Question

In a team of 10 persons there is a married couple,from them a committee of 5 is to be made,then the probability for the couple being either included or excluded is

• A. $\frac{{}^8{C}_3}{{}^{10}{C}_5}$
• B. $\frac{{}^8{C}_5}{{}^{10}{C}_5}$
• C. $\frac{2{(}^8{C}_5)}{{}^{10}{C}_5}$
• D. $\frac{2{(}^8{P}_3)}{{}^{10}{P}_5}$

Answer 1

The correct option is C $\frac{2{(}^8{C}_5)}{{}^{10}{C}_5}$

Number of ways of forming a committee with the couple included is : ${}^8{C}_3$, since the couple has already been chosen in the committee.

Number of ways of forming a committee with the couple excluded is : ${}^8{C}_5$ since we have to choose members from the other people remaining.

Hence, total number of ways of choosing a committee in which the couple is either included or excluded is ${}^8{C}_5{+}^8{C}_3=2{\times }^8{C}_5$ ( ${\because }^8{C}_3{=}^8{C}_5$)
Total ways of choosing any $5$ out of $10$ is ${}^{10}{C}_5$

So, probability$=\frac{2{\times }^8{C}_5}{{}^{10}{C}_5}$