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 $\frac{8}{15}$

Case(i): If $S_1,S_2$ are in the same pair, then exactly one of them wins.

Total number of ways to divide 16 into 8 pairs = $\frac{16!}{(2!{)}^{8}8!}$

If $S_1,S_2$ are in one pair, ways to divide 14 into 7 pairs

$=\frac{14!}{(2!{)}^{7}7!}$

So, probability = $\frac{\frac{14!}{(2!{)}^{7}7!}}{\frac{16!}{(2!{)}^{8}8!}}\times 1\to$ either $S_1$ or $S_2$ wining = $\frac{1}{15}$

Case (ii): If in different, one must lose and one must win.

number of ways to divide such that they are different

pair =$\frac{16!}{(2!{)}^{8}8!}-\frac{14!}{(2!{)}^{7}7!}$

$S_1$ lose, $S_2$

So, probability $=(1-\frac{1}{15})\times \frac{1}{2}\times \frac{1}{2}\times 2=\frac{7}{15}$

$\therefore$ Total probability = $\frac{1}{15}+\frac{7}{15}=\frac{8}{15}$