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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) = $\frac{1}{4}$ *  $\frac{3}{7}$ =  $\frac{3}{4}$ *  $\frac{1}{7}$

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(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) = $\frac{3}{4} * \frac{1}{2^{n}-1}$

selected

@sushant,though u solved perfectly,still not getting this part:-(

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

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

quite good explanation. @sushant! keep it up! and akriti just take a small sample space of maybe 5 players and follow his comment next to his answer. you will get it!
@Akriti. I hope you got the initial sample space P5, Pi, Pj, Pk............(1)

Now, In order that P1 wins tournament, P1 must also reach semis.

So, this is situation: P1, P5, Pj, Pk..................(2)

So, whats the probablity of situation in statement (2) given situation in statement (2) is sample space?

So, sample space = 7C3 since you need to select Pi,Pj,Pk

Events that account for situation in statement (2) = 6C2

since, you need to select Pj,Pk

So, prob = 6C2 / 7C3

@sushant,you are finding the probability that P1 is in semis-right?

then you need to select 3 personns for semis which can be done in 7C3 ways..i got till here.

but how favourable cases are 6C2??does this mean,you have chosen P1 and you are selectig rest 2 in semis..am i right??

Yup. As you know, whenever you are consider favourable events, you should assume that situation.