JEE Main 2025 Jan 22 Shift 1, Mathematics Q19: Conditional Probability

JEE Main2025Jan 22, Shift 1Mathematics

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$ , where $g cd (m, n) = 1$ , then $m + n$ is equal to :

A
14
B
4
C
11
D
13

Solution

Let $B_{i}$ denote the event that the $i$ -th ball drawn is black. We want $P (B_{1} ∣ B_{2}) = \frac{P ( B _{1} \cap B _{2} )}{P ( B _{2} )}$ .

$P (B_{1} \cap B_{2}) = \frac{6}{10} \cdot \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ .

$P (B_{2}) = P (B_{1}) P (B_{2} ∣ B_{1}) + P (W_{1}) P (B_{2} ∣ W_{1}) = \frac{6}{10} \cdot \frac{5}{9} + \frac{4}{10} \cdot \frac{6}{9} = \frac{30 + 24}{90} = \frac{6}{10} = \frac{3}{5}$ .

$P (B_{1} ∣ B_{2}) = \frac{1/3}{3/5} = \frac{5}{9}$ .

So $m = 5$ , $n = 9$ , and $m + n = 14$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q19: Conditional Probability

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