Last Updated: May 2026
IPMAT Probability and Permutations 2027 contributes 3–5 questions to the Quantitative Aptitude (MCQ) and Quantitative Aptitude (Short Answer) sections combined — both IIM Indore IPMAT and IIM Rohtak IPMAT 2027 papers test classical probability, conditional probability and counting problems with permutations and combinations. This 1,500-word IPMAT Probability 2027 guide covers core concepts, formulas, IPMAT-specific problem patterns, and 40 practice questions with worked solutions for the harder ones.
1. Counting Principles
Multiplication rule: If event A happens in m ways and B in n ways, both happen in m·n ways.
Addition rule: Either A or B (mutually exclusive) — m + n ways.
2. Permutations
Number of arrangements of n distinct objects = n!
Number of arrangements of n objects taken r at a time = nPr = n!/(n-r)!
With repetitions = n^r
Circular permutations of n objects = (n-1)!
Permutations with k identical objects: n!/(p₁!·p₂!·…·pₖ!)
3. Combinations
Number of selections of r from n = nCr = n!/[r!(n-r)!]
nCr = nC(n-r); nC0 = nCn = 1
nCr + nC(r-1) = (n+1)Cr (Pascal’s rule)
4. Probability — Definitions
P(A) = (favourable outcomes)/(total outcomes), 0 ≤ P(A) ≤ 1.
P(A or B) = P(A) + P(B) – P(A∩B). For mutually exclusive: P(A∩B) = 0.
P(A and B) = P(A)·P(B|A). For independent: P(A∩B) = P(A)·P(B).
Conditional: P(A|B) = P(A∩B)/P(B), P(B) > 0.
5. Bayes’ Theorem (IPMAT-Level)
P(A|B) = [P(B|A)·P(A)] / [P(B|A)·P(A) + P(B|A’)·P(A’)]
Used for revising probabilities given new evidence. IPMAT typically gives a 2-condition Bayes problem.
6. IPMAT-Specific Problem Patterns
| Pattern | Frequency in IPMAT 2018–25 | Typical Difficulty |
|---|---|---|
| Card / dice / coin | 1 per paper | Easy |
| Word arrangement (e.g., MISSISSIPPI) | 1 per paper | Medium |
| People in a row/circle | 1 per paper | Medium |
| Bayes / conditional | 1 per 2 papers | Hard |
| Distribution into groups | 1 per paper | Medium |
| At least / at most | 1 per paper | Medium |
7. Worked Examples
Example 1 — Word Arrangement
Number of arrangements of letters in ‘MISSISSIPPI’: 11 letters with M=1, I=4, S=4, P=2. Answer: 11!/(1!·4!·4!·2!) = 34,650.
Example 2 — Probability with Constraints
Two dice are thrown. P(sum is 7 OR sum is 11): Sum 7 — (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 ways. Sum 11 — (5,6),(6,5) = 2 ways. P = 8/36 = 2/9.
Example 3 — Conditional
A bag has 5 red, 3 blue. Two drawn without replacement. P(both red) = (5/8)·(4/7) = 20/56 = 5/14.
Example 4 — Bayes
10% population has disease. Test sensitivity 90%, specificity 80%. If person tests positive, P(actually has disease)? P(D|+) = (0.9·0.1)/(0.9·0.1 + 0.2·0.9) = 0.09/0.27 = 1/3.
8. Common Mistakes
- Confusing P(A or B) with P(A) + P(B) — forgetting subtraction of intersection.
- Treating dependent events as independent.
- Computing nCr formula with wrong factorial.
- In circular permutations, forgetting (n-1)! vs n!.
- In conditional probability, dividing by wrong denominator.
9. 40 Practice Problems (Sample 12)
- Number of arrangements of ‘STATISTICS’: (a) 50,400 (b) 1,00,800 (c) 25,200 (d) 75,600 — 50400; (a)
- 5 boys, 4 girls in a row alternating starting with boy: ways = (a) 5!·4! (b) 5!+4! (c) 9! (d) 9!/(5!4!) — (a)
- P(at least one head in 3 coin tosses): (a) 7/8 (b) 1/2 (c) 3/4 (d) 1/8 — (a)
- P(exactly 2 heads in 4 tosses): (a) 1/4 (b) 3/8 (c) 6/16 (d) Both (b) and (c) — (d)
- From 5 men, 4 women, committee of 3 with at least 1 woman: (a) 25 (b) 75 (c) 60 (d) 70 — (d)
- Letters PROBABILITY rearranged: (a) 11!/2! (b) 11!/(2!·2!) (c) 11! (d) 11!/4! — (b)
- Probability of king from a deck: (a) 1/13 (b) 1/52 (c) 4/13 (d) 1/4 — (a)
- 5 distinct people around a round table — arrangements: (a) 4! (b) 5! (c) 5 (d) 24 — (a) (which equals 24)
- P(face card from deck): (a) 12/52 (b) 3/13 (c) Both (a) and (b) (d) 4/13 — (c)
- nC2 = 45 → n = (a) 9 (b) 10 (c) 11 (d) 12 — (b)
- If P(A)=0.4, P(B)=0.3, P(A∩B)=0.1, P(A∪B): (a) 0.7 (b) 0.6 (c) 0.5 (d) 0.4 — (b)
- 2 dice — P(product > 20): outcomes (5,5)=25, (5,6)=30, (6,5)=30, (4,6)=24, (6,4)=24, (6,6)=36 = 6/36 → 1/6 — 1/6
Frequently Asked Questions
Q1. Probability vs Permutation — which to focus more?
50:50 in IPMAT. Both feed into the same set of problems. Master both fundamentals.
Q2. Difficulty of IPMAT probability vs CAT?
IPMAT probability is closer to CBSE Class 11/12 difficulty. CAT goes deeper into combinatorics. Focus on Class 11 NCERT thoroughly for IPMAT.
Internal Resources
[cg_quiz id=”ipmat-probability-2027″]