Visual SolutionPYQ 2023 · Jan Shift 1Standard

Question

There are three bags. Bag I contains 2 red and 3 black balls, Bag II contains 3 red and 2 black balls, and Bag III contains 4 red and 1 black ball. A bag is chosen at random and a ball drawn from it is found to be red. The probability that the ball came from Bag III is:

(A)

\frac{4}{9}

(B)

\frac{1}{3}

(C)

\frac{2}{9}

(D)

\frac{5}{9}

Solution Path

Bayes' theorem: compute likelihoods

P(R|\text{each bag})

, total probability

P(R) = 3/5

, then

P(\text{III}|R) = \frac{4/15}{3/5} = 4/9

01Question Setup

1/4

Three bags with different compositions: Bag I (2R, 3B), Bag II (3R, 2B), Bag III (4R, 1B). A bag is chosen at random and a red ball is drawn. Find

P(\text{Bag III} | \text{Red})

Prior:

P(\text{Bag I}) = P(\text{Bag II}) = P(\text{Bag III}) = \frac{1}{3}

02Build the Probability TreeKEY INSIGHT

2/4

Compute likelihoods:

P(R|\text{I}) = 2/5

P(R|\text{II}) = 3/5

P(R|\text{III}) = 4/5

. Each branch of the tree multiplies the prior by the likelihood.

P(R|\text{I}) = \frac{2}{5}

P(R|\text{II}) = \frac{3}{5}

P(R|\text{III}) = \frac{4}{5}

03Total Probability of Red

3/4

By the law of total probability:

P(R) = \frac{1}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{5} = \frac{1}{3} \cdot \frac{9}{5} = \frac{3}{5}

P(R) = \frac{3}{5}

04Apply Bayes' Theorem

4/4

Bayes' theorem gives

P(\text{III}|R) = \frac{P(R|\text{III}) \cdot P(\text{III})}{P(R)} = \frac{\frac{4}{5} \cdot \frac{1}{3}}{\frac{3}{5}} = \frac{4}{9}

P(\text{Bag III} | \text{Red}) = \frac{4}{9}

, Answer: (A)

Concepts from this question2 concepts unlocked

★

Bayes' Theorem

STANDARD

Reverse conditional probability: find P(cause|effect) from P(effect|cause) using prior probabilities.

P(A_i \mid B) = \frac{P(A_i) \cdot P(B \mid A_i)}{\sum_j P(A_j) \cdot P(B \mid A_j)}

JEE loves 'defective item from which factory' questions. Bayes flips the conditional -a pattern that feels unintuitive without practice.

Factory defectiveDisease testingMulti-source selection

Practice (12 Qs) →

★

Law of Total Probability

STANDARD

If events A₁, A₂, … partition the sample space, then P(B) = Σ P(Aᵢ) × P(B|Aᵢ).

P(B) = \sum_{i=1}^{n} P(A_i) \cdot P(B \mid A_i)

The denominator of Bayes' theorem. Also essential for any problem that says 'from one of several sources'.

Bayes denominatorMulti-urn problemsMixed populations

Practice (11 Qs) →

Want more practice?

Try more PYQs from this chapter →