A card from a pack of 52 cards is lost. From | vedclass

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(B) Let $S$ be the event that the lost card is a spade,and $E$ be the event that $n$ cards drawn from the remaining $51$ cards are all spades.We are g

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$2$

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(B) Let $S$ be the event that the lost card is a spade,and $E$ be the event that $n$ cards drawn from the remaining $51$ cards are all spades.We are given $P(S|E) = \frac{11}{50}$.By Bayes' Theorem,$P(S|E) = \frac{P(E|S)P(S)}{P(E|S)P(S) + P(E|S^c)P(S^c)}$.If the lost card is a spade $(S)$,there are $12$ spades left in $51$ cards. $P(E|S) = \frac{\binom{12}{n}}{\binom{51}{n}}$.If the lost card is not a spade $(S^c)$,there are $13$ spades left in $51$ cards. $P(E|S^c) = \frac{\binom{13}{n}}{\binom{51}{n}}$.$P(S) = \frac{13}{52} = \frac{1}{4}$ and $P(S^c) = \frac{39}{52} = \frac{3}{4}$.Substituting these values:$P(S|E) = \frac{\frac{\binom{12}{n}}{\binom{51}{n}} \cdot \frac{1}{4}}{\frac{\binom{12}{n}}{\binom{51}{n}} \cdot \frac{1}{4} + \frac{\binom{13}{n}}{\binom{51}{n}} \cdot \frac{3}{4}} = \frac{\binom{12}{n}}{\binom{12}{n} + 3 \cdot \binom{13}{n}} = \frac{11}{50}$.Using $\binom{13}{n} = \frac{13}{13-n} \binom{12}{n}$,we get $\frac{1}{1 + 3 \cdot \frac{13}{13-n}} = \frac{11}{50}$.$\frac{13-n}{13-n + 39} = \frac{11}{50} \implies \frac{13-n}{52-n} = \frac{11}{50}$.$50(13-n) = 11(52-n) \implies 650 - 50n = 572 - 11n$.$39n = 78 \implies n = 2$.

$A$ 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 being a spade is $\frac{11}{50}$,then $n$ is equal to

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