Question

In a game "odd man out" each of $m\ge 2$ persons, tosses a coin to determine who will buy refreshment for the entire group. The odd man out is the one with a different outcome from the rest. The probability that there is a loser in any game is

• A. $\frac{1}{2^{m-1}}$
• B. $\frac{m-1}{2^{m-1}}$
• C. $\frac{m}{2^{m-1}}$
• D. None of these

Answer 1

The correct option is C $\frac{m}{2^{m-1}}$

Let $A$ denote the event that there is an odd man out in a game.
The total number of possible cases $=2m$
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 ${}^m{C}_1=m$
Then the number of favorable ways in $m+m=2^m$
$\therefore P\left(A\right)=\frac{2m}{2^m}=\frac{m}{2^{m-1}}$