# Question 5 Exercise 6.4

Solutions of Question 5 of Exercise 6.4 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

A committee of five persons is to be selected at random form $6$ men and $4$ women. Find the probability that the committee will consist of $3$ men and $2$ women.

Total number of persons $=6+4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10)} C_5 &=\dfrac{10 !}{(10-5) ! 5 !}\\ &=252\\ n(S)&=252\end{align} When $3$ men and $2$ women.

By multiplication principle the total number of ways, selecting $3$ men and $2$ worncn are: \begin{align}{ }^6 \mathrm{C}_3\cdot{ }^{4} \mathrm{C}_2&=\dfrac{6 !}{(6-3) ! 3 !} \cdot \dfrac{4 !}{(4-2) ! 2 !}\\ &=120\end{align} Hence the probability of getting $3$ men and $2$ women in commitlee is: $$=\dfrac{120}{252}=\dfrac{10}{21}$$

A committee of five persons is to be selected at random form $6$ men and $4$ women. Find the probability that the committee will consist of $2$ men and $3$ women.

Total number of persons $=6+4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10} C_5 &=\dfrac{10 !}{(10-5) ! 5 !}\\ &=252\\ n(S)&=252\end{align} When $2$ men and $3$ wornen.

By multiplication principle the total number of ways, selecting $2$ men and $3$ women are: \begin{align}{ }^6 C_3 2\cdot{ }^4 C_3&=\dfrac{6 !}{(6-2) ! 2 !} \cdot \dfrac{4 !}{(4-3) ! 3 !}\\ &=15.4\\ &=60\end{align} Hence the probability of getting $2$ men and $3$ women in committee is: $$=\dfrac{60}{252}=\dfrac{5}{21}$$