JEE Mathematics 2025 Question with Solution

Given: A bag contains $4$ white balls and $6$ black balls. Two balls are selected one by one without replacement.

Find: $P(\text{first ball is black} \mid \text{second ball is black}) = \frac{m}{n}$ and then find $m+n$.

Let $A$ be the event that the first ball is black and $B$ be the event that the second ball is black. Using conditional probability,

Step 1: Find $P(A \cap B)$.

Step 2: Find $P(B)$. The second ball is black in two cases:

Step 3: Compute the conditional probability.

So, $m=5$ and $n=9$ with $\gcd(5,9)=1$.

Therefore, $m+n=5+9=14$. The correct option is A.

Given: $B_1$ is the event that the first ball is black and $B_2$ is the event that the second ball is black.

Find: $P(B_1\mid B_2)$.

Using the formula,

First, probability that both balls are black:

Now find $P(B_2)$ by splitting into two cases:

1. First black, second black

2. First white, second black

Hence,

Finally,

Thus, $\frac{m}{n}=\frac{5}{9}$ and so $m+n=14$.