Question

Twelve players $S_1,S_2,...,S_{12}$ play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that both $S_1$ and $S_2$ are among the six winners is :

• A. $\frac{12}{33}$
• B. $\frac{13}{33}$
• C. $\frac{5}{22}$
• D. None of these

Answer 1

The correct option is C $\frac{5}{22}$

Consider the probability for the given statement where S${}_1$ and S${}_2$ both are winners

Events where both S${}_1$ and S${}_2$ are there in 6 winners = ${}^{10}{C}_4$

Number of ways to select the 6 winners = ${}^{12}{C}_6$

Now, lets evaluate the probability

P(Both are winners)=$\frac{{}^{10}{C}_4}{{}^{12}{C}_6}=\frac{210}{924}=\frac{5}{22}$