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Brahma

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

Jay 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}$

$=\dfrac{1}{13}$

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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