Question : In a game called ‘odd man out’ m(m>2) persons toss a coin to determine who will buy 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
(A) 1/2m
(B) \dfrac{m}{2^{m-1}}
(C) 2/m
(D) none of these
Answer : option (B)
Solution :
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 mC_{1}=m.
Thus, the number of favourable ways is m + m = 2m.
Therefore, P\left( A\right) =\dfrac{2m}{2^{m}}=\dfrac{m}{2^{m-1}}