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A. \[\dfrac{5}{14}\]

B. \[\dfrac{1}{21}\]

C. $\dfrac{3}{14}$

D. $\dfrac{8}{21}$

Answer

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Hint: In this question we will find the probability of men in the committee of 4 members and then we will find the probability of women. Then, in the final step, we will find the probability of exactly 2 men in the committee by using the formula $=\dfrac{\left[ \text{Probability of 2 men }\times \text{ Probability of 1 women} \right]}{\text{Total Probability}}$.

Complete step by step answer:

It is given in the question that there are 4 men and 5 women out of which we have to make a committee of 3 members.

Also, it is given that there are always two men and only one woman in the 3 member committee.

Now, we know that the formula of probability $=\dfrac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}$.

So, the probability of men to be on the committee is \[={}^{4}{{C}_{2}}\].

As, we know that the formula for ${}^{n}{{C}_{r}}$ is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write as,

\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}\]

\[{}^{4}{{C}_{2}}=\dfrac{4\times 3\times 2!}{2!\times 2!}\]

Cancelling \[2!\] from numerator and denominator, we get,

\[\begin{align}

& {}^{4}{{C}_{2}}=\dfrac{4\times 3}{2!} \\

& {}^{4}{{C}_{2}}=\dfrac{12}{2} \\

& {}^{4}{{C}_{2}}=6 \\

\end{align}\]

Therefore, there are 6 combinations possible for men to be in the committee of 3 members.

Similarly, the probability of women to be on the committee is $={}^{5}{{C}_{1}}$.

As, we know that the formula for ${}^{n}{{C}_{r}}$is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write that,

\[{}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}\]

\[\begin{align}

& {}^{5}{{C}_{1}}=\dfrac{5!}{1!\times 4!} \\

& {}^{5}{{C}_{1}}=\dfrac{5\times 4!}{1\times 4!} \\

\end{align}\]

Cancelling \[4!\] from numerator and denominator, we get,

\[{}^{5}{{C}_{1}}=5\]

Therefore, there are 5 combinations possible for a woman to be in the committee of 3 members.

Also, the total probability $={}^{9}{{C}_{3}}$.

As, we know that the formula for ${}^{n}{{C}_{r}}$ is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write as, ${}^{9}{{C}_{3}}=\dfrac{9!}{3!\left( 9-3 \right)!}$

${}^{9}{{C}_{3}}=\dfrac{9\times 8\times 7\times 6!}{3!\times 6!}$

Cancelling \[6!\] from numerator and denominator, we get,

\[\begin{align}

& {}^{9}{{C}_{3}}=\dfrac{9\times 8\times 7}{3\times 2\times 1} \\

& {}^{9}{{C}_{3}}=84 \\

\end{align}\]

Thus, total probability = 84.

So, probability of exactly 2 men in the committee of 3 members will be $=\dfrac{\text{Probability of 2 men }\times \text{ Probability of 1 women}}{\text{Total Probability}}$

We can substitute the value of probability of 2 men as 6 and probability of 1 woman as 5. Also, w can substitute the total probability = 84 in the above equation and we will get,

Probability of exactly 2 men $=\dfrac{6\times 5}{84}=\dfrac{30}{84}$

Dividing numerator and denominator by 6, we get,

$\dfrac{30}{84}=\dfrac{5}{14}$

Thus, the probability exactly for 2 men in the committee of 3 members is $\dfrac{5}{14}$.

Note: This question can be solved in just few lines if you know the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Please note that the formula ${}^{n}{{C}_{r}}$ is different from the formula ${}^{n}{{P}_{r}}$. We use ${}^{n}{{P}_{r}}$ in finding permutation and ${}^{n}{{C}_{r}}$ in finding combinations.

Complete step by step answer:

It is given in the question that there are 4 men and 5 women out of which we have to make a committee of 3 members.

Also, it is given that there are always two men and only one woman in the 3 member committee.

Now, we know that the formula of probability $=\dfrac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}$.

So, the probability of men to be on the committee is \[={}^{4}{{C}_{2}}\].

As, we know that the formula for ${}^{n}{{C}_{r}}$ is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write as,

\[{}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}\]

\[{}^{4}{{C}_{2}}=\dfrac{4\times 3\times 2!}{2!\times 2!}\]

Cancelling \[2!\] from numerator and denominator, we get,

\[\begin{align}

& {}^{4}{{C}_{2}}=\dfrac{4\times 3}{2!} \\

& {}^{4}{{C}_{2}}=\dfrac{12}{2} \\

& {}^{4}{{C}_{2}}=6 \\

\end{align}\]

Therefore, there are 6 combinations possible for men to be in the committee of 3 members.

Similarly, the probability of women to be on the committee is $={}^{5}{{C}_{1}}$.

As, we know that the formula for ${}^{n}{{C}_{r}}$is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write that,

\[{}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}\]

\[\begin{align}

& {}^{5}{{C}_{1}}=\dfrac{5!}{1!\times 4!} \\

& {}^{5}{{C}_{1}}=\dfrac{5\times 4!}{1\times 4!} \\

\end{align}\]

Cancelling \[4!\] from numerator and denominator, we get,

\[{}^{5}{{C}_{1}}=5\]

Therefore, there are 5 combinations possible for a woman to be in the committee of 3 members.

Also, the total probability $={}^{9}{{C}_{3}}$.

As, we know that the formula for ${}^{n}{{C}_{r}}$ is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we can write as, ${}^{9}{{C}_{3}}=\dfrac{9!}{3!\left( 9-3 \right)!}$

${}^{9}{{C}_{3}}=\dfrac{9\times 8\times 7\times 6!}{3!\times 6!}$

Cancelling \[6!\] from numerator and denominator, we get,

\[\begin{align}

& {}^{9}{{C}_{3}}=\dfrac{9\times 8\times 7}{3\times 2\times 1} \\

& {}^{9}{{C}_{3}}=84 \\

\end{align}\]

Thus, total probability = 84.

So, probability of exactly 2 men in the committee of 3 members will be $=\dfrac{\text{Probability of 2 men }\times \text{ Probability of 1 women}}{\text{Total Probability}}$

We can substitute the value of probability of 2 men as 6 and probability of 1 woman as 5. Also, w can substitute the total probability = 84 in the above equation and we will get,

Probability of exactly 2 men $=\dfrac{6\times 5}{84}=\dfrac{30}{84}$

Dividing numerator and denominator by 6, we get,

$\dfrac{30}{84}=\dfrac{5}{14}$

Thus, the probability exactly for 2 men in the committee of 3 members is $\dfrac{5}{14}$.

Note: This question can be solved in just few lines if you know the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Please note that the formula ${}^{n}{{C}_{r}}$ is different from the formula ${}^{n}{{P}_{r}}$. We use ${}^{n}{{P}_{r}}$ in finding permutation and ${}^{n}{{C}_{r}}$ in finding combinations.