Mathematics · JEE Main · JEE Advanced

Probability: concepts, formulas & solved PYQ-style examples

Probability is a scoring JEE chapter once you write the sample space correctly. Bayes’ theorem and binomial distribution questions repeat in a few fixed patterns.

Define events in words first, then translate: “given that” means conditional probability.

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Key concepts

Classical probability

Favourable outcomes over equally likely total outcomes.

Addition and multiplication rules

Union of events and independent events.

Conditional probability

P(A|B) = P(A∩B)/P(B).

Total probability and Bayes’ theorem

Finding which bag, machine or person an outcome came from.

Binomial distribution

Exactly r successes in n independent trials.

Formula sheet

Addition rule

P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B) = P(A) + P(B) - P(A\cap B)

Conditional probability

P(A∣B)=P(A∩B)P(B)P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}

Bayes’ theorem

P(Ei∣A)=P(Ei)P(A∣Ei)∑jP(Ej)P(A∣Ej)P(E_i\mid A) = \dfrac{P(E_i)P(A\mid E_i)}{\sum_j P(E_j)P(A\mid E_j)}

Binomial distribution

P(X=r)=(nr)prqn−rP(X = r) = \binom{n}{r} p^r q^{n-r}

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Bayes’ theorem with two or three bags, urns or machines.
  • Dice and coin problems with conditions on the sum or the outcome.
  • Binomial probability of “at least one” or “exactly r” successes.
  • Mean and variance of a probability distribution.

Solved examples

Example 1

Two fair dice are thrown. Find the probability that the sum is 7.

  1. 1Total outcomes: 6×6=366 \times 6 = 36.
  2. 2Favourable pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes.
  3. 3P=636P = \dfrac{6}{36}.
Answer: 16\dfrac{1}{6}
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Example 2

Bag A has 3 red and 2 black balls; bag B has 2 red and 3 black. A bag is chosen at random and a ball drawn is red. Find the probability it came from bag A.

  1. 1P(A)=P(B)=12P(A) = P(B) = \tfrac12, P(R∣A)=35P(R\mid A) = \tfrac35, P(R∣B)=25P(R\mid B) = \tfrac25.
  2. 2Total probability of red: 12⋅35+12⋅25=12\tfrac12 \cdot \tfrac35 + \tfrac12 \cdot \tfrac25 = \tfrac12.
  3. 3Bayes: P(A∣R)=12⋅3512=35P(A\mid R) = \dfrac{\tfrac12 \cdot \tfrac35}{\tfrac12} = \dfrac{3}{5}.
Answer: 35\dfrac{3}{5}
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Common traps

  • Treating (1,6) and (6,1) as the same outcome when dice are distinguishable.
  • Confusing P(A|B) with P(B|A) in Bayes questions.
  • Using the binomial formula when trials are not independent (drawing without replacement).

Practise with

Definite Integration2 Statistics & Probability practice questionsVideo solutions

FAQ

When do I use Bayes’ theorem?

When you know the result (a red ball, a defective item) and must find the probability that it came from a particular source.

How do I solve “at least one” questions?

Use the complement: P(at least one) = 1 − P(none). It is almost always quicker.

Is probability important for JEE Main?

Yes — it is a regular, high-scoring chapter with predictable Bayes and binomial questions.

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