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players P1,P2,P3……… P16 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 pairs. Assuming that all the players are of equal strength, the probability that exactly one of the two players P1 and P2 is among the eight winners is ________.ans  8/15

asked in Probability by Active (2.1k points)   | 98 views

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Best answer

Lets get it by contradiction.

The negation of the 

Either P1 is there and not P2  in the 8 winners or vice-versa

is 

Either both P1 and P2 are there in 8 winners or none of P1 and P2 is there is 8 winers

.................................(1)

 

Consider the probablity for the negation.

Events where both P1 and P2 are not there in 8 winners = $\binom{14}{8}$

Events where both P1 and P2 are there in the 8 winners   = $\binom{14}{6}$

#ways to select the 8 winners = $\binom{16}{8}$

 

Now, lets evaluate the probablity for the statement (1).

P(  statement (1)  ) = ( $\binom{14}{8}$ + $\binom{14}{6}$ ) /  (  $\binom{16}{8}$  )..............(2)

 

P(Either P1 is there and not P2 or vice-versa) = 1 - answer in statement (2)

                                                                  = 8/15

answered by Veteran (15.2k points)  
selected by

it should have been like these,na ?

Events where both P1 and P2 are not there in 8 winners =  14C8

Events where both P1 and P2 are there in the 8 winners   =  14C6

yes.will edit.


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