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 P_{5}, P_{i }, P_{j} , P_{k} (i.e. P_{5}, P_{i }, P_{j} , P_{k }are 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_{i }, P_{j} , P_{k}).

Now, P(P1 is either one of the P_{i }, P_{j} , P_{k})

= ( select any two for P_{j} , P_{k }) / ( select any three for P_{i }, P_{j} , P_{k})

= $\frac{\binom{6}{2}}{\binom{7}{3}}$ **.......................................(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 P_{5}, P_{1}_{ }, P_{j} , P_{k}

P(P1 wins the tournament given that P_{5}, P_{1}_{ }, P_{j} , P_{k }are 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.

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