Question

In a game called "odd man out 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 three is a loser in any game is

• A. $1/2m$
• B. $m/2^{m-1}$`
• C. $2/m$
• D. $1/2^{m-1}$

Answer 1

Answer: (A), (B)

The correct options are
A $1/2m$
B $m/2^{m-1}$`

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