Question

From a group of $7$ men and $4$ ladies a committee of $6$ persons is formed, then the probability that the committed contains $2$ ladies is:

• A. $\frac{5}{13}$
• B. $\frac{5}{11}$
• C. $\frac{4}{11}$
• D. $\frac{3}{11}$

Answer 1

The correct option is B

$\frac{5}{11}$

Explanation for the correct option:

Step:1 Calculate the different ways $6$ persons can be selected as mentioned.

Here, the total number of persons is

$$\begin{gathered} =7+4 \\ =11 \end{gathered}$$

Therefore, different ways the committee of $6$ persons can be formed out of $11$ people is

$$\begin{gathered} =C_6^{11} \\ =\frac{11!}{6!} \\ =\frac{11\times 10\times 9\times 8\times 7\times 6!}{6!\times 5\times 4\times 3\times 2} \\ =462 \end{gathered}$$

Now, different ways $2$ girls can be selected out of $4$ girls is

$$\begin{gathered} =C_2^4 \\ =\frac{4!}{2!} \\ =6 \end{gathered}$$

Again, different ways $2$ girls can be selected out of $4$ girls is

$$\begin{gathered} =C_4^7 \\ =\frac{7!}{4!} \\ =\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2} \\ =35 \end{gathered}$$

Step: 2 Calculate the required probability.

Hence, the probability that there will be $2$ girls in the committee of $6$ is

$$\begin{gathered} =\frac{35\times 6}{462} \\ =\frac{5}{11} \end{gathered}$$

Therefore, Option(B) is the correct answer.