Question

In the game called ‘odd man out’, 4 people toss a coin to decide 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. Find the probability that there is a loser in any game?
(a) $ \dfrac{1}{3} $ .
(b) $ \dfrac{1}{2} $ .
(c) $ \dfrac{1}{5} $ .
(d) $ \dfrac{1}{6} $ .

Answer 1

Hint: We start solving the problem by solving for the total outcomes obtained if 4 persons are tossed the coin. We then find the possibilities of getting a loser by finding a total no. of possibilities that one person getting a tail and other getting heads and vice-versa. We then find the probability by dividing total possibilities with total outcomes to get the desired result.

Complete step-by-step answer:
Given that we have a game called “odd man out”. 4 persons toss a coin to decide who will buy refreshments for the group. A person who gets an outcome different from that of the rest of the group is called the odd man out. We need to find the probability that there is a loser in any game.
We know that there are two outcomes to tossing a coin. So, there are two outcomes for each person. The outcomes of each person are independent to each other. So, the total outcomes of each person will be multiplied to each other.
So, total no. of outcomes = \[{{\left( outcomes\text{ }per\text{ }each\text{ }person \right)}^{total persons}}\].
Total no. of outcomes = $ {{2}^{4}} $ .
Total no. of outcomes = 16 ---(1).
We get a loser when one person gets tails and others get heads or one person get heads and others get tails.
We try to get no. of outcomes for one person getting tails and others getting heads and the outcomes are as follows:
The outcomes that one person gets tails and others do not are = (THHH) + (HTHH) + (HHTH)+ (HHHT).
Here T = person getting tails, H = person getting heads.
No. of possibilities that one person gets tails and others are not = 4.
Now, we try to get no. of outcomes for one person getting heads and others getting tails and the outcomes are as follows:
The possibilities that one person gets tails and others are not = (HTTT) + (THTT) + (TTHT)+ (TTTH).
No. of possibilities that one person gets heads and others are not = 4.
So, total no. of possibilities that we get a loser = 4 + 4.
Total no. of possibilities that we get a loss = 8.
We know that probability is defined as $ probability=\dfrac{\text{total no}\text{. of possibilities}}{\text{total no}\text{. of outcomes}} $ .
Now, we have a probability that there is a loser in any game.
So, required probability = $ \dfrac{8}{16} $ .
Required probability = $ \dfrac{1}{2} $ .
∴ The required probability is $ \dfrac{1}{2} $ .
So, the correct answer is “Option B”.

Note: This problem can be assumed as a single person throwing four different coins to get three same outcomes and a different one. We should not just assume that one person getting tails and others getting heads is the only possibility of getting an odd man out. We always follow $ probability=\dfrac{\text{total no}\text{. of possibilities}}{\text{total no}\text{. of outcomes}} $ to find the probability.