JEE Main 2024 — Probability Question with Solution

Let's denote the events as follows :

Let $$A_1,A_2,$$ and $$A_3$$ be the events that urns A, B, and C are chosen, respectively.

Let $$B$$ be the event that a black ball is drawn.

We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $$P(A_1|B)$$.

Using Bayes' theorem, we have :

$$P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B)}$$

First, we calculate each term individually :

1. The probability of choosing any urn, since they are chosen at random :

$$P(A_1)=P(A_2)=P(A_3)=\frac{1}{3}$$

2. The probability of drawing a black ball from each urn :

Urn A: $$P(B|A_1)=\frac{5}{12}$$

Urn B: $$P(B|A_2)=\frac{7}{12}$$

Urn C: $$P(B|A_3)=\frac{6}{12}=\frac{1}{2}$$

3. The total probability of drawing a black ball, $$P(B)$$ :

$$P(B)=P(B|A_1)P(A_1)+P(B|A_2)P(A_2)+P(B|A_3)P(A_3)$$

$$P(B)=\frac{5}{12}\cdot \frac{1}{3}+\frac{7}{12}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}$$

$$P(B)=\frac{5}{36}+\frac{7}{36}+\frac{6}{36}$$

$$P(B)=\frac{18}{36}=\frac{1}{2}$$

Now, substitute these into Bayes' theorem :

$$P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B)}$$

$$P(A_1|B)=\frac{\frac{5}{12}\cdot \frac{1}{3}}{\frac{1}{2}}$$

$$P(A_1|B)=\frac{\frac{5}{36}}{\frac{1}{2}}$$

$$P(A_1|B)=\frac{5}{18}$$

Therefore, the probability that the black ball drawn is from urn A is option C :

$$\frac{5}{18}$$