Question

A committee of two persons is selected from two men and two women. What is the probability that the committee will have one man?

Answer 1

Hint: We are asked in the above problem to select two persons for a committee from two men and two women. The possibilities in which two persons are selected as follows: The two persons can be one man and one woman, two men or two women. And then add these three selections of two persons for the committee. As we have asked to find the probability that the committee will have one man so for that we require total possibilities and favorable possibilities. Total possibilities are the total ways of selecting two persons for the committee and favorable possibilities are the ones in which only one man is there. And then to get the probability, we are going to divide favorable possibilities by total possibilities.

Complete step-by-step solution:
There are 2 men and 2 women and we have to select two persons from these 4 persons. Now, two persons can be one man and one woman, two men or two women so a total of three cases are there.
Now, selecting one man and one woman from the 4 people which we are going to do by selecting one man from 2 men and one woman from two women by the combinatorial method and then multiplying both these selections.
${}^{2}{{C}_{1}}\left( {}^{2}{{C}_{1}} \right)$
We know that ${}^{n}{{C}_{1}}=n$ so using this relation to simplify the above expression we get,
$\begin{align}
  & 2\left( 2 \right) \\
 & =4 \\
\end{align}$
Now, selecting two men from 4 persons which is going to be done by selecting 2 men from 2 men by combinatorial method we get,
${}^{2}{{C}_{2}}$
We know that ${}^{n}{{C}_{n}}=1$ so using this relation in the above expression we get 1
Now, selecting 2 women from 4 persons which is going to be done by selecting 2 women from 2 women by combinatorial method we get ${}^{2}{{C}_{2}}$
We know that ${}^{n}{{C}_{n}}=1$ so using this relation in the above expression we get 1
Now, adding the three cases we get the total possibilities of selecting 2 persons.
$\begin{align}
  & 4+1+1 \\
 & =6 \\
\end{align}$
Hence, we got the total possibilities as 6.
We are asked to find the probability that the committee will have one man so the formula for probability is equal to:
$\dfrac{\text{Favorable possibilities}}{\text{Total possibilities}}$
In the above expression, we have already found the total possibilities as 6. Now, favorable possibilities are the ones in which the committee will have only one man which is the case when one man and one woman are the two persons and that number is 4. Hence, the favorable possibilities are 4. Now, substituting the favorable possibilities as 4 and total possibilities as 6 in the above expression we get,
$\begin{align}
  & \dfrac{4}{6} \\
 & =\dfrac{2}{3} \\
\end{align}$
Hence, the probability that the committee will have one man is $\dfrac{2}{3}$.

Note: You might think that favorable outcomes may also contain the possibilities when two men are there. The answer is no because the question has emphasized that we have to find the probability in which the committee will have one man.
Another mistake that could happen is that as the question is asking to find the probability in which the committee will have one man so in the total possibilities you might have not considered the case when two women are there so make sure you will consider that case also because total possibilities contain all the ways in which we can form a committee of 2 persons from 4 persons so don’t get biased here.