Question

Sixteen players $P_1.P_2.\dots \dots \dots P_{16}$ 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.Assuming that all the players are of equal strength, the probability that exactly one of the two players $P_1$ and $P_2$ is among the eight winners is

• A. $\frac{4}{15}$
• B. $\frac{7}{15}$
• C. $\frac{8}{15}$
• D. $\frac{17}{30}$

Answer 1

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

Let $E_1\left(E_2\right)$ denote the event that $P_1$ and $P_2$ are paired (not paired) together and let A denote the event that one of two players $P_1$ and $P_2$is among the winners.
Since, P1 can be paired with any of the remaining 15 players.
We have, $P\left(E_1\right)=\frac{1}{15}$
and $P\left(E_2\right)=1-P\left(E_1\right)=1-\frac{1}{15}=\frac{14}{15}$
In case $E_1$ occurs, it is certain that one of $P_1$ and $P_2$ will be among the winners. In case $E_2$ occurs, the probability that exactly of $P_1$ and $P_2$ is among the winners is
$P\left\{\right(P_1\cap \overline{P_2})\cup (\overline{P_1}\cap P_2\left)\right\}=P(P_1\cap \overline{P_2})+P(\overline{P_1}\cap P_2)=P\left(P_1\right)P\left(\overline{P_2}\right)+P\left(\overline{P_1}\right)P\left(P_2\right)=\left(\frac{1}{2}\right)(1-\frac{1}{2})+(1-\frac{1}{2})\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
ie, $P\left(\frac{A}{E_1}\right)=1$ and $P\left(\frac{A}{E_2}\right)=\frac{1}{2}$
By the total probability Rule,
$P\left(A\right)=P\left(E_1\right).P\left(\frac{A}{E_1}\right)+P\left(E_1\right).P\left(\frac{A}{E_2}\right)=\frac{1}{15}\left(1\right)+\frac{14}{15}\left(\frac{1}{2}\right)=\frac{8}{15}$