Eight players P₁, P₂, . . . . . . . . . . . . . P₈ play a knock – out tournament It is known that whenever the players P_i and P_j play, the player P_i will win if i ₄ reaches the final?

The number of ways in which P₁, P₂ . . . . . . . . . . . . . . . . . . . , P₈ can be paired in four pairs.

= 1/4! x ⁸ C₂ x ⁶ C₂ x ⁸ C₂ x ⁴ C₂ x ² C₂ = 105

Now, at least two players certainly reach the second round in between P₁, P₂, and P₃ and P₄ can reach in final if exactly two players play against each other in between P₁, P₂, P₃ and remaining player will play against one of the players from P₅, P₆, P₇, P₈ and P₄ plays against one of the remaining three from P₅, P₆, P₇, P₈

This can be possible in ³ C₂ x ^{4 C1} x ³ C₁ = 36 ways

∴ Prob. that P₄ and exactly one of P₅ . . . . . . . . . . . P₈ reach second round

= 36/105 = 12/35

If P₁, P_i, P₄ and P_j where i = 2 or 3 and j = 5 or 6 or 7 reach the second round then they can be paired in 2 pairs in

1/2! x ⁴ C₂ x ² C₂ = 3 ways

But P₄ will reach the final if P₁ plays against P_i and P₄ plays against P_j

Hence the prob. that P₄ reach the final round from the second = 1/3.

∴ prob. that P₄ reach the final is 12/35 x 1/3 = 4/35.