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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 by

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

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

Please explain this part once more

The initial sample space is P5, Pi, Pj, Pk because we need to find probablity that P1 wins tournament given P5 reaches semis, right? I hope this clear.

Now, when will P1 win tournament? This is possible if P1 also reaches semis, right?

So, if we want to find the P(P1 wins tournament/ P5 reaches semis), then P1 has to find place in players P5, Pi, Pj, Pk , right?

So, P1 can be Pi or Pj or Pk.

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So, whats the probablity that P1 is Pi or Pj or Pk?

the sample space is (2n-1)C3  since excluding P1, you can select any 3 players from remaining ones in place of Pi, Pj and Pk.

Favourable events are those where P1 is also in semis alongwith P5 i.e. P1, P5, Pj, Pk  for some j,k.

So, total favourable events = (2n-2)C2  i.e you are choosing for Pj and Pk.

Thats it.

Crystal Clear now!! Thanks!

@GateSet  which test series ?

Testbook
free or purchase ?
Purchased on discount
OK then keep posting good QS
Surely!!

@Debashish.

OK then keep posting good QS

Good comment ;)

But, really @Gateset that was wonerful question.

+1 vote