JEE Main 2025 Apr 4 Shift 2, Mathematics Q23: Total Probability Theorem And Bayes' Theorem

JEE Main2025Apr 4, Shift 2Mathematics

A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then $n$ is equal to

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Solution

Let $S$ be the event "the lost card is a spade" and $E$ be the event " $n$ spades are drawn from the remaining 51 cards".

$P (S) = \frac{1}{4}, P (S^{c}) = \frac{3}{4}$ .

Given $S$ : 12 spades remain among 51 cards, so $P (E ∣ S) = \frac{( n 12 )}{( n 51 )}$ .

Given $S^{c}$ : 13 spades remain among 51 cards, so $P (E ∣ S^{c}) = \frac{( n 13 )}{( n 51 )}$ .

By Bayes' theorem,

$P (S ∣ E) = \frac{P ( S ) P ( E ∣ S )}{P ( S ) P ( E ∣ S ) + P ( S ^{c} ) P ( E ∣ S ^{c} )} = \frac{( n 12 )}{( n 12 ) + 3 ( n 13 )} = \frac{11}{50}$ .

Using $\frac{( n 12 )}{( n 13 )} = \frac{13 - n}{13}$ , this reduces to

$\frac{13 - n}{13 - n + 39} = \frac{11}{50} \Rightarrow \frac{13 - n}{52 - n} = \frac{11}{50}$ .

Cross-multiplying: $50 (13 - n) = 11 (52 - n) \Rightarrow 650 - 50 n = 572 - 11 n \Rightarrow 39 n = 78 \Rightarrow n = 2$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q23: Total Probability Theorem And Bayes' Theorem

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