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Counting and Probability for CUET General Aptitude

About 15 min to learn · 65 questions

CUET's counting and probability questions cover the addition and multiplication rules, numbers formed from digits, permutations and combinations, repeated letters and seating, classical probability, at least one, independent events, drawing with and without replacement, and coins, dice and cards. The questions are short but punish a misread word: with or without replacement, order or group, repeats allowed or not. This lesson works through each with examples and the traps beside them.

Syllabus area
Reasoning
On this page · 9 sections
  1. 01The core idea
  2. 02Counting rules
  3. 03Permutations and combinations
  4. 04Probability
  5. 05Coins, dice and cards
  6. 06Exam shortcuts
  7. 07Common mistakes
  8. 08Quick revision
  9. 09Check your understanding

The core idea

Every counting question asks how many outcomes there are, and every probability question divides one count by another. So the first job is always the same: decide what one outcome looks like.

Then read the stem twice. "With replacement" or "without", "order matters" or "a group", "digits may repeat" or "no digit repeated": each word changes the answer, and the answer the other reading gives is usually among the options.

Counting rules

What is the difference between the addition rule and the multiplication rule?

Add when you make one choice from several separate groups, an "either … or". Multiply when you make several choices one after another and every choice goes with every other, an "and then".

SituationRuleExample
One item from group A or group Ba+ba + bOne sweet from 7 kinds of barfi or 4 of ladoo: 11
One from A and one from Ba×ba \times bOne barfi and one ladoo: 28
The same choice made kk times, nn options eachnkn^k3 letters, each any of 26: 26326^3

Example 1 · Add or multiply

A café sells 6 kinds of sandwich and 7 kinds of juice. In how many ways can a customer buy one item only, and in how many ways one sandwich with one juice?

  1. One item, from either list: 6+7=136 + 7 = 13.

  2. A sandwich and a juice: 6×7=426 \times 7 = 42.

Answer

13 and 42

Exam shortcut

Many questions mix the two. Split the situation into cases that cannot happen together, multiply inside each case, then add the cases.

How do you count numbers formed from given digits?

Draw one box for each place, fill the most restricted box first, and multiply the number of choices for each box. The usual restrictions are that the first digit cannot be 0, that the number must be even or odd, and that no digit may repeat.

  1. A leading digit can never be 0.
  2. "Even" or "odd" fixes the units digit. "Divisible by 5" means the units digit is 0 or 5.
  3. With no repetition, each box has one choice fewer than the box filled before it.
  4. If 0 is among the digits and the units digit is restricted, split into cases, units digit 0 and units digit not 0, because the leading-digit rule works differently in each.

Example 2 · Four-digit numbers

How many four-digit numbers can be formed from 0, 1, 3, 5, 7, 9 if no digit is repeated?

  1. Thousands place: any digit except 0, so 5 choices.

  2. Hundreds place: the remaining 5 digits, since 0 is now allowed.

  3. Tens: 4. Units: 3.

  4. Total: 5×5×4×3=3005 \times 5 \times 4 \times 3 = 300.

Answer

300

With repetition allowed it would be 5×6×6×6=1,0805 \times 6 \times 6 \times 6 = 1{,}080. One word in the stem moves the answer that far.

Watch out

Letting 0 lead is the commonest slip. A three-digit number cannot start with 0, so the first box never has all 10 digits.

Permutations and combinations

When do you use permutations and when do you use combinations?

Use a permutation when order matters: ranks, posts, positions, codes, arrangements in a row. Use a combination when only the group matters: teams, committees, handshakes, choosing items.

Arrangements and selections

nPr=n!(n−r)!nCr=n!r! (n−r)!nPr=nCr×r!{}^nP_r = \frac{n!}{(n - r)!} \qquad {}^nC_r = \frac{n!}{r!\,(n - r)!} \qquad {}^nP_r = {}^nC_r \times r!

Also 0!=10! = 1, nC0=nCn=1{}^nC_0 = {}^nC_n = 1, nC1=n{}^nC_1 = n, nC2=n(n−1)2{}^nC_2 = \tfrac{n(n - 1)}{2}, and nCr=nCn−r{}^nC_r = {}^nC_{n - r}.

Example 3 · Posts or a team

From 11 students, in how many ways can we choose a captain and a vice-captain, and in how many ways a two-member team?

  1. Posts, so order matters: 11P2=11×10=110{}^{11}P_2 = 11 \times 10 = 110.

  2. A team, so order does not: 11C2=110÷2=55{}^{11}C_2 = 110 \div 2 = 55.

Answer

110 and 55

The ÷2\div 2 is exactly the 2!2! ways of ordering the same pair.

Exam shortcut

Swap two of the people chosen. If that gives a different outcome, as when a captain and a vice-captain swap jobs, order matters. If nothing changes, as with two members of a committee, it does not.

Several standard counts follow from nC2{}^nC_2 and nC3{}^nC_3.

QuestionCount
Handshakes among nn peoplenC2{}^nC_2
Straight lines through nn points, no three in a linenC2{}^nC_2
Diagonals of an nn-sided polygonnC2−n=n(n−3)2{}^nC_2 - n = \tfrac{n(n - 3)}{2}
Triangles from nn points, no three in a linenC3{}^nC_3
Shortest grid routes, aa steps across and bb upa+bCa{}^{a + b}C_a

How do you count selections with a condition such as "at least 2"?

List the cases that satisfy the condition, count each as a product of combinations, and add. Or count every selection and subtract the ones you do not want. Doing both is a good check.

Example 4 · At least 2 women

A committee of 3 is chosen from 5 men and 4 women. In how many ways can it have at least 2 women?

  1. Exactly 2 women and 1 man: 4C2×5C1=6×5=30{}^4C_2 \times {}^5C_1 = 6 \times 5 = 30.

  2. Exactly 3 women: 4C3=4{}^4C_3 = 4.

  3. Total: 30+4=3430 + 4 = 34.

  4. Check by subtraction: all committees 9C3=84{}^9C_3 = 84, with no woman 5C3=10{}^5C_3 = 10, with one woman 4×5C2=404 \times {}^5C_2 = 40. So 84−10−40=3484 - 10 - 40 = 34.

Answer

34

How do you arrange letters when some of them repeat?

Divide the number of arrangements by the factorial of each repeated letter's count. Swapping two identical letters gives the same word, so every arrangement was counted that many times.

Arrangements with repeats

n!p! q! ⋯\frac{n!}{p!\,q!\,\cdots}

Here nn is the number of letters, and one letter appears pp times, another qq times, and so on.

Example 5 · Repeated letters

How many arrangements of all the letters of COMMITTEE are there?

  1. 9 letters: M twice, T twice, E twice, and C, O and I once each.

  2. Arrangements: 9!÷(2!×2!×2!)=3,62,880÷8=45,3609! \div (2! \times 2! \times 2!) = 3{,}62{,}880 \div 8 = 45{,}360.

Answer

45,360

Watch out

Dividing by only one of the repeated letters. Write down every letter's count before you start.

How do you handle "together", "never together" and round-table arrangements?

For "together", tie the group into one block, arrange the blocks, then multiply by the arrangements inside the block. For "never together", subtract the "together" count from the total. For a round table, fix one person and arrange the rest, so nn people give (n−1)!(n - 1)! seatings.

Example 6 · Together, apart and round

Seven people sit in a row, and 4 of them are friends who must sit together. Then, separately: in how many ways can 5 people sit in a row if A and B must not sit together? And in how many ways can 6 people sit at a round table?

  1. Together: the 4 friends form one block, so there are 1+3=41 + 3 = 4 units. That gives 4!×4!=24×24=5764! \times 4! = 24 \times 24 = 576, the second 4!4! arranging the friends inside the block.

  2. Never together: all rows 5!=1205! = 120, with A and B together 4!×2=484! \times 2 = 48. So 120−48=72120 - 48 = 72.

  3. Round table: (6−1)!=5!=120(6 - 1)! = 5! = 120.

Answer

576, 72 and 120

Watch out

Forgetting the arrangements inside a block. Two people who sit together can do so in 2 orders.

Necklaces and garlands

A necklace or garland can be flipped over, so clockwise and anticlockwise orders are the same: nn beads give (n−1)!2\tfrac{(n - 1)!}{2} arrangements. Use this only when the stem says the two directions count as one. It is never true of people at a table.

Probability

How do you find a probability?

When all outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes. It always lies between 0, impossible, and 1, certain.

Probability rules

P(A)=favourable outcomestotal outcomesP(A) = \frac{\text{favourable outcomes}}{\text{total outcomes}}

Complement: P(not A)=1−P(A)P(\text{not } A) = 1 - P(A). Either: P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B), and the last term is 0 when AA and BB cannot happen together.

Example 7 · A bag of beads

A bag has 8 yellow and 5 black beads. One bead is picked at random. Find the probability that it is black.

  1. There are 13 beads, each equally likely.

  2. P(black)=513P(\text{black}) = \tfrac{5}{13}.

Answer
513\tfrac{5}{13}

Watch out

The method needs equally likely outcomes. The 11 possible totals of two dice are not equally likely, and neither are the letters of a word in which one letter appears four times. Count the underlying outcomes, the ordered pairs or the individual cards, not the labels.

How do you find the probability of "at least one"?

Use the complement: the probability of at least one is 1 minus the probability of none. "None" is one case and "at least one" is many, so the complement is almost always faster.

Example 8 · At least one head

A fair coin is tossed 5 times. Find the probability of at least one head.

  1. P(no head)=P(all tails)=(12)5=132P(\text{no head}) = P(\text{all tails}) = (\tfrac{1}{2})^5 = \tfrac{1}{32}.

  2. P(at least one head)=1−132=3132P(\text{at least one head}) = 1 - \tfrac{1}{32} = \tfrac{31}{32}.

Answer
3132\tfrac{31}{32}

Watch out

Adding the separate probabilities for "at least one" is a classic error. It can even give an answer above 1.

How do you combine independent events?

Two events are independent when one happening does not change the chance of the other, such as two different people shooting, or a coin and a die. For independent events, the probability that both happen is the product of their probabilities.

Event, AA and BB independentProbability
Both happenP(A)×P(B)P(A) \times P(B)
Neither happens(1−P(A))×(1−P(B))(1 - P(A)) \times (1 - P(B))
At least one happens1−(1−P(A))(1−P(B))1 - (1 - P(A))(1 - P(B))
Exactly one happensP(A)(1−P(B))+(1−P(A))P(B)P(A)(1 - P(B)) + (1 - P(A))P(B)

Example 9 · Exactly one passes

Rohan passes a driving test with probability 0.7 and Sneha with probability 0.9, independently. Find the probability that exactly one of them passes.

  1. Rohan passes and Sneha fails: 0.7×0.1=0.070.7 \times 0.1 = 0.07.

  2. Rohan fails and Sneha passes: 0.3×0.9=0.270.3 \times 0.9 = 0.27.

  3. Exactly one: 0.07+0.27=0.340.07 + 0.27 = 0.34.

Answer

0.34

Both pass with 0.63 and neither with 0.03, and 0.63+0.34+0.03=10.63 + 0.34 + 0.03 = 1, as it must.

What changes when you draw without replacement?

Without replacement, each draw leaves one item fewer, so the second probability comes from a smaller total and the draws are no longer independent. With replacement, every draw is a fresh start.

Example 10 · Two balls without replacement

A bag holds 5 red and 3 green balls. Two are drawn without replacement. Find the probability that both are green.

  1. First green: 38\tfrac{3}{8}. Second green, with 2 green left among 7: 27\tfrac{2}{7}.

  2. Both green: 38×27=656=328\tfrac{3}{8} \times \tfrac{2}{7} = \tfrac{6}{56} = \tfrac{3}{28}.

  3. Check with combinations: 3C2÷8C2=328{}^3C_2 \div {}^8C_2 = \tfrac{3}{28}.

Answer
328\tfrac{3}{28}

With replacement it would be (38)2=964(\tfrac{3}{8})^2 = \tfrac{9}{64}.

Key idea

"Drawn together" and "drawn at once" both mean without replacement.

Coins, dice and cards

What should you know about coins, dice and playing cards?

They are equally likely sample spaces of known size: 2 outcomes per coin, 6 per die, and 52 cards in a deck. Count the outcomes, and keep ordered outcomes separate.

Coins. nn coins, or nn tosses of one coin, give 2n2^n outcomes, and the number with exactly kk heads is nCk{}^nC_k.

CoinsOutcomesExactly 0, 1, 2, 3, … heads
241, 2, 1
381, 3, 3, 1
4161, 4, 6, 4, 1

Dice. Two dice give 36 ordered outcomes, so (2, 5) and (5, 2) are different. Each total's count reads off the grid: 1 way for 2, rising by one to 6 ways for 7, then falling back to 1 way for 12. There are 6 doubles, and 15 outcomes each with the first die higher or with the second die higher.

A six by six grid of the totals of two dice, first die down the side and second die across the top. The five cells with total 8 are shaded: 2 and 6, 3 and 5, 4 and 4, 5 and 3, 6 and 2.
The 36 outcomes of two dice. A total of 8 comes up in 5 of them.

Example 11 · A dice total

Two dice are rolled. Find the probability that the total is 8.

  1. A total of 8 comes in 5 ways: (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).

  2. P=536P = \tfrac{5}{36}.

Answer
536\tfrac{5}{36}

Cards. A deck has 52 cards in 4 suits of 13: ace, 2 to 10, jack, queen and king. Hearts and diamonds are red, clubs and spades black. The jack, queen and king are face cards, so there are 12 face cards, 4 of each rank, and 26 of each colour.

Example 12 · A card

One card is drawn. Find the probability that it is a black face card.

  1. Black face cards: the jack, queen and king of clubs and of spades, so 6.

  2. P=652=326P = \tfrac{6}{52} = \tfrac{3}{26}.

Answer
326\tfrac{3}{26}

Watch out

For "A or B" with cards, ask what is counted twice: the red kings in "red or king", the face hearts in "heart or face card". Subtract them once.

Exam shortcuts

  • Fill the restricted box first. Leading digit, units digit, then the rest.
  • Swap test. If swapping two chosen items changes the outcome, use nPr{}^nP_r. If not, use nCr{}^nC_r.
  • "At least", "not", "neither". Go straight to the complement.
  • Small numbers. Listing outcomes is often faster than a formula, and the best check on one.
  • Exactly one, both and neither add to 1. Use that to check an answer.

Common mistakes

  • Adding where you should multiply. "And then" multiplies. "Either … or" adds.
  • Letting 0 lead a number.
  • Ignoring "no repetition", or assuming it.
  • nPr{}^nP_r for a committee, or nCr{}^nC_r for posts.
  • Dividing by only one repeated letter.
  • Forgetting the arrangements inside a block.
  • Treating unequal outcomes as equal. Dice totals, and "two heads, one head, no head" for two coins, are not equally likely.
  • Counting (3, 6) and (6, 3) as one outcome.
  • Adding probabilities for "at least one". Use 1−P(none)1 - P(\text{none}).
  • Double counting in "A or B". Subtract P(A and B)P(A \text{ and } B).
  • With-replacement numbers for a draw without replacement.
  • Answering the complement. If you found P(none)P(\text{none}) on the way, do not pick it.

Quick revision

Counting

  • Either … or: add. And then: multiply
  • Fill restricted places first, and 0 never leads
  • kk choices from nn options, repeats allowed: nkn^k

Permutations and combinations

  • nPr=n!(n−r)!{}^nP_r = \tfrac{n!}{(n - r)!}, nCr=n!r!(n−r)!{}^nC_r = \tfrac{n!}{r!(n - r)!}
  • Repeated letters: n!p! q!⋯\tfrac{n!}{p!\,q!\cdots}
  • Together: block, then inside the block. Round table: (n−1)!(n - 1)!
  • Diagonals n(n−3)2\tfrac{n(n - 3)}{2}, triangles nC3{}^nC_3

Probability

  • Favourable ÷ total, for equally likely outcomes
  • P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
  • At least one =1−P(none)= 1 - P(\text{none})
  • Independent: multiply. Without replacement: the total shrinks

Coins, dice, cards

  • nn coins: 2n2^n outcomes
  • Two dice: 36 ordered outcomes, 7 the commonest total
  • Cards: 52, 13 per suit, 12 face cards, 26 of each colour

A probability is a fraction of a whole, so ratio and proportion and percentages carry straight over, and counting factors by multiplying one more than each prime's power links this chapter to the number system.

Next · 08Quick revision

Check your understanding

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

The counting principleQ1 of 65Q1

The counting principle

16 questions

Question 1

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Tenzin packs 4 shirts, 3 pairs of trousers and 2 pairs of shoes for a trip. An outfit is one shirt, one pair of trousers and one pair of shoes. How many different outfits can he make?

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AnswerD · 24

  1. Each choice is made independently, so multiply: 4×3×24 \times 3 \times 2.

  2. 4×3=124 \times 3 = 12, and 12×2=2412 \times 2 = 24 outfits.

Worth noting

9 comes from adding 4+3+24 + 3 + 2; addition is for choosing ONE item from several groups. 12 forgets the shoes.

Question 2

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A college canteen has 5 different vegetarian thalis and 3 different non-vegetarian thalis on its menu. Gurpreet will order exactly one thali. In how many ways can she choose?

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AnswerC · 8

  1. She picks one thali from the veg list OR one from the non-veg list, never both.

  2. For 'either … or' choices, add: 5+3=85 + 3 = 8.

Worth noting

15 multiplies 5×35 \times 3, which would count a veg thali AND a non-veg thali together. She orders only one.

Question 3

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Three different roads connect Kochi to Thrissur, and four different roads connect Thrissur to Palakkad. In how many ways can Joseph drive from Kochi to Palakkad through Thrissur?

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AnswerA · 12

  1. The first leg can be done in 3 ways, and each of these can be followed by any of the 4 roads on the second leg.

  2. Routes =3×4=12= 3 \times 4 = 12.

Worth noting

24 doubles the answer as if the trip were a round trip. 64 and 81 are 434^3 and 343^4: powers are for repeating the SAME choice, not for two different legs.

Question 4

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How many three-digit numbers can be formed using only the digits 1, 2, 3, 4 and 5, if a digit may be repeated?

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AnswerC · 125

  1. Each of the three places can be filled by any of the 5 digits, because repetition is allowed.

  2. Numbers =5×5×5=125= 5 \times 5 \times 5 = 125.

Worth noting

60 is 5×4×35 \times 4 \times 3, the count when digits may NOT repeat. 243 is 353^5, the powers the wrong way round.

Question 5

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A ship signals by hoisting 3 flags one above another, chosen from 5 flags of different colours. No flag is used twice in a signal, and a different order gives a different signal. How many signals are possible?

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AnswerC · 60

  1. Top flag: 5 choices; middle: 4; bottom: 3.

  2. Signals =5×4×3=60= 5 \times 4 \times 3 = 60.

Worth noting

10 is the number of ways to CHOOSE 3 flags, ignoring order; here order changes the signal. 125 lets a flag repeat.

Question 6

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A teacher asks her class to list every two-digit number whose tens digit and units digit are both even. How many numbers are on a complete list?

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AnswerA · 20

  1. Tens digit: 2, 4, 6 or 8 (0 would not give a two-digit number), so 4 choices.

  2. Units digit: 0, 2, 4, 6 or 8, so 5 choices.

  3. Numbers =4×5=20= 4 \times 5 = 20.

Worth noting

25 is 5×55 \times 5, which lets the tens digit be 0. 40 checks only the tens digit, and 45 counts every even two-digit number, where only the units digit need be even.

Question 7

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Four roads join Jaipur to Ajmer and three roads join Ajmer to Udaipur. Farhan drives from Jaipur to Udaipur through Ajmer and back again through Ajmer, and he does not use any road a second time. How many different round trips are possible?

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AnswerC · 72

  1. Going: 4×3=124 \times 3 = 12 ways.

  2. Returning: one Udaipur–Ajmer road and one Ajmer–Jaipur road are already used, so 2×3=62 \times 3 = 6 ways.

  3. Round trips =12×6=72= 12 \times 6 = 72.

Worth noting

144 is 12×1212 \times 12, which allows a road to be reused. 12 counts only the outward journey.

Question 8

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How many three-digit numbers can be made from the digits 0, 2, 4, 6 and 8 if no digit is used more than once in a number?

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AnswerA · 48

  1. The hundreds digit cannot be 0, so it has 4 choices.

  2. The tens digit can be any of the remaining 4 digits (0 is now allowed).

  3. The units digit has the remaining 3 choices, so the count is 4×4×3=484 \times 4 \times 3 = 48.

Worth noting

60 is 5×4×35 \times 4 \times 3, which lets 0 lead (giving two-digit 'numbers' like 024). 80 is 4×5×44 \times 5 \times 4, which forgets that the hundreds digit is already used when filling the tens place. 100 allows repetition.

Question 9

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Using the digits 1 to 6, with no digit repeated, how many three-digit EVEN numbers can be formed?

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AnswerB · 60

  1. Fill the restricted place first: the units digit must be 2, 4 or 6, so 3 choices.

  2. The hundreds digit then has 5 choices and the tens digit 4.

  3. Even numbers =3×5×4=60= 3 \times 5 \times 4 = 60.

Worth noting

120 counts all three-digit numbers without the even condition. 108 is 6×6×36 \times 6 \times 3, which allows repetition.

Question 10

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A locker code is a string of 4 digits from 0 to 9. Digits may repeat, but the code must not begin with 0. How many codes are possible?

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AnswerC · 9,000

  1. First digit: 9 choices (1 to 9).

  2. Each of the other three digits: 10 choices, since repetition is allowed.

  3. Codes =9×10×10×10=9000= 9 \times 10 \times 10 \times 10 = 9000.

Worth noting

10,000 ignores the rule about 0. 4,536 is 9×9×8×79 \times 9 \times 8 \times 7, which wrongly forbids repetition.

Question 11

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A restaurant's lunch offer is one main course from 4 choices together with EITHER one starter from 3 choices OR one dessert from 2 choices (not both). How many different lunches are possible?

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AnswerC · 20

  1. Main with a starter: 4×3=124 \times 3 = 12 lunches.

  2. Main with a dessert: 4×2=84 \times 2 = 8 lunches.

  3. The two cases cannot overlap, so add: 12+8=2012 + 8 = 20.

Worth noting

24 is 4×3×24 \times 3 \times 2, which gives every lunch both a starter and a dessert. 12 counts only the starter case.

Question 12

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A quiz has 5 questions, and each question has 4 answer choices. A student marks exactly one choice for every question. In how many different ways can the answer sheet be filled?

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AnswerD · 1,024

  1. Each question is answered in 4 ways, independently of the others.

  2. Ways =4×4×4×4×4=45=1024= 4 \times 4 \times 4 \times 4 \times 4 = 4^5 = 1024.

Worth noting

625 is 545^4: the base must be the number of choices per question (4), raised to the number of questions (5). 20 multiplies 5 by 4 once.

Question 13

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How many positive whole numbers divide 360 exactly (including 1 and 360)?

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AnswerD · 24

  1. Prime factorisation: 360=23×32×5360 = 2^3 \times 3^2 \times 5.

  2. A divisor takes the power of 2 from 0 to 3 (4 ways), of 3 from 0 to 2 (3 ways), of 5 from 0 to 1 (2 ways).

  3. Divisors =4×3×2=24= 4 \times 3 \times 2 = 24.

Worth noting

6 multiplies the exponents 3×2×13 \times 2 \times 1, forgetting that each power can also be 0. 9 adds the choices instead of multiplying. 22 leaves out 1 and 360.

Question 14

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In the numbers from 1 to 500, both included, how many contain the digit 7 at least once?

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AnswerB · 95

  1. Count the numbers with NO 7. Write 0 to 499 as three digits (000 to 499): hundreds digit 0 to 4 (5 ways), tens and units each 9 ways (any digit except 7).

  2. That gives 5×9×9=4055 \times 9 \times 9 = 405 strings. Drop 000 and add 500 (which has no 7): still 405 numbers from 1 to 500 without a 7.

  3. Numbers with at least one 7 =500−405=95= 500 - 405 = 95.

Worth noting

Adding 50 numbers with 7 in the units place and 50 with 7 in the tens place gives 100, but 77, 177, 277, 377 and 477 are counted twice. 405 is the count WITHOUT a 7.

Question 15

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How many three-digit numbers that are divisible by 5 have all three digits different?

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AnswerC · 136

  1. Split by the units digit, since the rule for 0 changes with it.

  2. Units digit 0: hundreds digit has 9 choices, tens 8, so 9×8=729 \times 8 = 72.

  3. Units digit 5: hundreds digit cannot be 0 or 5, so 8 choices; tens then has 8 choices (0 is allowed), so 8×8=648 \times 8 = 64.

  4. Total =72+64=136= 72 + 64 = 136.

Worth noting

128 treats both cases like the units-5 case; 180 counts every three-digit multiple of 5, with repeats allowed. The two cases must be counted separately because 0 is barred from the front only in one of them.

Question 16

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A college issues parking stickers made of 2 letters from A to Z followed by 3 digits. The two letters must be different, the digits may repeat, and the first digit cannot be 0. How many stickers can be issued?

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AnswerA · 5,85,000

  1. Letters: 26×25=65026 \times 25 = 650 ordered pairs of different letters.

  2. Digits: first digit 9 ways, then 10 and 10, so 900900.

  3. Stickers =650×900=5,85,000= 650 \times 900 = 5{,}85{,}000.

Worth noting

6,08,400 is 26×26×90026 \times 26 \times 900, letting the letters repeat. 6,50,000 lets the digits start with 0, and 6,76,000 makes both mistakes.

Arrangements and selections

16 questions

Question 17

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Seven runners reach the final of a school 400 m race. In how many ways can the gold, silver and bronze medals be awarded, if no two runners tie?

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AnswerB · 210

  1. Gold can go to any of 7 runners, silver to any of the other 6, bronze to any of the remaining 5.

  2. Ways =7×6×5=7P3=210= 7 \times 6 \times 5 = {}^{7}P_{3} = 210.

Worth noting

35 is 7C3{}^{7}C_{3}, which ignores which medal each runner gets. 343 is 737^3, which lets one runner win two medals.

Question 18

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Eight volunteers offer to help at a blood-donation camp, and the organiser needs a group of 3 of them. The 3 have identical duties. How many different groups can be chosen?

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AnswerC · 56

  1. Order does not matter, so this is a combination.

  2. 8C3=8×7×63×2×1=3366=56{}^{8}C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56.

Worth noting

336 is 8P3{}^{8}P_{3}, which counts the same three people 6 times over, once for each order. 24 is 8×38 \times 3.

Question 19

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At the first meeting of a 12-member student council, every member shakes hands exactly once with every other member. How many handshakes take place?

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AnswerA · 66

  1. Each handshake is a pair of members, and the pair (A, B) is the same as (B, A).

  2. Handshakes =12C2=12×112=66= {}^{12}C_{2} = \frac{12 \times 11}{2} = 66.

Worth noting

132 is 12×1112 \times 11, counting every handshake twice. 144 lets a member shake hands with himself or herself, and 78 is 13C2{}^{13}C_{2}.

Question 20

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In how many different ways can all the letters of the word PLANET be arranged in a row?

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AnswerD · 720

  1. PLANET has 6 letters, all different.

  2. Arrangements =6!=6×5×4×3×2×1=720= 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.

Worth noting

360 halves the answer as if one letter were repeated, but no letter repeats. 120 is 5!5! and 36 is 6×66 \times 6.

Question 21

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Nine points are marked on a circle. How many triangles can be drawn with all three corners among these points?

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AnswerC · 84

  1. No three points on a circle lie on a straight line, so any 3 points make a triangle.

  2. Triangles =9C3=9×8×76=84= {}^{9}C_{3} = \frac{9 \times 8 \times 7}{6} = 84.

Worth noting

36 is 9C2{}^{9}C_{2}, the number of chords. 504 is 9P3{}^{9}P_{3}, which counts each triangle 6 times.

Question 22

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How many diagonals does a polygon with 10 sides have?

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AnswerA · 35

  1. Any 2 of the 10 corners give a line segment: 10C2=45{}^{10}C_{2} = 45.

  2. 10 of these segments are the sides, not diagonals.

  3. Diagonals =45−10=35= 45 - 10 = 35.

Worth noting

45 forgets to remove the sides. 70 is 10×710 \times 7, counting each diagonal from both ends.

Question 23

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How many different arrangements of all the letters of the word BANANA are possible?

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AnswerA · 60

  1. BANANA has 6 letters: A three times, N twice, B once.

  2. Swapping identical letters gives nothing new, so divide by the repeats.

  3. Arrangements =6!3!×2!=72012=60= \frac{6!}{3! \times 2!} = \frac{720}{12} = 60.

Worth noting

720 treats the letters as all different. 120 divides only by 3!3! for the A's, and 360 only by 2!2! for the N's.

Question 24

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How many distinct arrangements can be made using all the letters of the word COLLEGE?

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AnswerB · 1,260

  1. COLLEGE has 7 letters: L twice, E twice, and C, O, G once each.

  2. Arrangements =7!2!×2!=50404=1260= \frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260.

Worth noting

5040 ignores the repeats, and 2520 corrects for only one of the two repeated letters. 630 divides by an extra 2.

Question 25

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A residents' welfare committee of 3 men and 2 women is to be formed from 6 men and 5 women who have volunteered. In how many ways can the committee be formed?

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AnswerB · 200

  1. Men: 6C3=20{}^{6}C_{3} = 20 ways. Women: 5C2=10{}^{5}C_{2} = 10 ways.

  2. Both choices are needed together, so multiply: 20×10=20020 \times 10 = 200.

Worth noting

30 adds 20+1020 + 10; the committee needs the men AND the women. 462 is 11C5{}^{11}C_{5}, which ignores the 3-and-2 split. 2400 uses permutations, but committee members have no order.

Question 26

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Five friends, including Anjali and Bhavesh, stand in a row for a photograph. In how many arrangements do Anjali and Bhavesh stand next to each other?

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AnswerB · 48

  1. Tie Anjali and Bhavesh into one block. Now there are 4 units to arrange: 4!=244! = 24 ways.

  2. Inside the block they can stand in 2 orders.

  3. Arrangements =24×2=48= 24 \times 2 = 48.

Worth noting

24 forgets that the pair can swap places. 72 is the number of arrangements where they are NOT together.

Question 27

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Six speakers, including Ritu and Sandeep, will speak one after another at a seminar. The organisers do not want Ritu and Sandeep to speak one immediately after the other (in either order). How many speaking orders are possible?

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AnswerD · 480

  1. All orders: 6!=7206! = 720.

  2. Orders with the two together: treat them as a block, 5!×2=2405! \times 2 = 240.

  3. Orders with them apart =720−240=480= 720 - 240 = 480.

Worth noting

240 is the number of orders where they ARE together. 360 simply halves 720, and 120 is 5!5! without the factor 2.

Question 28

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Four-letter codes are made from the letters of the word PENCIL, with no letter used twice. How many of these codes begin with a vowel?

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AnswerB · 120

  1. The vowels in PENCIL are E and I, so the first letter has 2 choices.

  2. The other three places are filled from the remaining 5 letters: 5×4×3=605 \times 4 \times 3 = 60.

  3. Codes =2×60=120= 2 \times 60 = 120.

Worth noting

360 is 6P4{}^{6}P_{4}, every code with no condition. 60 forgets that there are two vowels to start with.

Question 29

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The letters of the word EQUATION are arranged in a row. In how many arrangements do all five vowels appear together?

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AnswerC · 2,880

  1. EQUATION has 8 different letters: vowels E, U, A, I, O and consonants Q, T, N.

  2. Treat the 5 vowels as one block. With Q, T and N that makes 4 units: 4!=244! = 24 orders.

  3. The vowels can be arranged inside the block in 5!=1205! = 120 ways.

  4. Arrangements =24×120=2880= 24 \times 120 = 2880.

Worth noting

120 counts only the order inside the block, and 720 multiplies 5!5! by 3!3!, forgetting that the block itself moves among the consonants. 40,320 is 8!8!, with no condition.

Question 30

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A quiz team of 4 is to be picked from 5 boys and 4 girls. The team must include at least 2 girls. How many different teams are possible?

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AnswerB · 81

  1. Split into cases by the number of girls.

  2. 2 girls: 4C2×5C2=6×10=60{}^{4}C_{2} \times {}^{5}C_{2} = 6 \times 10 = 60. 3 girls: 4C3×5C1=4×5=20{}^{4}C_{3} \times {}^{5}C_{1} = 4 \times 5 = 20. 4 girls: 4C4=1{}^{4}C_{4} = 1.

  3. Teams =60+20+1=81= 60 + 20 + 1 = 81.

  4. Check: total teams 9C4=126{}^{9}C_{4} = 126, minus 5 with no girl and 40 with one girl, gives 81.

Worth noting

60 is only the 'exactly 2 girls' case. 86 subtracts the one-girl teams but forgets the all-boy teams. 126 ignores the condition.

Question 31

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Seven members of a book club sit around a round table. Two seatings count as the same if one can be turned into the other by everyone moving the same number of places around the table. How many different seatings are possible?

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AnswerC · 720

  1. In a row, 7 people can sit in 7!=50407! = 5040 ways.

  2. Around a table, each seating appears 7 times in that count (once for each rotation).

  3. Seatings =50407=6!=720= \frac{5040}{7} = 6! = 720.

Worth noting

5040 treats the table as a row. 360 also divides by 2 for mirror images, which applies to a necklace, not to people who can tell left from right.

Question 32

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Streets in a town form a grid. To get from home to school, Lalremsiami must walk 4 blocks east and 3 blocks north, in any order, never going west or south. How many different shortest routes can she take?

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AnswerB · 35

  1. Every shortest route is a string of 7 moves: 4 E's and 3 N's.

  2. A route is fixed once you choose which 3 of the 7 moves are N.

  3. Routes =7C3=7×6×56=35= {}^{7}C_{3} = \frac{7 \times 6 \times 5}{6} = 35.

Worth noting

12 multiplies 4×34 \times 3, which is not how routes combine. 128 is 272^7, which allows any number of E's and N's.

Basic probability

17 questions

Question 33

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A bag contains 5 red, 3 blue and 2 green marbles. Kavya takes out one marble without looking. What is the probability that it is blue?

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AnswerD · 310\frac{3}{10}

  1. Total marbles =5+3+2=10= 5 + 3 + 2 = 10, all equally likely to be picked.

  2. Favourable outcomes: 3 blue marbles.

  3. Probability =310= \frac{3}{10}.

Worth noting

37\frac{3}{7} divides by the number of NON-blue marbles instead of the total. 13\frac{1}{3} treats the three colours as equally likely, and 710\frac{7}{10} is the chance of NOT blue.

Question 34

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A raffle has tickets numbered 1 to 30, and one ticket is drawn at random. What is the probability that the number on it is a prime number?

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AnswerD · 13\frac{1}{3}

  1. Primes up to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, which is 10 numbers.

  2. Probability =1030=13= \frac{10}{30} = \frac{1}{3}.

Worth noting

Forgetting that 2 is prime gives 930=310\frac{9}{30} = \frac{3}{10}; counting 1 as prime gives 1130\frac{11}{30}. 23\frac{2}{3} is the chance of NOT a prime.

Question 35

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The probability that a student picked at random from a college wears spectacles is 0.28. What is the probability that a student picked at random does not wear spectacles?

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AnswerC · 0.72

  1. 'Wears spectacles' and 'does not' are complementary: their probabilities add to 1.

  2. Probability =1−0.28=0.72= 1 - 0.28 = 0.72.

Worth noting

0.28 is the chance of wearing spectacles, the event itself. 0.62 and 0.82 are subtraction slips; check by adding back: 0.72+0.28=10.72 + 0.28 = 1.

Question 36

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Each letter of the word MAHARASHTRA is written on its own card, and the 11 cards are shuffled. One card is drawn at random. What is the probability that it shows the letter A?

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AnswerB · 411\frac{4}{11}

  1. There are 11 cards, and A appears 4 times: M-A-H-A-R-A-S-H-T-R-A.

  2. Probability =411= \frac{4}{11}.

Worth noting

16\frac{1}{6} treats the 6 different letters as equally likely, but A is on more cards than any other letter. 211\frac{2}{11} is the chance for H or for R.

Question 37

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A jar holds 24 sweets, each either mango or lemon flavoured. The probability of picking a mango sweet at random is 38\frac{3}{8}. How many lemon sweets are in the jar?

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AnswerD · 15

  1. Mango sweets =38×24=9= \frac{3}{8} \times 24 = 9.

  2. Lemon sweets =24−9=15= 24 - 9 = 15.

Worth noting

9 is the number of MANGO sweets. 8 and 3 come from dividing 24 by 3 or by 8 instead of multiplying by the fraction.

Question 38

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A two-digit number is chosen at random. What is the probability that it is a perfect square?

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AnswerD · 115\frac{1}{15}

  1. There are 90 two-digit numbers (10 to 99).

  2. The perfect squares among them are 16, 25, 36, 49, 64 and 81, which is 6.

  3. Probability =690=115= \frac{6}{90} = \frac{1}{15}.

Worth noting

There are 90 two-digit numbers, not 99 or 100: using 99 gives 233\frac{2}{33} and using 100 gives 350\frac{3}{50}. Wrongly including 100 (or 09) as a seventh square gives 790\frac{7}{90}.

Question 39

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Arjun hits a target with probability 23\frac{2}{3} and Nandini hits it with probability 35\frac{3}{5}. Each fires one shot, independently of the other. What is the probability that both of them hit the target?

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AnswerD · 25\frac{2}{5}

  1. For independent events, P(both) = product of the probabilities.

  2. P(both hit) =23×35=615=25= \frac{2}{3} \times \frac{3}{5} = \frac{6}{15} = \frac{2}{5}.

Worth noting

215\frac{2}{15} is the chance that both MISS (13×25)\left(\frac{1}{3} \times \frac{2}{5}\right), and 1315\frac{13}{15} is the chance that at least one hits. 415\frac{4}{15} is P(Arjun hits and Nandini misses).

Question 40

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Meera can solve a puzzle with probability 34\frac{3}{4} and Joseph with probability 25\frac{2}{5}. They try it independently. What is the probability that the puzzle gets solved, that is, at least one of them solves it?

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AnswerB · 1720\frac{17}{20}

  1. The puzzle stays unsolved only if BOTH fail: 14×35=320\frac{1}{4} \times \frac{3}{5} = \frac{3}{20}.

  2. P(solved) =1−320=1720= 1 - \frac{3}{20} = \frac{17}{20}.

Worth noting

320\frac{3}{20} is the chance it is NOT solved. 310\frac{3}{10} is P(both solve) and 1120\frac{11}{20} is P(exactly one solves); 'at least one' includes both of those cases.

Question 41

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Two machines in a factory work independently. On a given day, machine X breaks down with probability 15\frac{1}{5} and machine Y with probability 14\frac{1}{4}. What is the probability that exactly one of the two machines breaks down that day?

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AnswerC · 720\frac{7}{20}

  1. Exactly one = (X fails and Y works) or (X works and Y fails).

  2. 15×34=320\frac{1}{5} \times \frac{3}{4} = \frac{3}{20} and 45×14=420\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}.

  3. These cases cannot both happen, so add: 720\frac{7}{20}.

Worth noting

25\frac{2}{5} is P(at least one fails), which also includes both failing (120)(\frac{1}{20}). 35\frac{3}{5} is P(neither fails).

Question 42

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A box holds 4 red and 6 white balls. Two balls are drawn one after the other WITHOUT replacement. What is the probability that both are red?

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AnswerD · 215\frac{2}{15}

  1. First ball red: 410\frac{4}{10}.

  2. With one red ball gone, 3 red remain among 9: 39\frac{3}{9}.

  3. P(both red) =410×39=1290=215= \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}.

Worth noting

425=(410)2\frac{4}{25} = \left(\frac{4}{10}\right)^2 is the answer WITH replacement. Without replacement the second draw has one ball fewer of each kind drawn.

Question 43

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A number is picked at random from 1 to 50. What is the probability that it is divisible by 3 or by 5 (or by both)?

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AnswerC · 2350\frac{23}{50}

  1. Multiples of 3 up to 50: 16. Multiples of 5: 10.

  2. Multiples of both, i.e. of 15: 15, 30, 45, which is 3, and they were counted twice.

  3. Favourable =16+10−3=23= 16 + 10 - 3 = 23, so probability =2350= \frac{23}{50}.

Worth noting

825\frac{8}{25} counts only the multiples of 3, and 15\frac{1}{5} only the multiples of 5. 350\frac{3}{50} counts only those divisible by both.

Question 44

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March 2027 has 31 days, and 1 March 2027 is a Monday. A date in March 2027 is chosen at random. What is the probability that it is a Sunday?

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AnswerA · 431\frac{4}{31}

  1. The first Sunday is 7 March; the Sundays are 7, 14, 21 and 28 March.

  2. 29, 30 and 31 March are Monday, Tuesday and Wednesday, so there is no fifth Sunday.

  3. Probability =431= \frac{4}{31}.

Worth noting

17\frac{1}{7} assumes each weekday is equally common, but a 31-day month has 5 of three weekdays and 4 of the rest. 531\frac{5}{31} is right for a Monday, Tuesday or Wednesday.

Question 45

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Five cards numbered 1, 2, 3, 4 and 5 are in a hat. Two cards are drawn together (without replacement). What is the probability that the sum of the two numbers is even?

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AnswerD · 25\frac{2}{5}

  1. Pairs possible: 5C2=10{}^{5}C_{2} = 10.

  2. An even sum needs two odd cards (from 1, 3, 5: 3 pairs) or two even cards (2 and 4: 1 pair).

  3. Probability =3+110=25= \frac{3 + 1}{10} = \frac{2}{5}.

Worth noting

1325\frac{13}{25} is the answer if the first card is put back, since then a card can pair with itself. 12\frac{1}{2} assumes odd and even sums are equally likely; with three odd cards and two even, they are not.

Question 46

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A carton of 12 bulbs contains 3 defective ones. An electrician picks 2 bulbs at random, without replacement. What is the probability that at least one of the two is defective?

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AnswerB · 511\frac{5}{11}

  1. Ways to pick 2 of 12: 12C2=66{}^{12}C_{2} = 66.

  2. Ways to pick 2 good bulbs (none defective): 9C2=36{}^{9}C_{2} = 36.

  3. P(at least one defective) =1−3666=3066=511= 1 - \frac{36}{66} = \frac{30}{66} = \frac{5}{11}.

Worth noting

716=1−(34)2\frac{7}{16} = 1 - \left(\frac{3}{4}\right)^2 treats the draws as WITH replacement. 14\frac{1}{4} is the chance for a single bulb, and 122\frac{1}{22} is the chance that both are defective.

Question 47

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Three archers aim at the same target, each shooting once and independently. Their chances of hitting are 12\frac{1}{2}, 13\frac{1}{3} and 14\frac{1}{4}. What is the probability that the target is hit at least once?

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AnswerB · 34\frac{3}{4}

  1. The target escapes only if all three miss: 12×23×34=624=14\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}.

  2. P(hit at least once) =1−14=34= 1 - \frac{1}{4} = \frac{3}{4}.

Worth noting

Adding the three chances gives 1312\frac{13}{12}, more than 1, which is impossible. 124\frac{1}{24} is P(all hit), 14\frac{1}{4} is P(all miss) and 1124\frac{11}{24} is P(exactly one hits).

Question 48

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Five friends, including Nikhil and Farah, sit in a row of 5 seats in random order. What is the probability that Nikhil and Farah get the two end seats?

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AnswerD · 110\frac{1}{10}

  1. All seatings: 5!=1205! = 120.

  2. Nikhil and Farah at the ends: 2 ways (either can take the left end); the other three fill the middle in 3!=63! = 6 ways, giving 12.

  3. Probability =12120=110= \frac{12}{120} = \frac{1}{10}.

Worth noting

120\frac{1}{20} forgets that Nikhil and Farah can swap ends. 25\frac{2}{5} is the chance that they sit NEXT to each other, a different event.

Question 49

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A class committee of 3 is chosen at random from 5 boys and 4 girls. What is the probability that the committee has exactly 2 girls?

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AnswerA · 514\frac{5}{14}

  1. All committees: 9C3=84{}^{9}C_{3} = 84.

  2. Exactly 2 girls: 4C2×5C1=6×5=30{}^{4}C_{2} \times {}^{5}C_{1} = 6 \times 5 = 30.

  3. Probability =3084=514= \frac{30}{84} = \frac{5}{14}.

Worth noting

1742\frac{17}{42} adds the all-girl committees (4 of them), which answers 'at least 2 girls'. 121\frac{1}{21} is the chance of 3 girls, and 49\frac{4}{9} is only the chance that one chosen person is a girl.

Coins, dice and cards

16 questions

Question 50

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Two fair coins are tossed together. What is the probability of getting at least one head?

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AnswerC · 34\frac{3}{4}

  1. The equally likely outcomes are HH, HT, TH and TT.

  2. Only TT has no head, so 3 of the 4 outcomes have at least one head.

  3. Probability =34= \frac{3}{4}.

Worth noting

23\frac{2}{3} lists only HH, 'one head' and TT, but 'one head' happens in two ways (HT and TH), so those three are not equally likely. 14\frac{1}{4} is P(no head).

Question 51

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A fair six-sided die is rolled once. What is the probability that the number shown is greater than 4?

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AnswerA · 13\frac{1}{3}

  1. Numbers greater than 4: 5 and 6, which is 2 outcomes out of 6.

  2. Probability =26=13= \frac{2}{6} = \frac{1}{3}.

Worth noting

'Greater than 4' does not include 4; counting it gives 12\frac{1}{2}. 23\frac{2}{3} is P(4 or less), and 16\frac{1}{6} counts only one of the two numbers.

Question 52

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Two fair dice are rolled together. What is the probability that the numbers shown add up to 9?

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AnswerD · 19\frac{1}{9}

  1. There are 6×6=366 \times 6 = 36 equally likely ordered outcomes.

  2. Sum 9: (3, 6), (6, 3), (4, 5), (5, 4), which is 4 outcomes.

  3. Probability =436=19= \frac{4}{36} = \frac{1}{9}.

Worth noting

111\frac{1}{11} treats the 11 possible sums (2 to 12) as equally likely. 221\frac{2}{21} counts unordered pairs, but (3, 6) and (6, 3) are different outcomes. 112\frac{1}{12} misses one of the four.

Question 53

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Three fair coins are tossed at the same time. What is the probability of getting exactly two heads?

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AnswerA · 38\frac{3}{8}

  1. Three coins give 23=82^3 = 8 equally likely outcomes.

  2. Exactly two heads: HHT, HTH, THH, which is 3 outcomes.

  3. Probability =38= \frac{3}{8}.

Worth noting

12\frac{1}{2} is P(at least two heads), which also counts HHH. 78\frac{7}{8} is P(at least one head), and 23\frac{2}{3} wrongly divides 2 heads by 3 coins.

Question 54

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A standard deck has 52 cards in 4 suits of 13 (ace, 2 to 10, jack, queen, king); hearts and diamonds are red, clubs and spades are black, and the jack, queen and king are the face cards. One card is drawn at random. What is the probability that it is a face card or an ace?

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AnswerA · 413\frac{4}{13}

  1. Face cards: 3 in each suit, so 12. Aces: 4. No card is both.

  2. Favourable =12+4=16= 12 + 4 = 16, so probability =1652=413= \frac{16}{52} = \frac{4}{13}.

Worth noting

313\frac{3}{13} counts only the face cards and 113\frac{1}{13} only the aces. 14\frac{1}{4} is the chance of any one suit.

Question 55

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Rukhsana tosses a fair coin and rolls a fair die at the same time. What is the probability that she gets a head on the coin and an even number on the die?

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AnswerA · 14\frac{1}{4}

  1. The coin and the die are independent: P(head) =12= \frac{1}{2}, P(even) =36=12= \frac{3}{6} = \frac{1}{2}.

  2. P(head and even) =12×12=14= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Worth noting

34\frac{3}{4} is P(head OR even). 112\frac{1}{12} is the chance of one particular pair, such as a head with a 4.

Question 56

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A standard deck has 52 cards in 4 suits of 13 (ace, 2 to 10, jack, queen, king); hearts and diamonds are red, clubs and spades are black, and the jack, queen and king are the face cards. A card is drawn at random. What is the probability that it is red or a king (or both)?

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AnswerB · 713\frac{7}{13}

  1. Red cards: 26. Kings: 4, of which 2 are red and already counted.

  2. Favourable =26+4−2=28= 26 + 4 - 2 = 28.

  3. Probability =2852=713= \frac{28}{52} = \frac{7}{13}.

Worth noting

1526=3052\frac{15}{26} = \frac{30}{52} counts the king of hearts and king of diamonds twice. 12\frac{1}{2} counts only the red cards.

Question 57

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Two fair dice are rolled. What is the probability that the product of the two numbers is even?

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AnswerA · 34\frac{3}{4}

  1. A product is odd only when BOTH numbers are odd: 3×3=93 \times 3 = 9 of the 36 outcomes.

  2. P(product even) =1−936=2736=34= 1 - \frac{9}{36} = \frac{27}{36} = \frac{3}{4}.

Worth noting

14\frac{1}{4} is P(product odd). 12\frac{1}{2} assumes even and odd products are equally likely, but one even number is enough to make the product even.

Question 58

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A standard deck has 52 cards in 4 suits of 13 (ace, 2 to 10, jack, queen, king); hearts and diamonds are red, clubs and spades are black, and the jack, queen and king are the face cards. Two cards are drawn one after the other without replacement. What is the probability that both are aces?

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AnswerB · 1221\frac{1}{221}

  1. First card an ace: 452\frac{4}{52}. Second card an ace, with 3 aces left in 51 cards: 351\frac{3}{51}.

  2. P(both aces) =452×351=122652=1221= \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}.

Worth noting

1169=(452)2\frac{1}{169} = \left(\frac{4}{52}\right)^2 is WITH replacement. 1442\frac{1}{442} divides 4C2=6{}^{4}C_{2} = 6 by the ordered count 52×5152 \times 51, mixing combinations with ordered draws.

Question 59

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Mohit rolls a fair die twice. What is the probability that the first number is greater than the second?

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AnswerD · 512\frac{5}{12}

  1. Of the 36 outcomes, 6 are doubles, where the numbers are equal.

  2. The other 30 split evenly: in 15 the first is bigger, in 15 the second is bigger.

  3. Probability =1536=512= \frac{15}{36} = \frac{5}{12}.

Worth noting

12\frac{1}{2} forgets the 6 ties. 712\frac{7}{12} is P(first is greater than or equal to the second), and 56\frac{5}{6} is P(the numbers differ).

Question 60

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At a board-game club, two fair dice are rolled together. What is the probability that the total is a prime number?

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AnswerC · 512\frac{5}{12}

  1. Possible prime totals: 2, 3, 5, 7, 11.

  2. Ways: 2 in 1 way, 3 in 2, 5 in 4, 7 in 6, 11 in 2, so 1+2+4+6+2=151 + 2 + 4 + 6 + 2 = 15.

  3. Probability =1536=512= \frac{15}{36} = \frac{5}{12}.

Worth noting

511\frac{5}{11} counts 5 prime totals out of 11 possible totals, but the totals are not equally likely. 12\frac{1}{2} is P(odd total), which includes 9 and leaves out 2. 712\frac{7}{12} is P(not prime).

Question 61

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A fair coin is tossed four times. What is the probability of getting exactly two heads?

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AnswerA · 38\frac{3}{8}

  1. Four tosses give 24=162^4 = 16 equally likely sequences.

  2. Sequences with exactly two heads: 4C2=6{}^{4}C_{2} = 6.

  3. Probability =616=38= \frac{6}{16} = \frac{3}{8}.

Worth noting

12\frac{1}{2} assumes 'half heads' is the typical result, so its chance must be a half. 1116\frac{11}{16} is P(at least two heads) and 58\frac{5}{8} is P(not exactly two).

Question 62

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Two fair dice, one red and one blue, are rolled. What is the probability that the two numbers differ by exactly 2?

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AnswerA · 29\frac{2}{9}

  1. Unordered pairs differing by 2: (1, 3), (2, 4), (3, 5), (4, 6).

  2. Each can appear with either die higher, so there are 8 outcomes out of 36.

  3. Probability =836=29= \frac{8}{36} = \frac{2}{9}.

Worth noting

19\frac{1}{9} counts only 4 outcomes, forgetting that red 3, blue 1 differs from red 1, blue 3. 518\frac{5}{18} is P(differ by exactly 1).

Question 63

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In a board game you may start only after rolling at least one six with two fair dice. What is the probability of getting at least one six on a single roll of the two dice?

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AnswerD · 1136\frac{11}{36}

  1. No six on either die: 5×5=255 \times 5 = 25 of the 36 outcomes.

  2. P(at least one six) =1−2536=1136= 1 - \frac{25}{36} = \frac{11}{36}.

Worth noting

518=1036\frac{5}{18} = \frac{10}{36} is P(exactly one six), which leaves out the double six. 16\frac{1}{6} is for one die only, and 136\frac{1}{36} is P(two sixes).

Question 64

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A standard deck has 52 cards in 4 suits of 13 (ace, 2 to 10, jack, queen, king); hearts and diamonds are red, clubs and spades are black, and the jack, queen and king are the face cards. One card is drawn at random. What is the probability that it is neither a heart nor a face card?

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AnswerB · 1526\frac{15}{26}

  1. Cards that are a heart or a face card: 13 hearts + 12 face cards − 3 face cards that are hearts = 22.

  2. Neither: 52−22=3052 - 22 = 30.

  3. Probability =3052=1526= \frac{30}{52} = \frac{15}{26}.

Worth noting

2752\frac{27}{52} removes all 12 face cards from the 39 non-hearts, but only 9 face cards are not hearts. 34\frac{3}{4} ignores the face-card condition, and 1013\frac{10}{13} ignores the heart condition.

Question 65

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A standard deck has 52 cards in 4 suits of 13 (ace, 2 to 10, jack, queen, king); hearts and diamonds are red, clubs and spades are black, and the jack, queen and king are the face cards. Two cards are drawn together at random (without replacement). What is the probability that one is red and the other is black?

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AnswerA · 2651\frac{26}{51}

  1. Ways to draw 2 cards: 52C2=1326{}^{52}C_{2} = 1326.

  2. One red and one black: 26×26=67626 \times 26 = 676.

  3. Probability =6761326=2651= \frac{676}{1326} = \frac{26}{51}.

Worth noting

12\frac{1}{2} is the answer WITH replacement (2×12×12)\left(2 \times \frac{1}{2} \times \frac{1}{2}\right). 2551\frac{25}{51} is P(same colour), and 1351\frac{13}{51} counts only the order red-then-black.

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