Question

A certain group has eight members. In January, three members are selected at random to serve on a committee. In February, four members are selected at random and independently of the first selection to serve on another committee. In March, five members are selected at random and independently of the previous two selections to serve on a third committee. Determine the probability that each of the eight members serves on at least one of the three committees.

Short answer

The probability of each of the eight members serving on at least one of the three committees is 0.279337

Step-by-step solution

Given information

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

Calculation of probability for eight members serves on at least one of the three committees.

The total number of ways to pick the committee members i.e., the sample space is,

\(\left( {\begin{aligned}{{}{}}8\\3\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}8\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}8\\5\end{aligned}} \right)\)

With all members at least once= The Total number of members – At least one member + At least two members -At least three members + At least four members-…

All members of at least one committee are,

\(\left( {\begin{aligned}{{}{}}8\\3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}8\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}8\\5\end{aligned}} \right) - 8\left( {\begin{aligned}{{}{}}7\\3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7\\5\end{aligned}} \right) + \left( {\begin{aligned}{{}{}}8\\2\end{aligned}} \right) \times \left( {\begin{aligned}{{}{}}6\\3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}6\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}6\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}8\\3\end{aligned}} \right) \times \left( {\begin{aligned}{{}{}}5\\3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5\\5\end{aligned}} \right) + 0...\)

\(\left( {\begin{aligned}{{}{}}8\\2\end{aligned}} \right) \times \left( {\begin{aligned}{{}{}}8\\3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}8\\4\end{aligned}} \right)\left( {\begin{aligned}{{}{}}8\\5\end{aligned}} \right)\)

Comes from first choosing, where two members do not appear.

Thus, the probability of each eight members at least one of the three committees is,

\(\begin{aligned}{}\frac{{\left( {\begin{aligned}{{}}8\\3\end{aligned}} \right)\left( {\begin{aligned}{{}}8\\4\end{aligned}} \right)\left( {\begin{aligned}{{}}8\\5\end{aligned}} \right) - 8\left( {\begin{aligned}{{}}7\\3\end{aligned}} \right)\left( {\begin{aligned}{{}}7\\4\end{aligned}} \right)\left( {\begin{aligned}{{}}7\\5\end{aligned}} \right) + \left( {\begin{aligned}{{}}8\\2\end{aligned}} \right) \times \left( {\begin{aligned}{{}}6\\3\end{aligned}} \right)\left( {\begin{aligned}{{}}6\\4\end{aligned}} \right)\left( {\begin{aligned}{{}}6\\5\end{aligned}} \right) - \left( {\begin{aligned}{{}}8\\3\end{aligned}} \right) \times \left( {\begin{aligned}{{}}5\\3\end{aligned}} \right)\left( {\begin{aligned}{{}}5\\4\end{aligned}} \right)\left( {\begin{aligned}{{}}5\\5\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}}8\\3\end{aligned}} \right)\left( {\begin{aligned}{{}}8\\4\end{aligned}} \right)\left( {\begin{aligned}{{}}8\\5\end{aligned}} \right)}}\\ = \frac{{613}}{{219520}}\\ = 0.279337\end{aligned}\)

The probability of each of the eight members serving on at least one of the three committees is 0.279337