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Quantitative Aptitude

Probability

Aptitude-level probability is counting with a fraction bar: count the outcomes you want, count all the equally likely outcomes, and divide. Almost every lost mark comes from a miscounted sample space or an overlap counted twice, not from hard theory.

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8practice questions
3worked examples
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Updated 11 October 2026 · Not in every paper; when it appears, expect one or two questions on dice, coins, cards or a bag of balls.

How to think about probability

Favourable over total, if outcomes are equally likely

P(E) = favourable outcomes / total outcomes, but only when every outcome in the total is equally likely. Two dice give 36 equally likely ordered pairs, not 11 equally likely totals: a total of 7 can happen 6 ways, a total of 2 only 1 way. Write the sample space as ordered outcomes — (3, 5) and (5, 3) are different — and the count stays honest.

Use the complement for "at least one"

"At least one" usually splits into several cases. Its opposite, "none", is a single case. So P(at least one) = 1 − P(none). For three coin tosses, P(at least one head) = 1 − P(no head) = 1 − 1/8 = 7/8. Whenever the direct count needs more than two cases, count the opposite instead.

AND multiplies, OR adds

For independent events, P(A and B) = P(A) × P(B). For events that cannot happen together, P(A or B) = P(A) + P(B). If they can overlap, subtract the overlap once: a card that is a queen or a red card is 4 + 26 − 2 = 28 cards, not 30, because the two red queens are in both groups.

With or without replacement

With replacement, the bag is the same for every draw, so the draws are independent. Without replacement, the second draw has one item fewer, and one fewer of the colour just drawn. For two reds from 3 red and 2 blue, multiply step by step (3/5 × 2/4) or count selections (³C₂ / ⁵C₂). Both give 3/10; combinations are quicker once three or more items are drawn.

Formulas and shortcuts

  • Basic probabilityP(E) = favourable outcomes / total outcomes

    Only for equally likely outcomes. The answer always lies between 0 and 1.

  • ComplementP(not E) = 1 − P(E)

    So P(at least one) = 1 − P(none).

  • A or B (addition rule)P(A or B) = P(A) + P(B) − P(A and B)

    If A and B cannot happen together, the last term is 0.

  • A and B, independent eventsP(A and B) = P(A) × P(B)

    Independent means one result does not change the other: separate dice, separate coins, or draws with replacement.

  • Size of the sample spacen coins → 2ⁿ outcomes; n dice → 6ⁿ outcomes

    Count ordered outcomes: HT and TH are two different results.

  • Drawing r items without replacementP = (ways to choose the wanted items) / ⁿCᵣ

    n = items in the bag, r = items drawn. Order does not matter, so count selections on top and bottom.

  • Standard pack of 52 cards4 suits × 13 cards; 26 red, 26 black; 12 face cards; 4 aces

    Hearts and diamonds are red; spades and clubs are black. Face cards are J, Q and K, so aces are not face cards.

Solved examples

Q1. Two fair dice are thrown together. What is the probability that the total is 8?

  1. Total outcomes = 6 × 6 = 36 ordered pairs, all equally likely.
  2. Pairs that add to 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) — 5 pairs.
  3. (3, 5) and (5, 3) are different outcomes; (4, 4) happens only once.
  4. P(total 8) = favourable / total = 5 / 36

Answer: 5/36

Totals 6 and 8 each come up 5 ways and 7 comes up 6 ways, the most of any total. If your count for 8 is not 5, you dropped a reversed pair or counted (4, 4) twice.

Q2. A bag has 5 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?

  1. Order does not matter, so count selections.
  2. All ways to choose 2 balls from 8: ⁸C₂ = (8 × 7) / 2 = 28
  3. Ways to choose 2 red from 5: ⁵C₂ = (5 × 4) / 2 = 10
  4. P(both red) = 10 / 28

Answer: 5/14

Step by step gives the same: first red is 5/8, then 4 reds are left in 7 balls, so 4/7. 5/8 × 4/7 = 20/56 = 5/14.

Q3. A can solve a problem with probability 1/3 and B with probability 1/4. If both try it independently, what is the probability that the problem gets solved?

  1. "Solved" means at least one of them solves it. Use the complement: it stays unsolved only if both fail.
  2. P(A fails) = 1 − 1/3 = 2/3. P(B fails) = 1 − 1/4 = 3/4.
  3. They work independently, so P(both fail) = 2/3 × 3/4 = 1/2
  4. P(solved) = 1 − P(both fail)

Answer: 1/2

The addition rule agrees: 1/3 + 1/4 − (1/3 × 1/4) = 4/12 + 3/12 − 1/12 = 6/12 = 1/2. Adding 1/3 + 1/4 = 7/12 without removing the overlap is the trap.

Practice questions with answers

1. A fair die is thrown once. What is the probability of getting a number greater than 4?

A. 1/6
B. 1/4
C. 1/3
D. 1/2

Answer: C. 1/3 — Greater than 4 means 5 or 6: 2 of the 6 outcomes. 2/6 = 1/3.

2. Three fair coins are tossed. What is the probability of getting at least two heads?

A. 1/2
B. 3/8
C. 7/8
D. 1/4

Answer: A. 1/2 — Three coins give 2³ = 8 outcomes. Exactly two heads: HHT, HTH, THH (3). Three heads: HHH (1). That is 4/8 = 1/2. 3/8 counts only "exactly two".

3. A card is drawn at random from a well-shuffled pack of 52 cards. What is the probability that it is a king or a heart?

A. 1/13
B. 1/4
C. 17/52
D. 4/13

Answer: D. 4/13 — Hearts: 13. Kings: 4. The king of hearts is in both groups, so count it once: 13 + 4 − 1 = 16. 16/52 = 4/13. 17/52 counts the king of hearts twice.

4. Two fair dice are thrown. What is the probability that the total is a prime number?

A. 1/3
B. 5/12
C. 7/18
D. 1/2

Answer: B. 5/12 — The prime totals possible are 2, 3, 5, 7 and 11. Ways: 1 + 2 + 4 + 6 + 2 = 15 out of 36. 15/36 = 5/12.

5. A bag has 4 red and 6 green balls. Two balls are drawn at random without replacement. What is the probability that one is red and the other is green?

A. 2/5
B. 4/15
C. 12/25
D. 8/15

Answer: D. 8/15 — Favourable = ⁴C₁ × ⁶C₁ = 4 × 6 = 24. Total = ¹⁰C₂ = 45. P = 24/45 = 8/15. 4/15 counts only red-then-green, and 12/25 is the answer if the first ball is put back.

6. A fair die is thrown twice. What is the probability of getting at least one six?

A. 11/36
B. 1/3
C. 5/18
D. 25/36

Answer: A. 11/36 — P(no six) = 5/6 × 5/6 = 25/36. So P(at least one six) = 1 − 25/36 = 11/36. Adding 1/6 + 1/6 = 1/3 is wrong because it counts (6, 6) twice.

7. What is the probability that a leap year chosen at random has 53 Sundays?

A. 1/7
B. 2/7
C. 3/7
D. 53/366

Answer: B. 2/7 — A leap year has 366 days: 52 full weeks plus 2 extra days. The 2 extra days are one of 7 equally likely pairs (Sun–Mon, Mon–Tue, …, Sat–Sun), and 2 of those pairs contain a Sunday. So the answer is 2/7. A non-leap year has only 1 extra day, which gives 1/7.

8. A speaks the truth in 3 out of 4 cases and B in 4 out of 5 cases. If they speak independently, what is the probability that they contradict each other on the same statement?

A. 3/5
B. 13/20
C. 7/20
D. 1/5

Answer: C. 7/20 — They contradict when exactly one of them tells the truth. A true, B false: 3/4 × 1/5 = 3/20. A false, B true: 1/4 × 4/5 = 4/20. Total = 3/20 + 4/20 = 7/20, which is 35% of cases.

Where marks get lost

  • Treating the 11 possible totals of two dice as equally likely. Count the 36 ordered pairs instead.
  • Adding overlapping events without subtracting the overlap: king or heart is 16 cards, not 17.
  • Working out "at least one" case by case when 1 − P(none) takes one line.
  • Keeping the bag unchanged on the second draw when the question says without replacement.

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