Question

A committee has to be made of $5$ members from $6$ men and $4$ women.

The probability that at least one woman is present in the committee is

• A. $\frac{1}{42}$
• B. $\frac{41}{42}$
• C. $\frac{2}{63}$
• D. $\frac{1}{7}$

Answer 1

The correct option is B

$\frac{41}{42}$

Explanation for the correct option:

Step 1: Find the possible outcomes.

We have been given that, a committee has to be made of $5$ members from $6$ men and $4$ women.

We need to find probability that at least one woman is present in the committee.

Let $S$be the total outcomes and $A$be the possible outcomes.

As a committee has to be made of $5$ members from $6$ men and $4$ women, total cases would be,

$n\left(S\right)={}^{4}C_1\times {}^{6}C_4+{}^{4}C_2\times {}^{6}C_3+{}^{4}C_3\times {}^{6}C_2+{}^{4}C_4\times {}^{6}C_1+{}^{6}C_5$

$$\begin{gathered} =(\frac{4!}{1!(4-1)!}\times \frac{6!}{4!(6-4!)})+(\frac{4!}{2!(4-2)!}\times \frac{6!}{3!(6-3)!})+(\frac{4!}{3!(4-3)!}\times \frac{6!}{2!(6-2)!})+(\frac{4!}{4!(4-4)!}\times \frac{6!}{1!(6-5)!})+\frac{6!}{5!(6-5)!} \\ =60+120+60+6+6 \\ =252 \end{gathered}$$

Since at least one woman is present on the committee. So possible outcomes would be

$n\left(A\right)={}^{4}C_1\times {}^{6}C_4+{}^{4}C_2\times {}^{6}C_3+{}^{4}C_3\times {}^{6}C_2+{}^{4}C_4\times {}^{6}C_1$

$$\begin{gathered} =(\frac{4!}{1!(4-1)!}\times \frac{6!}{4!(6-4!)})+(\frac{4!}{2!(4-2)!}\times \frac{6!}{3!(6-3)!})+(\frac{4!}{3!(4-3)!}\times \frac{6!}{2!(6-2)!})+(\frac{4!}{4!(4-4)!}\times \frac{6!}{1!(6-5)!}) \\ =60+120+60+6 \\ =246 \end{gathered}$$

Step 2: Find the required probability:

We know that $\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

$$\begin{gathered} \Rightarrow P\left(A\right)=\frac{246}{252} \\ \Rightarrow \mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{41}}{\mathbf{42}} \end{gathered}$$

Hence, option (B) is the correct option.