Question

Sixteen players $s_1,s_2,.........,s_{16}$ 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. Assume that all the players are of equal strength. The probability that "exactly one of the two players $s_1$ & $s_2$ is among the eight winners" is

• A. $\frac{4}{15}$
• B. $\frac{7}{15}$
• C. $\frac{8}{15}$
• D. $\frac{9}{15}$

Answer 1

The correct option is C $\frac{8}{15}$

Total number of ways of selecting winners ${=}^{16}{C}_8$
favourable ways ${=}^2{C}_1{.}^{14}{C}_7$ (out of $s_1$ & $s_2$ & rest)
so probability $=\frac{{}^2{C}_1{\cdot }^{14}{C}_7}{{}^{16}{C}_8}=\frac{8}{15}$