Question

From 4 children, 2 women and 4 men, 4 persons are selected. The probability that there are exactly 2 children among the selected is
A) $\dfrac{1}{7}$
B) $\dfrac{2}{7}$
C) $\dfrac{4}{7}$
D) $\dfrac{3}{7}$

Answer 1

We will find the total outcomes and the favourable outcomes using the formula for combination. We will find the probability by using the formula for probability. We will cancel out the common factors (if there are any) and finally choose the correct option.

Formula used: We will use the following formulas:
The number of ways of selecting $r$ objects from a set of $n$ objects is ${}^n{C_r}$
The formula for ${}^n{C_r}$ is ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
The formula for probability (P) is given by $P =$ number of favourable outcomes $\div$ total number of outcomes.

Complete step by step solution:
There are 4 children, 2 women and 4 men. We will find the total number of people.
Total number of people $= 4 + 2 + 4 = 10$
There are 10 people in total. We will find the total number of ways in which 4 people can be selected from 10 people.
Substituting 4 for $r$ and 10 for $n$ in ${}^n{C_r}$, we get
${}^{10}{C_4} = \dfrac{{10!}}{{\left( {10 - 4} \right)!4!}}$
Computing the factorial, we get
$\Rightarrow {}^{10}{C_4} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{6!4!}}$
Simplifying the terms, we get
$\begin{array}{l} \Rightarrow {}^{10}{C_4} = \dfrac{{10 \times 9 \times 8 \times 7}}{{24}}\\\ \Rightarrow {}^{10}{C_4} = 210\end{array}$
There are 210 ways in which 4 people can be selected from 10 people. So, the total number of outcomes is 210.
We want to select exactly 2 children. We will substitute 2 for $r$ and 4 for $n$ in ${}^n{C_r}$to find the number of ways in which 2 children can be selected out of 4 children.
Therefore,
${}^4{C_2} = \dfrac{{4!}}{{\left({4-2}\right)!2!}}$
Computing the factorial, we get
$\begin{array}{l} \Rightarrow {}^4{C_2} = \dfrac{{24}}{{2 \times 2}}\\\ \Rightarrow {}^4{C_2} = 6\end{array}$
There are 6 ways in which exactly 2 children can be selected from 4 children.
We have to select 4 persons out of which 2 should be children. The rest 2 can be men or women. There are 2 women and 4 men. We will find the number of ways of selecting 2 people from 6 people.
Substituting 2 for $r$ and 6 for $n$ in ${}^n{C_r}$, we get
$\Rightarrow {}^6{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)!2!}}$
Computing the factorial, we get
$\begin{array}{l} \Rightarrow {}^6{C_2} = \dfrac{{720}}{{24 \times 2}}\\\ \Rightarrow {}^6{C_2} = 15\end{array}$
The total number of ways of selecting 4 people such that exactly 2 children are selected is:
$\Rightarrow 6 \times 15 = 90$
So, the number of favourable outcomes is 90.
We will find the probability by substituting 90 in the numerator and 210 in the denominator in the formula for probability. So,
$P = \dfrac{{90}}{{210}}$
Breaking the terms, we get
$\begin{array}{l} \Rightarrow P = \dfrac{{3 \times 30}}{{7 \times 30}}\\\ \Rightarrow P = \dfrac{3}{7}\end{array}$

$\therefore$ Option D is the correct option.

Note:
We must know that the formula for combination (${}^n{C_r}$) is used in cases where the order of selection does not matter whereas in cases where the order of selection also matters, we should use the formula for permutations. The formula for permutations is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ where $n$ is the total number of objects and $r$ is the number of objects to be selected.