In a game called odd man out `m(m >2)` persons toss a coin to determine who will but refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is `1//2m` b. `m//2^(m-1)` c. `2//m` d. none of these

Options

A. `1//2m`
B. `m//2^(m-1)`
C. `2//m`
D. none of these

Answer

B

Explanation

Let A denote the event that there is an odd man out in a game. The total number of possible cases is `2^(m)`. A person is odd man out if he is alone in getting a head or a tail.
The number of ways in which there is exactly one tail (head) and the rest are heads (tails) is `.^(m)C_(1) = m`. Thus, the number of favourable ways is m + m = 2m. Therefore,
`P(A) = (2m)/(2^(m))= (m)/(2^(m - 1))`