A, B, C, D cut a pack of 52 well-shuffled pla | vedclass

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(D) Let $p$ be the probability of drawing a spade, $p = \frac{13}{52} = \frac{1}{4}$.Let $q$ be the probability of not drawing a spade, $q = 1 - p = \

Vedclass · Accepted Answer

$\frac{64}{175}$

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(D) Let $p$ be the probability of drawing a spade, $p = \frac{13}{52} = \frac{1}{4}$.Let $q$ be the probability of not drawing a spade, $q = 1 - p = \frac{3}{4}$.$A$ wins if $A$ draws a spade on the $1^{st}, 5^{th}, 9^{th}, \dots$ turn.The probability that $A$ wins is:$P(A) = p + q^4 p + q^8 p + \dots$This is an infinite geometric series with first term $a = p$ and common ratio $r = q^4$.The sum is $S = \frac{a}{1-r} = \frac{p}{1-q^4}$.Substituting the values:$P(A) = \frac{1/4}{1 - (3/4)^4} = \frac{1/4}{1 - 81/256} = \frac{1/4}{175/256} = \frac{1}{4} 	imes \frac{256}{175} = \frac{64}{175}$.

$A, B, C, D$ cut a pack of $52$ well-shuffled playing cards successively in the same order. If the person who cuts a spade first wins the game and the game continues until this happens, then the probability that $A$ wins the game is

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