Question

From a group of $8$ boys and $3$ girls, a committee of $5$ members to be formed. Find the probability that $2$ particular girls are included in the committee.

• A. $\frac{4}{11}$
• B. $\frac{2}{11}$
• C. $\frac{6}{11}$
• D. $\frac{8}{11}$

Answer 1

The correct option is B

$\frac{2}{11}$

Explanation for the correct option:

Step: 1 Calculate the different ways $5$members can be selected from a group of $11$.

The committee has a total of members $5$ and they must be selected from $8$ boys and $3$ girls.

The total members are $11$.

Therefore, $5$members can be selected from a group of $11$ members in different ways

$$\begin{gathered} C_5^{11} \\ =\frac{11!}{5!} \\ =\frac{11\times 10\times 9\times 8\times 7\times 6!}{6!\times 5\times 4\times 3\times 2} \\ =462 \end{gathered}$$

Step: 2 Calculate the different ways $2$ members can be selected for $2$ positions.

The committee must have $2$ girls in it and the other $3$ members could be anyone.

Therefore, the different ways $2$girls can be selected is

$$\begin{gathered} =C_2^2 \\ =1 \end{gathered}$$

Step: 3 Calculate the different ways $3$ members can be selected from a group of $9$people.

Now, the different ways $3$ other members can be selected from $9$ people is

$$\begin{gathered} =C_3^9 \\ =\frac{9!}{6!\times 3!} \\ =\frac{9\times 8\times 7\times 6!}{6!\times 3\times 2} \\ =84 \end{gathered}$$

$\therefore$Required probability $=\frac{{}^{2}C_2{}^{9}C_3}{{}^{11}C_5}$

$$\begin{gathered} =\frac{84}{462} \\ =\frac{\mathbf{2}}{\mathbf{11}} \end{gathered}$$

Hence, Option ‘B’ is Correct.