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) $ \dfrac{1}{{{2}^{m-1}}} $
(b) $ \dfrac{m-1}{{{2}^{m-1}}} $
(c) $ \dfrac{m}{{{2}^{m-1}}} $
(d) None of these

Answer 1

Hint: We will find the probability of $ m $ simultaneous coin tosses. Then we have to choose one coin out of $ m $ that will have a different outcome from the rest of the coins. There are two possibilities for this one coin's outcome. So, we will get an expression that will tell us the probability that there is one loser in any game.

Complete step by step answer:
There are $ m $ people playing the game, so we have $ m $ simultaneous coin tosses. Now, we know that one coin toss can have outcome either heads or tails. So, the probability of obtaining either of them as an outcome is $ \dfrac{1}{2} $ . Since we have $ m $ simultaneous coin tosses, the probability of an outcome of all the tosses is $ \dfrac{1}{2}\times \underbrace{\ldots }_{m\text{ times}}\times \dfrac{1}{2}={{\left( \dfrac{1}{2} \right)}^{m}} $ . For this outcome to be such that it has just one heads and the rest of them as tails is $ \left( \begin{align}
  & m \\
 & 1 \\
\end{align} \right){{\left( \dfrac{1}{2} \right)}^{m}} $ . Similarly, for the outcome to be such that it has just one tails and the rest of them as heads is $ \left( \begin{align}
  & m \\
 & 1 \\
\end{align} \right){{\left( \dfrac{1}{2} \right)}^{m}} $ .
Therefore, the probability of the outcome satisfying the conditions of the "odd man out" game is
 $ 2\left( \begin{align}
  & m \\
 & 1 \\
\end{align} \right){{\left( \dfrac{1}{2} \right)}^{m}}=\dfrac{m}{{{2}^{m-1}}} $
Hence, the correct option is C.

Note:
It is important to understand the conditions given in the game. This will help us list the outcomes needed and find the probability of the occurrence. We know that there are $ m $ ways of choosing one object out of $ m $ objects. Therefore, we see that $ \left( \begin{align}
  & m \\
 & 1 \\
\end{align} \right)=m $ . There are other similar identities that allow us to calculate the combinations conveniently.