Question

Sixteen players P1, P2, ….. P16 plays 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. Assuming that all the players are of equal strength, the probability that exactly one of the players P1 and P2 is among the eight winners is
1. 4/15
2. 8/15
3. 11/15
4. 1/3

Answer 1

Correct Answer - Option 2 : 8/15

Calculation:

⇒ Let E1 and E2 denote the event that P1 and P2 are paired or not paired together. Let A denote the event that one of the two players P1 and P2 is amongst the winners.

⇒ Since P1 can be paired with any of the remaining 15 players, so P(E1) = 1/15

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

⇒ In case E1 occurs, it is certain that one of P1 and P2 will be among the winners. In case E2 occurs, the probability that exactly one of P1 and P2 is among the winners is

⇒ P[(P1 ∩ P2) ∪ (P1 ∩ P2)] = P(P1 ∩ P2) + P(P1 ∩ P2) 

⇒ P(P1) P(P2) + P(P1) P(P2)

⇒ ½ (1 - 1/2) + (1 - 1/2)1/2 

⇒ 1/4 + 1/4

⇒ 1/2

⇒ i.e. P(A/E1) = 1 and P(A/E2) = 1/2

⇒ By the total probability rule,

⇒ P(A) = P(E1) . P(A/E1) + P(E2) P(A/E2)

⇒ 1/15 (1) + 14/15(1/2)

⇒ 8/15

⇒ If A and B are two events, then;

⇒ P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

⇒ P(A ∩ B) = P (B) ⋅ P(A | B)