Question

In a game called "odd man out" m(m>2) person toss a coin to determine who will buy refreshment for the entire group. A person who gets an outcome different from that of the rest of the numbers of the group is called the odd man out. The probability that there is loser in any game is

& A.\dfrac{1}{2m} \\\ & B.\dfrac{m}{{{2}^{m-1}}} \\\ & C.\dfrac{2}{m} \\\ & D.\text{None of these} \\\ \end{aligned}$$

Answer 1

In this question, we will first find the total number of outcomes possible by multiplying the number of outcomes for each person. After that, we will find favorable outcomes for when there is a loser in the game. Using the formula, $\text{Probability}=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$ we will find our required probability.

Complete step-by-step answer:
Let us represent m person in the game as ${{P}_{1}},{{P}_{2}},{{P}_{3}},\ldots \ldots {{P}_{m}}$. When a coin is tossed, we have only two outcomes which are heads or tails. So we can say, when every person tosses a coin, there are 2 outcomes for each person i.e. ${{P}_{1}}$ has 2 outcome, ${{P}_{2}}$ has 2 outcome, ..... similarly ${{P}_{m}}$ has 2 outcomes.
We know that, according to the multiplication principle, when two independent events access simultaneously the total outcome is equal to the multiplication of their individual outcomes. Here, each person is tossing coin individually. So, total outcome becomes equal to $\underset{\text{m times}}{\mathop{2\times 2\times 2\times \ldots \ldots \times 2}}\,$.
So we can say the total outcome ${{2}^{m}}$.
Now, let us find the number of favorable outcomes. We have to find a favorable outcome for when there is a loser in any game. Since a person only loses if he gets an outcome different from rest of group, so possible cases will be:
Case I: One person out of m person gets a tail and the rest people gets a head. So here any of the m man can get tails and then others will have heads. So 1 case arrives for each person. So total outcomes will be $\underset{\text{m times}}{\mathop{1+1+1+\ldots \ldots +1}}\,=m$.
Case II: One person out of m person gets a head and the rest people get a tail. Similar to the previous case, there will be 1 case for each person. So the total outcome will be $\underset{\text{m times}}{\mathop{1+1+1+\ldots \ldots +1}}\,=m$.
So from both cases, we see that favorable outcomes are m+m = 2m.
Hence we have a total outcome as ${{2}^{m}}$ and a favorable outcome as 2m.
We know that, the probability is given by,
$\text{Probability}=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$.
Therefore, $\text{Probability}=\dfrac{\text{2m}}{{{2}^{m}}}$.
We know that, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\text{ and }{{a}^{m}}=\dfrac{1}{{{a}^{-m}}}$ so we get:
$\text{Probability}={{\text{2}}^{1-m}}\cdot \text{m}=\dfrac{\text{m}}{{{2}^{m-1}}}$.
Hence our required probability is $\dfrac{\text{m}}{{{2}^{m-1}}}$.
Hence option B is the correct answer.

So, the correct answer is “Option B”.

Note: Students should note that, we have to take both the cases into consideration while counting favorable outcomes. At the end, we have simplified our answer just to match it with our options. Probability always lies between 0 and 1.