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 decided to 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 prabability that the player `S_(1)` is among the eight winners.
(b) Find the probability that exactly one of the two players `S_(1)and S_(2)` is among the eight winners.

Explanation

(a) The probability of `S_(1)` to be among the eight winners is equal to the probability of `S_(1)` winning in the group, which is given by 1/2.
(b) If `S_(1) and S_(2)` are in the same pair then exactly one wins. If `S_(1) and S_(2)` are in two separate pairs, then for exactly one of `S_(1)and S_(2)` to be among the eight winners, `S_(1)` wins and `S_(2)` loses or `S_(1)` loses and `S_(2)` wins.
Now the probability of `S_(1),S_(2)` being in the same pair and one wins is
(Probability of `S_(1),S_(2)` being in the same psir) `xx` (Probability of any one winning in the pair).
And the probability of `S_(1).S_(2)` being in the same pair is `=(n(E))/(n(S))`
The number of ways 16 players are divided into 8 pairs is
`n(S)=(16!)/((2!)^(8)xx8!)`
The number of ways in which 16 persons can be divided in 8 pairs to that `S_(1)and S_(2)` are in same pair is
`n(E)=(16!)/((2!)^(7)xx7!)`
Therefore, the probability of `S_(4)and S_(2)` being in the same pair is
`((14!)/((2!)^(7)xx7!))/((16!)/((2!)^(8)xx8!))=(2!xx8)/(16xx15)=1/15`
The probability of any one winning in the pair of `S_(1),S_(2)` is P (certain event) =1.
Hence, the probability that the pair of `S_(1),S_(2)` wins is given by the sum the probability of `S_(1),S_(2)` being in two pairs separately and `S_(1)` wins, `S_(2)` loses and the probability of `S_(1),S_(2)` being in two pairs separately and `S_(1)` loses, `S_(2)` wins. It is given by
`[1-1/15]xx1/2xx1/2+[1-1/15]xx1/2xx1/2=1/2xx14/15=7/15`
Therefore, the required probability is `(1//15)+(7//15)=(8//15).`