QJEEM24Mac4da: JEE Main 2024 Mathematics - Class 12 JEEM24Mac4da Solution: Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black…

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 :

$\frac{5}{18}$