InterviewSolution
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InterviewSolution
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Sixteen players S_(1),S_(2),…..S_(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. Assume that all the players are of equal strength. (i) Find the probability that the player S_(1) is among the eight winners. (ii) Find the probability that exactly one of the two playersS_(1) and S_(2) is among the eight winners. |
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Answer» <P> `=("Probability of"S_(1) "being a pair")` `xx` (Prbability f `S_(1)` winnig the GROUP) `=1xx(1)/(2)=(1)/(2)""["since " S_(1) " is definitely in a group" ]` (ii)If `S_(1)and S_(2)` are in two separately, then exactly one wins. IF `S_(1)and S_(2)` will be among the eight winners. IF `S_(1)` wins and `S_(2)` loses or `S_(1)` losses and `S_(2)` wins. Now, the prbability of `S_(1).S_(2)` being the same pair and one wins `=("Probability of " S_(1),S_(2)` " being in the same pair" )` `xx` (Probability of anyone winning in the pair). and the probability of `S_(1),S_(2)` being thesame pair `=(N(E))/(n(S))` where, `n(E)=` the number of ways in which 16persons can be divided in 8 pairs. `therefore n(E)=((14)!)/((2!)^(7)*7!)and n(S)=((16)!)/((2!)^(8)*8!)` `therefore` Probability of `S_(1) and S_(2)` being in the same pair `=((14)!*(2!)^(8)*8!)/((2!)^(7)*7!*(16)!)=(1)/(15)` The probability of any one wining in the pairs of `S_(1),S_(2)=P` (certain event)= 1 `therefore` Thepairs of `S_(1),S_(2)` being in two pairs separately and `S_(1)` wins, `S_(2)` losses + The probability of `S_(1),S_(2)` being in two pairs separately and `S_(1)` loses , `S_(2)` wins. `=[1-((((14)!)/((2!)^(7)*7!))/(((16)!)))/((2!)^(8)*8!)]xx(1)/(2)xx(1)/(2)+[1-((((14)!)/((2!)^(7)*7!))/(((16)!)))/((2!)^(8)*8!)]xx(1)/(2)xx(1)/(2)` `=(1)/(2)xx(14xx(14)!)/(15xx(14)!)=(7)/(15)` `therefore` Required probability `=(1)/(15)+(7)/(15)=(8)/(15)` |
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