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.

Answer

A, B

Explanation

(i) Probability of `S_(1)`to be among the eight winners
`=("Probability of"S_(1) "being a pair")`
`xx` (Prbability f `S_(1)` winnig the group)
`=1xx(1)/(2)=(1)/(2)" "["since " S_(1) " is definitely in a group" ]`
(ii)If `S_(1)and S_(2)` are in two separately, then exactly one wins.
IF `S_(1) and S_(2)` will be among the eight winners. IF `S_(1)` wins and `S_(2)` loses or `S_(1)` losses and `S_(2)` wins.
Now, the prbability of `S_(1).S_(2)` being the same pair and one wins
`=("Probability of " S_(1),S_(2)` " being in the same pair" )`
`xx` (Probability of anyone winning in the pair).
and the probability of `S_(1),S_(2)` being the same pair
`=(n(E))/(n(S))`
where, `n(E)=` the number of ways in which 16 persons can be divided in 8 pairs.
`therefore n(E)=((14)!)/((2!)^(7)*7!)and n(S)=((16)!)/((2!)^(8)*8!)`
`therefore` Probability of `S_(1) and S_(2)` being in the same pair
`=((14)!*(2!)^(8)*8!)/((2!)^(7)*7!*(16)!)=(1)/(15)`
The probability of any one wining in the pairs of `S_(1),S_(2)=P` (certain event)= 1
`therefore` The pairs of `S_(1),S_(2)` being in two pairs separately and `S_(1)` wins, `S_(2)` losses + The probability of `S_(1),S_(2)` being in two pairs separately and `S_(1)` loses , `S_(2)` wins.
`=[1-((((14)!)/((2!)^(7)*7!))/(((16)!)))/((2!)^(8)*8!)]xx(1)/(2)xx(1)/(2)+[1-((((14)!)/((2!)^(7)*7!))/(((16)!)))/((2!)^(8)*8!)]xx(1)/
(2)xx(1)/(2)`
`=(1)/(2)xx(14xx(14)!)/(15xx(14)!)=(7)/(15)`
`therefore` Required probability `=(1)/(15)+(7)/(15)=(8)/(15)`