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profileBrahma
asked 2017-03-05 15:50:52 
A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a queen. If A begins the game, then the probability that A wins the game is
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profileJay Patel
answered 2017-03-05 19:49:03 
P(A wins in his first draw)
$=\dfrac{1}{13}$

P(A does not win in his first draw
AND B does not win in his first draw
AND A wins in his second draw)
$=\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{1}{13}$


P(A does not win in his first draw
AND B does not win in his first draw
AND A does not win in his second draw
AND B does not win in his second draw
AND A wins in his third draw
$=\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{1}{13}$

so on

Required probability
$=\dfrac{1}{13}+\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{1}{13}+\dfrac{12}{13}.\dfrac{12}{13}.\dfrac{12}
{13}.\dfrac{12}{13}.\dfrac{1}{13}+\cdots\\
=\dfrac{\dfrac{1}{13}}{1-\dfrac{12}{13}×\dfrac{12}{13}}=\dfrac{13}{169-144}=\dfrac{13}{25}$
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