Question 12 Marks
Four persons are chosen at random from a group of 4 males, 3 females and 5 children. Find the probability that exactly two of them are children.
Answer
View full question & answer→Total number of people $=4+3+5=12$
4 people from 12 people can be chosen in ${ }^{12} C _4$ ways
$=\frac{12 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}=495$
There must be two children in every selection which can be done in ${ }^5 C _2$ ways. Remaining two persons will be chosen from $3$ males $+$ $4$ females $=7$ persons which can be done in ${ }^7 C _2$ ways. Therefore, favourable outcomes
$\begin{array}{l}={ }^5 C_2 \times{ }^7 C_2 \\=\frac{5 \times 4}{1 \times 2} \times \frac{7 \times 6}{1 \times 2}=210\end{array}$
Hence, required probability $=\frac{210}{495}=\frac{14}{33}$
4 people from 12 people can be chosen in ${ }^{12} C _4$ ways
$=\frac{12 \times 11 \times 10 \times 9}{1 \times 2 \times 3 \times 4}=495$
There must be two children in every selection which can be done in ${ }^5 C _2$ ways. Remaining two persons will be chosen from $3$ males $+$ $4$ females $=7$ persons which can be done in ${ }^7 C _2$ ways. Therefore, favourable outcomes
$\begin{array}{l}={ }^5 C_2 \times{ }^7 C_2 \\=\frac{5 \times 4}{1 \times 2} \times \frac{7 \times 6}{1 \times 2}=210\end{array}$
Hence, required probability $=\frac{210}{495}=\frac{14}{33}$