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In a boxing tournament 2n equally skilled players P1,P2,P3...........P$2^{n}$, are participating.

In each round players are divided in pairs at random and winner from each pair moves in next round. If P5 reaches the semifinals then what is the probability that P1 wins the tournament?

A. $(\frac{1}{2})^{logn}$

B. $\frac{2^{logn-1}}{2^{n}-1}$

C. $\frac{3}{4} * \frac{1}{2^{n}-1}$

D. $\frac{7}{8} * \frac{1}{2^{n}-1}$

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Lets take an example.

Suppose there 8 participants P1, P2..................P7, P8.

Now, they are saying find the probablity that P wins the tournamant given that P5 reaches semi-finals. 

So, sample space is P5, P, Pj , Pk (i.e. P5, P, Pj , Pare semi-finalists).

Now, we would like to find the probablity that P1 wins the tournament given the above sample space.

This means that P1 should reach the semifinals (i.e. P1 should be either one of the P, Pj , Pk).

Now, P(P1 is either one of the P, Pj , Pk)

= ( select any two for Pj , Pk  ) /  ( select any three for P, Pj , Pk)

= $\frac{\binom{6}{2}}{\binom{7}{3}}$          .......................................(1)

= $\frac{3}{7}$          

Now, we have got P1 in semifinals. So, we need to find probablity that P1 wins the tournament given that semifinalists are P5, P1 , Pj , Pk

P(P1 wins the tournament given that P5, P1 , Pj , Pare semi-finalists) =  $\frac{1}{4}$   .................(2)

$\therefore$ P(P1 wins the tournament / P5 reaches semis) = $\frac{1}{4}$ *  $\frac{3}{7}$ =  $\frac{3}{4}$ *  $\frac{1}{7}$

This answer matches with answer B.

----------------------------------------------------------------------

Lets generalize the answer.

(A) = $\frac{\binom{2^{n}-2}{2}}{\binom{2^{n}-1}{3}}$       ...........from statement(1) above

(B) = $\frac{1}{4}$   .......................from statement(2) above

$\therefore$ P(P1 wins the tournament / P5 reaches semis) = (B) * (A) = $\frac{3}{4} * \frac{1}{2^{n}-1}$

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