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 there is a loser in any game is.
(This question has multiple correct options)
\[\left( \text{a} \right)\text{ }\dfrac{1}{2m}\]
\[\left( \text{b} \right)\text{ }\dfrac{m}{{{2}^{m-1}}}\]
\[\left( \text{c} \right)\text{ }\dfrac{2}{m}\]
\[\left( \text{d} \right)\text{ }\dfrac{1}{{{2}^{m-1}}}\]

Answer 1

Hint: First of all, find the total outcomes by multiplying the total outcomes for each person. Now, find the favourable outcomes when there is a loser in the game or odd one out when he gets different outcomes from the rest of the people. There would be two cases. Finally, use the formula, \[\text{Probability}=\dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}\]

Complete step-by-step answer:
We are given that 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. We have to find the probability that there is a loser in any game.
Let us represent m persons as \[{{P}_{1}},{{P}_{2}},{{P}_{3}},{{P}_{4}}.....\] and so on upto \[{{P}_{m}}.\] In the toss of a coin, we know that we can either get a head or a tail. So, while tossing a coin, there are a total of 2 outcomes.
So, when the first person \[{{P}_{1}}\] will toss a coin, there would be a total of 2 outcomes.
So, when the second person \[{{P}_{2}}\] will toss a coin, there would be a total of 2 outcomes.
So, when the third person \[{{P}_{3}}\] will toss a coin, there would be a total of 2 outcomes.
This will continue for all m persons. So, similarly, when the \[{{m}^{th}}\] person \[{{P}_{m}}\] will toss the coin, there would be a total of 2 outcomes. We know that according to the multiplication principle when two independent events access simultaneously, the total outcomes are equal to the multiplication of their individual outcomes.
So, we know that each person is tossing the coin individually and it is independent of each other. So, we get,
\[\begin{align}
  & \text{Number of total outcomes }N\left( T \right)=\left( \text{outcomes of }{{\text{P}}_{1}} \right)\times \left( \text{outcomes of }{{\text{P}}_{2}} \right)\times \left( \text{outcomes of }{{\text{P}}_{3}} \right) \\
 & \times \left( \text{outcomes of }{{\text{P}}_{4}} \right)\times ......\times \left( \text{outcomes of }{{\text{P}}_{m}} \right) \\
\end{align}\]
\[N\left( T \right)=2\times 2\times 2.....m\text{ times}\]
\[N\left( T \right)={{2}^{m}}\]
Now, let us see the favourable outcomes. We have to find the favourable outcomes for when there is a loser in any game. We are given that a person loses when he/she gets an outcome different from the rest of the group. So, there would be two cases which are as follows.
Case 1: One who loses gets a tail and the rest of the people get a head.
In this case, \[{{P}_{1}}\] can get tail and rest can get head or \[{{P}_{2}}\] can get tail and the rest can get head or \[{{P}_{3}}\] can get a tail and rest can get head and this can be possible for every person up to when \[{{P}_{m}}\] will get tail and rest will get head. So, there is 1 case for each person. So, for m persons, there are total m favourable outcomes.
Case 2: One who loses gets head and the rest of the people get tail.
In this case, \[{{P}_{1}}\] can get head and the rest can get tail or \[{{P}_{2}}\] can get head and the rest can get tail or \[{{P}_{3}}\] can get head and rest can get tail and this can be possible for every person up to when \[{{P}_{m}}\] will get head and rest will get tail. So, there is 1 case for each person. So, for m persons, there are total m favourable outcomes.
So, we get the total favourable outcomes,
\[N\left( F \right)=\left( \text{No}\text{.of Favourable Outcomes of Case 1} \right)+\left( \text{No}\text{.of Favourable Outcomes of Case 2} \right)\]
\[\Rightarrow N\left( F \right)=m+m\]
\[\Rightarrow N\left( F \right)=2m\]
We know that the probability that there is a loser,
\[P\left( L \right)=\dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}\]
\[\Rightarrow P\left( L \right)=\dfrac{N\left( F \right)}{N\left( T \right)}\]
\[\Rightarrow P\left( L \right)=\dfrac{2m}{{{2}^{m}}}\]
We know that, \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}=\dfrac{1}{{{a}^{n-m}}}.\] By using this, we get,
\[P\left( L \right)=\dfrac{m}{{{2}^{m-1}}}\]
Hence, option (b) is the right answer.

Note: In this question, many students forget to consider both the cases and give the answer as just \[\dfrac{m}{{{2}^{m}}}.\] So this must be taken care of. Also, students must properly rearrange their final answer in order to get the correct option as we have written the final answer \[\dfrac{2m}{{{2}^{m}}}\] as \[\dfrac{m}{{{2}^{m-1}}}\] to match it with the options.