`2^n` players of equal strength are playing a knock out tournament. If they are paired at randomly in all rounds, find out the probability that out of two particular players `S_1a n dS_2,` exactly one will reach in semi-final `(n in N ,ngeq2)dot`

Explanation

Four players will reach in semi-final.
Since all the players are of equal strength, this is equivalent to selecting four players out of `2^(n)` players.
Total number of ways = `.^(2n)C_(4)`
Favorable number of ways = Selecting 4 players from `2^(n)` players of which 3 players are from `(2^(n) - 2)` players (other than `S_(1)` and `S_(2)`) and 1 from `S_(1)` and `S_(2)`
`therefore` Required probability = `(.^((2^(n)-2))C_(3)xx2)/(.^(2^n)C_(4))`
`= ((2^(n) - 2) xx (2^(n) - 3) xx (2^(n) - 4) xx 8)/(2^(n)(2^(n)-1)(2^(n) - 2)(2^(n) - 3))`
`=(8xx(2^(n)-4))/(2^(n)(2^(n)-1))`