JEE Main Mathematics · Chapter 18

Probability

Classical probability, addition and multiplication theorems, conditional probability, Bayes’ theorem and binomial distribution.

NCERT Class 11 & 12JEE Main: 1–2 Qs per shift 6 min read

1. Chapter at a Glance

Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). This chapter combines counting (permutations and combinations) with logical rules for combining events. It is a high-weightage chapter for JEE Main — one or two questions every shift — frequently on conditional probability, Bayes' theorem and the binomial distribution.

What JEE asks most
  • Classical probability using combinations (cards, dice, balls).
  • Conditional probability and independence.
  • Bayes' theorem: finding the probability of a cause after an event has happened.
  • Binomial distribution, mean and variance; "at least one" problems.

2. Basic Concepts

  • A random experiment has outcomes that cannot be predicted with certainty. The set of all outcomes is the sample space S; an event is a subset of S.
  • Classical probability: P(E) = (number of favourable outcomes)/(total number of equally likely outcomes).
  • 0 ≤ P(E) ≤ 1, P(S) = 1, P(∅) = 0, P(E′) = 1 − P(E).
  • Mutually exclusive events cannot occur together: A ∩ B = ∅. Exhaustive events together cover the whole sample space.
  • Odds in favour of E are P(E) : P(E′); odds against are P(E′) : P(E).

3. Addition and Multiplication Theorems

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∪ B ∪ C) = ΣP(A) − ΣP(A ∩ B) + P(A ∩ B ∩ C)
  • P(only A) = P(A) − P(A ∩ B); P(exactly one of A, B) = P(A) + P(B) − 2P(A ∩ B).
  • P(neither A nor B) = 1 − P(A ∪ B) = P(A′ ∩ B′).

Conditional probability

P(A | B) = P(A ∩ B)/P(B), P(B) ≠ 0

So P(A ∩ B) = P(B)·P(A | B) = P(A)·P(B | A) (multiplication theorem).

Independent events

A and B are independent if P(A ∩ B) = P(A)·P(B), i.e. knowing B happened does not change the probability of A. If A and B are independent, so are A′ and B, A and B′, and A′ and B′.

Do not confuse

Mutually exclusive means they cannot happen together (P(A ∩ B) = 0). Independent means one does not affect the other (P(A ∩ B) = P(A)P(B)). Two events with non-zero probabilities cannot be both.

4. Total Probability and Bayes' Theorem

If E1, E2, …, En are mutually exclusive and exhaustive events (the possible "causes"), then for any event A:

P(A) = Σ P(Ei) P(A | Ei)   (total probability)
P(Ei | A) = P(Ei) P(A | Ei) / Σ P(Ej) P(A | Ej)   (Bayes)
½½ Bag I (3R, 2B)Bag II (2R, 3B) Red 3/5Black 2/5 Red 2/5Black 3/5
Tree diagram for Bayes' theorem: multiply along a branch, add across branches.

5. Random Variables and the Binomial Distribution

A random variable X assigns a number to each outcome. Its mean (expectation) is E(X) = Σxipi and its variance is Var(X) = Σxi²pi − [E(X)]².

In n independent Bernoulli trials, each with success probability p (and q = 1 − p), the number of successes X follows the binomial distribution:

P(X = r) = ⁿCr pʳ qⁿ⁻ʳ,   r = 0, 1, …, n
  • Mean = np; variance = npq; so variance < mean.
  • P(at least one success) = 1 − qⁿ.
  • The most likely number of successes (mode) is near (n + 1)p.

6. Formula Master Table

ResultFormula
ComplementP(A′) = 1 − P(A)
Addition theoremP(A∪B) = P(A) + P(B) − P(A∩B)
Conditional probabilityP(A|B) = P(A∩B)/P(B)
IndependenceP(A∩B) = P(A)P(B)
Total probabilityP(A) = ΣP(Ei)P(A|Ei)
Bayes' theoremP(Ei|A) = P(Ei)P(A|Ei)/P(A)
Binomial probabilityⁿCrpʳqⁿ⁻ʳ
Binomial mean, variancenp, npq
At least one success1 − qⁿ
ExpectationE(X) = Σxp

7. Mind Map

Probability
Basics
  • Sample space, events
  • Classical P(E)
  • Complement, odds
Rules
  • Addition theorem
  • Conditional P(A|B)
  • Independence
Bayes
  • Total probability
  • Tree diagrams
  • Reverse probability
Distributions
  • E(X), Var(X)
  • Binomial ⁿCrpʳqⁿ⁻ʳ
  • np, npq

8. Tips and Tricks

Shortcuts
  • "At least one" → 1 − P(none). This is almost always quicker.
  • For Bayes questions, draw a tree: the answer is (the branch you want) ÷ (sum of all branches leading to the observed event).
  • For games where players take turns until someone wins, use an infinite GP: P(first player wins) = p/(1 − q²) when both have the same chance p per turn.
  • Two dice: remember the sums — 7 has 6 ways, 6 and 8 have 5 ways, 5 and 9 have 4 ways, and so on.
Mistakes students make
  • Adding probabilities of events that are not mutually exclusive.
  • Multiplying probabilities of events that are not independent.
  • Using P(A|B) when the question asks for P(B|A).

9. Important One-liners

  • The probability of an impossible event is 0; of a sure event, 1.
  • For two dice, there are 36 equally likely outcomes; for three coins, 8.
  • A leap year has 53 Sundays with probability 2/7; a non-leap year with probability 1/7.
  • In a binomial distribution, mean − variance = np², which is always positive.
  • If A ⊆ B, then P(A) ≤ P(B) and P(B|A) = 1.
  • For a fair die, E(X) = 3.5 and Var(X) = 35/12.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Two fair dice are thrown. The probability that the sum is 8 is:

  1. 1/6
  2. 5/36
  3. 1/9
  4. 7/36
Show solution

Favourable outcomes: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) — 5 out of 36.

Answer: (B)

Q2Easy

If P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, then P(A | B) is:

  1. 0.5
  2. 0.4
  3. 0.2
  4. 0.8
Show solution

P(A | B) = 0.2/0.5 = 0.4. (Since this equals P(A), A and B are independent.)

Answer: (B)

Q3Medium

Bag I has 3 red and 2 black balls; Bag II has 2 red and 3 black balls. A bag is chosen at random and a ball drawn from it is red. The probability that it came from Bag I is:

  1. 1/2
  2. 3/5
  3. 2/5
  4. 3/10
Show solution

P(red and Bag I) = ½ × 3/5 = 3/10; P(red and Bag II) = ½ × 2/5 = 2/10.

By Bayes: P(Bag I | red) = (3/10)/(3/10 + 2/10) = 3/5.

Answer: (B)

Q4Medium

Five fair coins are tossed. The probability of getting exactly 3 heads is:

  1. 5/16
  2. 3/8
  3. 1/4
  4. 10/16
Show solution

P(X = 3) = ⁵C3(½)³(½)² = 10/32 = 5/16.

Answer: (A)

Q5Numerical

For a binomial distribution with n = 10 and p = 1/2, find the variance.

Show solution

Variance = npq = 10 × ½ × ½ = 2.5.

Answer: 2.5

Q6Medium

A and B are independent events with P(A) = 1/2 and P(B) = 1/3. The probability that neither occurs is:

  1. 1/6
  2. 1/3
  3. 2/3
  4. 5/6
Show solution

A′ and B′ are also independent: P(A′ ∩ B′) = (1 − ½)(1 − ⅓) = ½ × ⅔ = 1/3.

Answer: (B)

Q7Hard

A and B throw a fair die alternately, A starting first. Whoever first throws a 6 wins. The probability that A wins is:

  1. 1/2
  2. 6/11
  3. 5/11
  4. 1/6
Show solution

A wins on throw 1, 3, 5, …: P = (1/6) + (5/6)²(1/6) + (5/6)⁴(1/6) + …

This is an infinite GP: (1/6)/(1 − 25/36) = (1/6)/(11/36) = 6/11.

Answer: (B)

Q8Numerical

A fair coin is tossed n times. Find the least n for which the probability of getting at least one head is greater than 0.9.

Show solution

P(at least one head) = 1 − (½)ⁿ > 0.9 ⇒ (½)ⁿ < 0.1 ⇒ 2ⁿ > 10.

2³ = 8 is not enough; 2⁴ = 16 works. So n = 4.

Answer: 4

11. Quick Revision Summary

Remember before the exam
  • P(A ∪ B) = P(A) + P(B) − P(A ∩ B); P(A|B) = P(A ∩ B)/P(B).
  • Independent: P(A ∩ B) = P(A)P(B) (different from mutually exclusive).
  • Bayes: required branch ÷ total of branches (use a tree).
  • Binomial: ⁿCrpʳqⁿ⁻ʳ, mean np, variance npq.
  • At least one = 1 − P(none); turn-taking games → infinite GP.
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