IPMAT Probability 2027 — Conditional Probability, Bayes Theorem and 30 Practice MCQs

Last Updated: May 2026

Probability is a high-yield Quantitative Aptitude topic in IPMAT and contributes 4-5% of IPMAT QA section weightage, with 2-3 direct questions in IPMAT Indore 2024 and 2025 across both Multiple-Choice (MCQ) and Short Answer (SA) formats. With IPMAT’s tight 90-minute window and 60 questions, mastering Probability — especially classical probability, conditional probability, and Bayes’ Theorem — is crucial for IIM Indore selection.

This IPM Gurukul guide gives you all formula derivations, IPMAT-specific shortcuts, and 30 IPMAT-pattern problems with detailed solutions. Use this as your single revision document before IPMAT 2027 attempts.

Why Probability Matters for IPMAT 2027

IPMAT Indore tests 60 questions in 90 minutes (40 QA + 20 VA), with sectional cutoffs at IIM Indore typically 30-35% per section for shortlisting. The QA section is heavy on Algebra, Number System, P&C, and Probability — the latter accounts for 2-3 questions consistently. IIM Rohtak IPM has 120 questions in 120 minutes with similar Probability emphasis.

1. Classical Probability Definition

For a random experiment with finite, equally likely outcomes:

P(E) = n(E) / n(S)

where n(E) = number of favorable outcomes, n(S) = total outcomes in sample space.

• 0 ≤ P(E) ≤ 1
• P(impossible event) = 0
• P(certain event) = 1
• P(E) + P(E’) = 1, where E’ is complement

2. Sample Spaces (IPMAT Standard)

• Tossing one coin: S = {H, T}, |S| = 2
• Tossing two coins: S = {HH, HT, TH, TT}, |S| = 4
• Tossing n coins: |S| = 2ⁿ
• Rolling one die: |S| = 6
• Rolling two dice: |S| = 36
• Drawing one card from 52-card deck: |S| = 52
• Standard deck: 4 suits (Hearts, Diamonds, Spades, Clubs) × 13 cards = 52. 26 red + 26 black. 12 face cards (J, Q, K each suit). 4 Aces.

3. Addition Rule

For any two events A and B:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

For mutually exclusive events: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).

4. Conditional Probability

Probability of A given B has occurred:

P(A|B) = P(A ∩ B) / P(B) (provided P(B) > 0)

Equivalent: P(A ∩ B) = P(B) · P(A|B) = P(A) · P(B|A).

5. Independent Events

Events A, B are independent if P(A ∩ B) = P(A)·P(B), equivalently P(A|B) = P(A).

• Independence ≠ Mutual Exclusivity
• For independent A, B: P(A ∪ B) = P(A) + P(B) − P(A)·P(B)
• If A and B are independent, so are A and B’, A’ and B, A’ and B’

6. Bayes’ Theorem

For partition E₁, E₂, …, E_n of sample space, and event A with P(A) > 0:

P(E_i|A) = P(E_i)·P(A|E_i) / Σ P(E_j)·P(A|E_j)

IPMAT Indore loves 2-event Bayes’ problems involving box-ball or disease-test scenarios.

7. Total Probability Theorem

P(A) = Σ P(E_i)·P(A|E_i)

8. Common IPMAT Probability Distributions

Coin Toss (n times)

P(exactly k heads) = ⁿC_k · (½)ⁿ

Example: P(exactly 2 heads in 5 tosses) = ⁵C₂ · (½)⁵ = 10/32 = 5/16.

Dice (two dice sum)

P(sum = 7) = 6/36 = 1/6 (highest probability sum).

P(sum = 2 or 12) = 1/36 each (lowest).

P(sum even) = 18/36 = 1/2.

Cards (one card from 52)

• P(Ace) = 4/52 = 1/13
• P(face card) = 12/52 = 3/13
• P(red) = 26/52 = 1/2
• P(spade) = 13/52 = 1/4

9. IPMAT 2027 Predicted Question Stems

• P(at least one head in n tosses) — use complement 1 − (1/2)ⁿ
• Drawing 2 cards without replacement (use multiplication rule)
• Box A vs Box B Bayes — classic “ball drawn from one of two boxes” setup
• Conditional probability with Venn diagram
• Two dice — sum, product, difference conditions
• Independent vs dependent event distinction

10. 30 IPMAT-Pattern Practice Problems with Solutions

1. Probability of getting a tail in single coin toss: (a) 0 (b) 1/2 (c) 1 (d) 1/4
2. Probability of sum 7 with two dice: (a) 1/12 (b) 1/6 (c) 1/4 (d) 1/3 — 6 ways out of 36.
3. P(at least one head in 3 coin tosses): (a) 1/8 (b) 3/8 (c) 7/8 (d) 1/2 — 1 − (1/2)³.
4. P(drawing a red queen from deck): (a) 1/13 (b) 1/26 (c) 1/52 (d) 2/52 — 2 red queens.
5. P(both heads when 2 coins tossed): (a) 1/2 (b) 1/4 (c) 3/4 (d) 1
6. Two dice rolled, P(sum > 9): (a) 1/4 (b) 1/6 (c) 1/3 (d) 5/36 — sums 10,11,12: 3+2+1 = 6/36.
7. P(face card from deck): (a) 1/13 (b) 3/13 (c) 1/4 (d) 12/13
8. Bag has 5 red, 4 blue balls. P(red ball drawn): (a) 4/9 (b) 5/9 (c) 1/2 (d) 5/4
9. If P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2, then P(A∪B) = (a) 0.5 (b) 0.6 (c) 0.7 (d) 0.9
10. Two events A, B are independent. P(A) = 1/3, P(B) = 1/4. P(A∩B) = (a) 7/12 (b) 1/2 (c) 1/12 (d) 1/7
11. P(A|B) = P(A∩B)/P(B). If P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2, P(A|B) = (a) 0.4 (b) 0.5 (c) 0.8 (d) 0.25
12. 2 cards drawn from deck without replacement. P(both Aces): (a) 1/169 (b) 1/221 (c) 1/52 (d) 4/221 — (4/52)(3/51) = 1/221.
13. 3 coins tossed. P(exactly 2 heads): (a) 1/2 (b) 3/8 (c) 1/4 (d) 1/8 — ³C₂(½)³ = 3/8.
14. A box has 3 white, 5 black balls. 2 drawn at random. P(both white): (a) 3/56 (b) 5/14 (c) 3/28 (d) 1/4 — (3/8)(2/7) = 6/56 = 3/28.
15. P(odd sum with 2 dice): (a) 1/4 (b) 1/3 (c) 1/2 (d) 18/36 — both options c & d valid; choose c.
16. Bayes: 2 boxes — A has 4R, 6B; B has 5R, 5B. Box chosen at random; ball drawn is red. P(box A | red) = (a) 4/9 (b) 4/9 (c) 5/9 (d) 1/2 — P(R|A)=2/5, P(R|B)=1/2; P(A|R) = (1/2·2/5)/[(1/2)(2/5+1/2)] = (1/5)/(9/20) = 4/9.
17. P(king OR heart from deck): (a) 4/13 (b) 4/13 (c) 17/52 (d) 5/13 — 4/52+13/52−1/52 = 16/52 = 4/13.
18. If P(A∪B) = 0.7 and P(A) = 0.4, P(B) = 0.5, P(A∩B) = (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 — addition rule rearranged.
19. 3 dice rolled. P(all show same number): (a) 1/36 (b) 1/36 (c) 1/216 (d) 6/216 — 6 favorable / 216 = 1/36.
20. P(no head in 4 tosses): (a) 1/4 (b) 1/8 (c) 1/16 (d) 1/2 — (1/2)⁴ = 1/16.
21. Card drawn at random from deck. P(card is neither king nor queen): (a) 11/13 (b) 11/13 (c) 8/13 (d) 4/13 — (52−8)/52 = 11/13.
22. 2 cards drawn with replacement. P(both kings): (a) 1/169 (b) 1/169 (c) 1/221 (d) 4/52 — (4/52)² = 1/169.
23. If A and B mutually exclusive, P(A) = 0.3, P(B) = 0.5, P(A∪B) = (a) 0.15 (b) 0.8 (c) 0.65 (d) 1.0
24. P(getting a prime number on a die): (a) 1/6 (b) 1/3 (c) 1/2 (d) 2/3 — primes: 2, 3, 5 → 3/6 = 1/2.
25. P(sum > 8 with 2 dice): (a) 1/4 (b) 1/6 (c) 5/18 (d) 1/3 — sums 9,10,11,12: 4+3+2+1 = 10/36 = 5/18.
26. 3 students solve a problem with prob 1/2, 1/3, 1/4. P(none solves): (a) 1/24 (b) 1/4 (c) 5/24 (d) 1/8 — (1/2)(2/3)(3/4) = 6/24 = 1/4.
27. From above, P(at least one solves): (a) 1/4 (b) 1/2 (c) 3/4 (d) 23/24 — 1 − 1/4 = 3/4.
28. Bag: 2 red, 3 white, 5 black. P(NOT drawing white): (a) 3/10 (b) 7/10 (c) 2/10 (d) 1/2 — 7/10.
29. P(drawing king of hearts in 1 draw): (a) 1/26 (b) 1/13 (c) 4/52 (d) 1/52
30. Two dice rolled. P(both even): (a) 1/4 (b) 1/4 (c) 1/9 (d) 1/2 — (3/6)(3/6) = 1/4.

Frequently Asked Questions

Q1. How many Probability questions appear in IPMAT Indore?

2-3 questions consistently across QA MCQ and SA sections, contributing 8-12 marks. With IPMAT’s tight 90-minute window for 60 questions, accuracy on Probability is critical for the 99+ percentile.

Q2. Is calculator allowed in IPMAT?

No — IPMAT Indore and Rohtak both ban calculators. All Probability arithmetic must be done by hand. Practice fraction simplification and basic combinatorics for speed.

Q3. Best book for IPMAT Probability?

NCERT Class 11 Chapter 16 Probability and Class 12 Chapter 13 Probability cover 90% of the IPMAT syllabus. Supplement with Arihant IPMAT QA and IPM Gurukul daily practice tests.

Q4. Common IPMAT trap?

Confusing “with replacement” vs “without replacement” — always re-read the question. Also: assuming events are independent without verification; never assume — check P(A∩B) = P(A)·P(B).

Q5. How much time per Probability question?

IPM Gurukul recommends 90-120 seconds per Probability MCQ. Bayes’ Theorem problems can take up to 3 minutes — approach last if time-constrained.

Conclusion

Probability is one of the most predictable scoring topics in IPMAT QA. With 2-3 questions consistently per paper and the syllabus capped at NCERT Class 11-12 level, mastery delivers 8-12 marks per attempt. The 30 problems above cover every IPMAT pattern from 2018-2025.

For comprehensive IPMAT 2027 preparation including QA + VA + LR mastery, daily mock tests, video lectures by IIM Indore alumni, and personalized doubt support, explore IPM Gurukul Courses. Take a Free IPMAT Mock Test or visit IPMAT FAQ for exam pattern, syllabus, and selection process.

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