Question

From a group of $3$ man and $2$ women, two person are selected at random. Find the probability that at least one women is selected.

• A. $\frac{1}{5}$
• B. $\frac{7}{10}$
• C. $\frac{2}{5}$
• D. $\frac{5}{6}$

Answer 1

The correct option is B

$\frac{7}{10}$

Explanation of the correct option:

Finding the sample space $S$:

Total no. of persons $=3+2=5$

No. of persons to be selected $=4$

Let $S$ be the set of all possible outcomes. As $2$ persons are to be selected from $5$,

$\Rightarrow n\left(S\right)={}^{5}C_2$

$=\frac{5!}{2!3!}$ $\left[{}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$$\begin{gathered} =\frac{5\times 4\times 3!}{2\times 1\times 3!} \\ =10 \end{gathered}$$

Finding the required probability:

Let $A$ be the event of selecting at least one woman. For this we have two options, either one man and one woman will be selected or two women will be selected.

$\Rightarrow n\left(A\right)=\left({}^{3}C_1\times {}^{2}C_1\right)+\left({}^{2}C_2\right)$

$=\left(\frac{3!}{1!2!}\times \frac{2!}{1!1!}\right)+\left(\frac{2!}{2!0!}\right)$ $\left[{}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$=\left(\frac{3\times 2!}{2!}\times \frac{2\times 1}{1}\right)+\frac{2!}{2!}$ [as $0!=1$]

$$\begin{gathered} =\left(3\times 2\right)+1 \\ =7 \end{gathered}$$

Therefore, the required probability:

$\begin{array}{rcl}P\left(A\right)& =& \frac{n\left(A\right)}{n\left(S\right)}\\ & =& \frac{7}{10}\end{array}$.

Hence, the correct answer is option B.