# Probability , kindly solve this question ?

## Sixteen players S1, S2 . . . . . . . . . . . . . . . . . . . S16 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:- (a) Find the probability that the player S1 is among the eight winners. (b) Find the probability that exactly one of the two players S1 and S2 is among the eight winners.

May 2, 2017

See below:

#### Explanation:

a) The probability of player $S 1$ being a winner is $\frac{1}{2}$. This is because the probability of a player (1 person) winning in a pair (2 people) is 50% since the odds are "50-50." In more detail, the answer is derived from the following:

"P" = ("probability of outcome")/("sample size")

In this case, it's:

(1 " player (S1)")/(2 " players per pair")= 1/2 = 50%

b) This is a slightly more complex probability question. The probability of just one of $S 1$ or $S 2$ winning is $\frac{1}{2}$ each, as can be determined from part "a". The probability both winning is, then, $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

However, also consider the possibility of both S1 and S2 being in the same group. There's only one combination of pairs in which $S 1$ and $S 2$ would be together which is $\frac{1}{8}$. Since $\frac{1}{4} - \frac{1}{8} = \frac{1}{8}$, then the answer to this question must be $\frac{1}{8}$ probability, or 12.5%.