Probability : 16 players S1 , S2 ....... , S16 play in a tournament. They are ..................?

Since all the players are of the same strength and skill, with whom S1 is paired will have no bearing on the probability of his winning.

T number of ways of choosing 8 winners out of 16 is ¹⁶C₈ The number of ways of choosing S₁ and 7 other winners out of 15 is ¹⁵C₇

=> Probability that S1 will win = ¹⁵C₇ / ¹⁶C₈ = ½

Let E1(E2) denote the event that S1 and S2 are paired ( not paired ) together and let A denote the event that one of the two players S1 and S2 is among the winners. Since S1 can be paired with any of the remaining 15 players, we have –

P ( E1) = 1/15 and P (E2) = 14/15.

In case E1 occurs, it is certain that one of S1 and S2 will be among the winners.

In case E2 occurs, the probability that exactly one of S1 and S2 is among the winners. is –

P [ ( S1 ∩ S’ ₂ ) U (S’1 ∩ S2) ] = P ( S1 ∩ S’ ₂) U (S’1 ∩ S2) + P ( S’1 ∩ S2)

=> P(S1) P(S’2) + P(S1’) . P (S2) = ( ½ ) [ 1 – (1/2) ] + [ 1 – (1/2) ] ( ½ ) = ½

That is P ( A|E1) = 1 and P ( A|E2) = ½

By the total probability rule ---

P (A) = P(E1) P(A|E1) + P(E2) P ( A|E2)

=> P (A) = ( 1/15 ) + ( 14/15 ) ( ½ ) = 8/15 …… Answer

Dr N

.

a) 8/16=1/2

b) 1/2 since the chance that S1 won and S2 lost is 1/2*1/2=1/4, and the chance that S2 won and S1 lost is the same.

(a) 1/2

(b) 3/4