The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+1 vote
135 views

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 (2.4k points) | 135 views

1 Answer

+3 votes
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 (18k 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.


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

29,138 questions
36,959 answers
92,026 comments
34,803 users