8n players P₁, P₂, P₃, .... P_8n play a knockout tournament. It is known that all the players are of equal strength. The tournament is held in three rounds where the players are paired at random in each round. If it is given that P₁ wins in the third round. If the probability that P₂ looses in the second
round is 8/31, then the value of n is ____

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answer is 4.

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Detailed Solution

Probability that 2 P wins in first round is given by
$P_{1} wins = \frac{^{8 n - 2} C_{4 n - 2}}{^{8 n} C_{4 n}} = \frac{4 n - 1}{8 n - 1}$
In the second round probability that P₂ looses in second round is given by
$P_{1} wins = 1 - \frac{2 n - 1}{4 n - 1} = \frac{2 n}{4 n - 1}$
Hence, probability that P₂ looses in second round. Given P₁ wins in third round is
$\frac{4 n - 1}{8 n - 1} \times \frac{2 n}{4 n - 1} = \frac{2 n}{8 n - 1} . \frac{2 n}{8 n - 1} . = \frac{8}{31} \Rightarrow 31 n = 4 (8 n - 1) \Rightarrow 31 n = 32 n - 4 \Rightarrow n = 4$