Sixteen players  $P_1,P_2,\mathrm{.........},P_{16}$ 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  

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,  $P_1$ 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 one of  $P_1$ and $P_2$   is among the winners is
$P\{(P_1\cap \overline{P_2})\cup (\overline{P_1}\cap P_2)\}=P(P_1\cap \overline{P_2})+P(\overline{P_1}\cap P_2)$

$=\left(\frac{1}{2}\right)(1-\frac{1}{2})+(1-\frac{1}{2})\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

i.e,  $P(A/E_1)=1$ and  $P(A/E_2)=\frac{1}{2}$
By the total probability Rule,  $P\left(A\right)=P\left(E_1\right).P\left(\frac{A}{E_1}\right)+P\left(E_2\right).P\left(\frac{A}{E_2}\right)$
$=\frac{1}{15}\left(1\right)+\frac{14}{15}\left(\frac{1}{2}\right)=\frac{8}{15}$