Sixteen players P1.P2.………P16 play in a tournament.

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Sixteen players

P_{1} . P_{2} . \dots \dots \dots P_{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 of the pair.Assuming that all the players are of equal strength, the probability that exactly one of the two players

P_{1}

and

P_{2}

is among the eight winners is

Options:

A . 415

B . 715

C . 815

D . 1730

Answer: Option C
:
C
Let

E_{1} (E_{2})

denote the event that

P_{1}

and

P_{2}

are paired (not paired) together and let A denote the event that one of two players

P_{1}

and

P_{2}

is among the winners.
Since, P1 can be paired with any of the remaining 15 players.
We have,

P (E_{1}) = \frac{1}{15}

and

P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{1}{15} = \frac{14}{15}

In case

E_{1}

occurs, it is certain that one of

P_{1}

and

P_{2}

will be among the winners. In case

E_{2}

occurs, the probability that exactly of

P_{1}

and

P_{2}

is among the winners is

P {(P_{1} \cap_{2}) \cup (_{1} \cap P_{2})} = P (P_{1} \cap_{2}) + P (_{1} \cap P_{2}) = P (P_{1}) P (_{2}) + P (_{1}) P (P_{2}) = (\frac{1}{2}) (1 - \frac{1}{2}) + (1 - \frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

ie,

P (\frac{A}{E_{1}}) = 1

and

P (\frac{A}{E_{2}}) = \frac{1}{2}

By the total probability Rule,

P (A) = P (E_{1}) . P (\frac{A}{E_{1}}) + P (E_{1}) . P (\frac{A}{E_{2}}) = \frac{1}{15} (1) + \frac{14}{15} (\frac{1}{2}) = \frac{8}{15}

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