JEE Mains Maths · Probability

Total Probability and Bayes' Theorem

Splitting an event over the ways it can happen (total probability), and reversing a conditional to find which way it most likely happened (Bayes' theorem), including cases where the contents of a bag are unknown.

All Probability notes JEE Mains Maths strategy

Why this matters

Twenty-six PYQs, all but three of them multiple choice. Five only need the total probability of an outcome; sixteen then reverse it with Bayes' theorem; five infer an unknown bag or a lost card from what was drawn. Three ideas cover the page.

Concept 1 of 3: Total probability

When an outcome can happen through several mutually exclusive routes — which bag, which machine, which ball was transferred — add the route probabilities:

P(E)=\sum P(H_i)\,P(E\mid H_i)

. A transfer between bags is a two-stage experiment: the transferred ball is the route.

Definition

$P(E)=\sum_iP(H_i)\,P(E\mid H_i)$ , with $H_i$ mutually exclusive and covering all cases.
A tree: multiply along a branch, add across branches.
Transfer from bag 1 to bag 2, then draw: the routes are the colours transferred.

Total probability

P(E)=\sum_{i}P(H_i)\,P(E\mid H_i)

Worked example

Bag I has 2 red and 3 blue balls, bag II has 4 red and 1 blue. A bag is chosen at random and a ball drawn. Find

P(\text{red})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q61Moderate

Example 1 · Probability · Total Probability and Bayes' Theorem

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A . Then a ball is randomly drawn from the bag A . If the probability, that the ball drawn is white, is

\frac{p}{q},gcd(p,q) = 1

, then

p + q

is equal to

The receiving bag has one more ball

After a transfer, the second bag holds one extra ball. Using its original total in the denominator is the usual mistake.

Drill 4 more on total probability

Concept 2 of 3: Bayes' theorem

Given that the outcome happened, the chance it came through route

H_k

is that route's share of the total:

P(H_k\mid E)=\frac{P(H_k)P(E\mid H_k)}{\sum_iP(H_i)P(E\mid H_i)}

. Compute every route product once; the answer is one product divided by their sum. With equal priors, the priors cancel.

Definition

$P(H_k\mid E)=\frac{P(H_k)\,P(E\mid H_k)}{\sum_iP(H_i)\,P(E\mid H_i)}$ .
Equal priors: $P(H_k\mid E)=\frac{P(E\mid H_k)}{\sum P(E\mid H_i)}$ .
Percentages: work with the products directly, e.g. $0.5\times0.02$ .

Bayes' theorem

P(H_k\mid E)=\frac{P(H_k)\,P(E\mid H_k)}{\sum_iP(H_i)\,P(E\mid H_i)}

Worked example

Machines A and B make 70% and 30% of the items; 2% of A's and 5% of B's are faulty. A faulty item is found. Find the probability it came from B.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q68Moderate

Example 2 · Probability · Total Probability and Bayes' Theorem

In a bolt factory, machines

A,B

and

C

manufacture respectively

20\%,30\%

and

50\%

of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine

C

Priors are not always equal

When bags or machines are chosen with different chances, keep

P(H_i)

in every product. Dropping them is only allowed when they are equal.

Drill 15 more on bayes' theorem

Concept 3 of 3: Inferring an unknown bag or a lost card

When the number of black balls in a bag is unknown, each possible number is a route, usually taken as equally likely. The draw's probability under each route is a ratio of combinations, and Bayes' theorem weights the routes by those ratios. A lost card works the same way: the routes are 'the lost card was a spade' and 'was not'.

Definition

Routes: every possible composition, equally likely unless stated.
Likelihood of drawing $r$ of a kind from $k$ of that kind among $n$ : $\frac{\binom kr}{\binom nr}$ .
Common denominators cancel, so work with $\binom kr$ as weights.
Lost card: priors $\frac14$ (spade) and $\frac34$ (not).

Weights for an unknown bag

P(k\mid\text{draw})=\frac{\binom kr}{\sum_j\binom jr}\quad(\text{equal priors})

Worked example

A bag has 4 balls, each black or white, all compositions equally likely. One ball is drawn and it is black. Find the probability that all 4 are black.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q67Moderate

Example 3 · Probability · Total Probability and Bayes' Theorem

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Include the impossible compositions

Compositions that could not produce the draw have weight 0 but still count when setting equal priors. The denominator sums over all of them; the zeros simply add nothing.

Drill 4 more on inferring an unknown bag or a lost card

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Total probability
Total probability
$P(E)=\sum_{i}P(H_i)\,P(E\mid H_i)$
Bayes' theorem
Bayes' theorem
$P(H_k\mid E)=\frac{P(H_k)\,P(E\mid H_k)}{\sum_iP(H_i)\,P(E\mid H_i)}$
Inferring an unknown bag or a lost card
Weights for an unknown bag
$P(k\mid\text{draw})=\frac{\binom kr}{\sum_j\binom jr}\quad(\text{equal priors})$

Watch out for (3)

The receiving bag has one more ball
→ Total probability
Priors are not always equal
→ Bayes' theorem
Include the impossible compositions
→ Inferring an unknown bag or a lost card

Test yourself on Probability

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