\{TTT, HTT, THT, TTH\}$.

AnswerA · 'a number less than 3' and 'a number greater than 4'

1. Mutually exclusive events have no outcome in common: they cannot happen together.

2. $\{1, 2\}$ and $\{5, 6\}$ share nothing. The others share 2, then 3, then 3 again.

AnswerC · Both I and II

1. Equal numbers give an even sum, never 7, so the two events cannot happen together: I is true.

2. Every sum is even or odd, so the two events cover the whole sample space: II is true.

AnswerB · I, II and IV only

1. $E$ and $E'$ share no outcome (I) and together make up the whole sample space (II). If $A$ lies inside $B$, their common part is $A$ itself (IV).

2. III is false: on a die, $\{1\}$ and $\{2\}$ are mutually exclusive but leave out 3, 4, 5 and 6.

AnswerA · $2$

1. $B = \{HHT, HTH, THH\}$.

2. Of these, the ones that start with $H$ are $HHT$ and $HTH$: 2 outcomes.

AnswerC · $\{1, 2\}, \{2, 3, 4\}, \{4, 5, 6\}$

1. Events are exhaustive when together they cover the whole sample space $\{1, 2, 3, 4, 5, 6\}$.

2. $\{1, 2\} \cup \{2, 3, 4\} \cup \{4, 5, 6\}$ is everything; the others miss 6, 5, and 4 and 6.

AnswerA · Only I

1. I: not 'A or B' means neither $A$ nor $B$, that is $A' \cap B'$. True.

2. II: not 'A and B' means at least one fails, which is $A' \cup B'$, not $A' \cap B'$. False.

AnswerB · $(A \cap B') \cup (A' \cap B)$

1. Exactly one: either $A$ without $B$, $A \cap B'$, or $B$ without $A$, $A' \cap B$.

2. Join the two: $(A \cap B') \cup (A' \cap B)$.

AnswerD · 'a king' and 'a queen'

1. Mutually exclusive events cannot happen together: no card is in both.

2. No card is both a king and a queen. The king of hearts, every heart, and the face cards of spades break the other pairs.

AnswerD · $\frac{7}{6}$

1. A probability always lies between 0 and 1, both included.

2. $\frac{7}{6}$ is more than 1, so it cannot be a probability. 0 is the probability of an impossible event.

AnswerA · $0.4$

1. The probabilities of all the outcomes of a sample space add up to 1.

2. $0.1 + 0.3 + p + 0.2 = 1$, so $p = 0.4$.

AnswerA · $\frac{3}{8}$

1. Odds 3 : 5 mean 3 favourable cases for every 5 unfavourable ones, 8 in all.

2. So the probability is $\frac{3}{8}$.

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

1. R is one of the rules every probability obeys: $0 \le P(E) \le 1$.

2. 1.2 is more than 1, so A is true, and it is true because of R.

AnswerA · $\frac{2}{9}$

1. Odds against 7 : 2 mean 7 unfavourable cases for every 2 favourable ones, 9 in all.

2. So the probability that it happens is $\frac{2}{9}$.

AnswerD · 0.2, 0.5, 0.3

1. A valid assignment gives every outcome a probability from 0 to 1, and the probabilities add up to 1.

2. 0.2 + 0.5 + 0.3 = 1. The others add to 1.2, have a negative value, or add to 0.9.

AnswerC · I, II and III only

1. I and II are axioms. III: $B$ contains every outcome of $A$ and possibly more.

2. IV is false: on a die, 'even' has probability $\frac{1}{2}$ and 'less than 5' has $\frac{2}{3}$, which add to more than 1, since the events overlap.

AnswerD · $\frac{2}{3}$

1. Mutually exclusive and exhaustive: $P(A) + P(B) = 1$.

2. $2P(B) + P(B) = 1$, so $P(B) = \frac{1}{3}$ and $P(A) = \frac{2}{3}$.

AnswerD · $\frac{6}{11}$

1. Write $P(\omega_1) = p$; then $P(\omega_2) = \frac{p}{2}$ and $P(\omega_3) = \frac{p}{3}$.

2. $p + \frac{p}{2} + \frac{p}{3} = \frac{11p}{6} = 1$, so $p = \frac{6}{11}$.

AnswerB · $0.3$

1. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1.3 - P(A \cap B)$.

2. $P(A \cup B) \le 1$, so $P(A \cap B) \ge 0.3$: the events must overlap by at least 0.3.

AnswerC · $\frac{1}{6}$

1. There are $6 \times 6 = 36$ equally likely outcomes.

2. The sum is 7 in 6 of them: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). So the probability is $\frac{6}{36} = \frac{1}{6}$.

AnswerC · $\frac{3}{8}$

1. There are $2^{3} = 8$ equally likely outcomes.

2. 3 of them have exactly two heads, so the probability is $\frac{3}{8} = \frac{3}{8}$.

AnswerA · $\frac{2}{5}$

1. The primes up to 20 are 2, 3, 5, 7, 11, 13, 17 and 19: 8 numbers.

2. The probability is $\frac{8}{20} = \frac{2}{5}$.

AnswerB · $\frac{5}{14}$

1. Ways to choose 2 of the 8 balls: ${}^{8}C_2 = 28$. Ways to choose 2 of the 5 red: ${}^{5}C_2 = 10$.

2. So the probability is $\frac{10}{28} = \frac{5}{14}$.

AnswerB · $\frac{1}{6}$

1. There are 36 equally likely ordered pairs.

2. The doublets are $(1, 1)$ to $(6, 6)$: 6 of 36, so $\frac{6}{36} = \frac{1}{6}$.