Question

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

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

Answer 1

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

Let $S=$ the sample space, then $n\left(S\right)=2^m$
Let $E_1=$ the event that one persons gets a head whereas each of the remaining $(m-1)$ persons get a tail and
$E_2=$ the event that one persons gets a tail whereas each of the remaining $(m-1)$ persons get a head.
The required event $E=E_1\cup E_2$
$\therefore n\left(E\right)=n\left(E_1\right)+n\left(E_2\right)$
${}^m{C}_1\times {{}^{m-1}C}_{m-1}1.1^{m-1}{+}^m{C}_1{\times }^{m-1}{C}_{m-1}1.1^{m-1}$
$=m+m=2m$
$\therefore$ Required probability, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{2m}{2^m}=\frac{m}{2^{m-1}}.$