Question

Four persons are chosen at random from a group of $3$ men, $2$ women and $4$ children. The chance that exactly $2$: of them are children, is:

• A. $\frac{10}{21}$
• B. $\frac{11}{13}$
• C. $\frac{13}{25}$
• D. $\frac{21}{32}$

Answer 1

The correct option is A

$\frac{10}{21}$

The explanation for the correct option:

Step1. Calculate the probability of selecting $4$ people from a group of $9$:

The number of people in that group are $(3+2+4)=9$.

The different ways $4$ people can be selected from a group of $9$ is ${}^{9}C_4$ways

$$\begin{gathered} =\frac{9!}{4!\times 5!}\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{]} \\ =\frac{9\times 8\times 7\times 6\times 5!}{5!\times 4\times 3\times 2} \\ =126 \end{gathered}$$.

Step2. Calculate the probability of selecting $2$ children from a group of $4$:

We can select $2$ children from a group of $4$children is ${}^{4}C_2$ways

$$\begin{gathered} =\frac{4!}{2!\times 2!} \\ =\frac{4\times 3\times 2}{2\times 2} \\ =6 \end{gathered}$$.

Step3. Calculate the probability of selecting $2$people from the group of $5$:

For the group of $4$people, other $2$people can be selected from the rest of the $5$ people in ${}^{5}C_2$ways

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

Step: 4 Calculating the required probability:

Hence, the required probability that exactly $2$ of them are children is

$$\begin{gathered} =\frac{6\times 10}{126} \\ =\frac{10}{21} \end{gathered}$$

Therefore, Option(A) is the correct answer.