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Q.

Sixteen players S₁, S₂, ... S₁₆ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assuming that all the palyers are to equal strength. The probability that the players S1 is among the eight winners is

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a

1/2

b

1/3

c

2/3

d

none of these

answer is A.

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Detailed Solution

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Since all player are of equal strength, with whom S1
is paired, will have no bearing on the probability of his winning.
The number of ways of choosing 8 winners out of 16 : $^{16} C_{8}$
The number of ways of choosing S₁and 7 other winners out of 15 is $^{15} C_{7}$
∴ Probability that S₁ will win $= \frac{^{15} C_{7}}{^{16} C_{8}} = \frac{15!}{7! 8!} \frac{8! 8!}{16!} = \frac{1}{2}$

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