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Question

Three players A,B,C in this order, cut a pack of cards, and the whole pack is reshuffled after each cut. If the winner is one who first draws a diamond then C′s chance of winning is

A
928
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B
937
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C
964
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D
2764
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Solution

The correct option is B 937
Let E= event of drawing a diamond from a well shuffled deck of 52 cards
Number of diamonds in a deck = 13
Thus, P(E)=1352=14 and ¯¯¯¯¯¯¯¯¯¯¯¯¯P(E)=1P(E)=114=34
Let P(A)= probability that A wins = A draws a diamond.
Then, ¯¯¯¯¯¯¯¯¯¯¯¯P(A)= probability that A losses.
Similarly, P(B)= probability that B wins = B draws a diamond.
Then, ¯¯¯¯¯¯¯¯¯¯¯¯¯P(B)= probability that B losses
and, P(C)= probability that C wins = C draws a diamond.
Then, ¯¯¯¯¯¯¯¯¯¯¯¯¯P(C)= probability that C losses.
Since, A,B,C wins if they draw a diamond hence,
P(A)=P(B)=P(C)=P(E)=14
and A,B,C losses if they do not draw a diamond hence,
¯¯¯¯¯¯¯¯¯¯¯¯P(A)=¯¯¯¯¯¯¯¯¯¯¯¯¯P(B)=¯¯¯¯¯¯¯¯¯¯¯¯¯P(C)=¯¯¯¯¯¯¯¯¯¯¯¯¯P(E)=34
Hence, probability of C winning the game
=¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).P(C)+¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).¯¯¯¯¯¯¯¯¯¯¯¯¯P(C).¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).P(C)+¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).¯¯¯¯¯¯¯¯¯¯¯¯¯P(C).¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).¯¯¯¯¯¯¯¯¯¯¯¯¯P(C).¯¯¯¯¯¯¯¯¯¯¯¯P(A).¯¯¯¯¯¯¯¯¯¯¯¯¯P(B).P(C)+......
=(34×34×14)+(34×34×34×34×34×14)+(34×34×34×34×34×34×34×34×14)+........
=14{(34)2+(34)5+(34)8+.....}
=14×(34)21(34)3 (Using the formula for sum of infinite GP series)
=14×(34)2×112764
=964×6437=937

Hence, the correct answer is option B.


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