# 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.

10 years ago
Hello Student,
(a) Prob. of S1 to among the eight winners = (prob. of S1 being in a pair) x (prob. of S1 winning in the group).
= 1 x 1/2 [∵ S1 is definitely in a group i.e. certain event] = 1/2
(b) If S1 and S­2 are in the same pair then exactly on wins. If S1 and S2 are in two pairs separately then exactly one of S1 and S2 will be among the eight winners if S1 win and S2 loses or Sloses and S2 wins.
Now the prob. of S1, S2 being in the same pair and one wins.
= (prob. of S1, S2 being in the same pair) x (prob. of any one winning in the pair)
And the prob. of S1, S2 being in the same pair
= n (E)/n (S), where n (S) = the no. of ways in which 16 person can be divided in 8 pairs; n (E) = the no. of ways in which S1, S2 are in same pair or 14 persons can be divided into 7 pairs.
∴ n (E) = 14!/(2!)7 .7! and n (S) = 16!/(2!)8 .8!
∴ Prob. of S1 and S2 being in the same pair
= $\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}$= 21.8/16.15 = 1/15
The prob. of any one winning in the pair of S1, S2 = P (certain event) = 1
∴ The pair of S1, S2 being in two pairs separately and any one of S1, S2 wins.
= the prob. of S1, S2 being in two pairs separately and S1 wins, S2 loses + the prob. of S1, S2 being in two pairs separately and S1 loses, S2 wins.
= $\left [ 1-\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}\right]$x 1/2 x 1/2 + $\left [ 1-\frac{\frac{14!}{(2!)^7 7!}}{\frac{16!}{(2!)^8 .8!}}\right]$x 1/2 x 1/2
= x 1/4 = 1/2 x 14 x 14!/15 x 14! = 7/15
∴ Required prob. = 1/15 + 7/15 = 8/15

Thanks