JEE Main 2025 Jan 29 Shift 2, Mathematics Q16: Total Probability Theorem And Bayes' Theorem

JEE Main2025Jan 29, Shift 2Mathematics

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains $n$ white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is $29/45$ , then $n$ is equal to:

A
$3$
B
$4$
C
$5$
D
$6$

Solution

Bag 2 ends up with $n + 4$ balls. Condition on the colour transferred:

$P (white from Bag 2) = \frac{4}{9} \cdot \frac{n + 1}{n + 4} + \frac{5}{9} \cdot \frac{n}{n + 4}$

$= \frac{4 ( n + 1 ) + 5 n}{9 ( n + 4 )} = \frac{9 n + 4}{9 ( n + 4 )} = \frac{29}{45}$ .

Cross-multiplying, $5 (9 n + 4) = 29 (n + 4) \Rightarrow 45 n + 20 = 29 n + 116 \Rightarrow 16 n = 96 \Rightarrow n = 6$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q16: Total Probability Theorem And Bayes' Theorem

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