91Ó°ÊÓ

Referring to the exercise 18

A group has eight members. In January three members are selected randomly on the committee. In February four members are selected randomly and independently of the first selection to serve on another committee. In March five members are selected randomly and independently of the previous two selections to be on a third committee.

Let,\({E_i}\)be the event that A and B are both selected for committee i\(\left( {i = 1,2,3} \right)\)

And let,

\(\Pr \left( {{E_i}} \right) = {p_i}\)

For first committee one member is to be selected from remaining 6 members as A and B is already selected.

The probability of selecting A and B is:

\(\begin{aligned}{}{p_1} = \frac{{\left( {\begin{aligned}{{}{}}6\\1\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}8\\3\end{aligned}} \right)}}\\ = \frac{6}{{56}}\\ = 0.107143\end{aligned}\)

For second committee two members need to be selected from remaining 6 members as A and B is already selected.

The probability of selecting A and B is:

\(\begin{aligned}{}{p_2} = \frac{{\left( {\begin{aligned}{{}{}}6\\2\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}8\\4\end{aligned}} \right)}}\\ = \frac{{15}}{{70}}\\ = 0.214286\end{aligned}\)

For third committee three members need to be selected from remaining 6 members as A and B is already selected.

The probability of selecting A and B is:

\(\begin{aligned}{}{p_3} = \frac{{\left( {\begin{aligned}{{}{}}6\\3\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}8\\5\end{aligned}} \right)}}\\ = \frac{{20}}{{56}}\\ = 0.357143\end{aligned}\)

Since\(E1,\,\,{E_2},\,\,and\,\,{E_3}\)are independent.

The required probability is,

\(\begin{aligned}{}\Pr \left( {{E_1} \cup {E_2} \cup {E_3}} \right) = {p_1} + {p_2} + {p_3} - {p_1}{p_2} - {p_2}{p_3} - {p_1}{p_3} + {p_1}{p_2}{p_3}\\ = 0.5490\end{aligned}\)

Therefore, the required probability is 0.5490