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We're going to have ten equally likely possibilities because we're picking at random. Before we go any further, suppose that instead of names, the initial letter of each name is written on a sheet of paper, with Cox having Cox and Cramer having Cr.

\(\begin{aligned}{{\rm{\{ A,B\} ,\{ A,Cox\} ,\{ A,Cr\} ,\{ A,F\} ,\{ B,Cox\} }}}\\{{\rm{\{ B,Cr\} ,\{ B,F\} ,\{ Cox,Cr\} ,\{ Cox,F\} ,\{ Cr,F\} }}}\end{aligned}\).

The probability of both Anderson and Box being chosen is represented by letters \({\rm{A}}\)and \({\rm{B}}\)

\({\rm{P(\{ A,B\} ) = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{,}}\)

because the outcomes are equally likely to occur (probability is \({\rm{1/10}}\) for each simple occurrence out of ten).

The likelihood of at least one of the two members whose names begin with \({\rm{C}}\) being chosen equals the probability of one of the seven outcomes including at least \({\rm{Cox}}\) or \({\rm{Cr}}\) (disjoint union of the seven events)

\(\begin{aligned}P(\{~starts\text{ }with\text{ }at\text{ }least\text{ }one\text{ }C~\})&=P(\{A,Cox\}+\{A,Cr\}+\{B,Cox\}+\{B,Cr\} \\+\{Cox,Cr\}+\{Cox,F\}+\{Cr,F\}) \\&=\frac{7}{10}, \\\end{aligned}\)

1. denotes the joining of discontinuous occurrences.
2. equally likely events.

We still obtain \({\rm{10}}\) outcomes if we replace the name with the number of years of teaching experience. There are six outcomes with a sum more than or equal to \({\rm{15}}\) years of teaching experience, and we want to find the likelihood that two chosen representatives have a total of at least \({\rm{15}}\) years of teaching experience.

\({\rm{\{ 3,14\} ,\{ 6,10\} ,\{ 6,14\} ,\{ 7,10\} ,\{ 7,14\} ,\{ 10,14\} }}\)

As a result, the likelihood is

\(\begin{aligned}P(~at\text{ }least~15~years\text{ }\exp erience~\})&=P(\{3,14\}+\{6,10\}+\{6,14\}+ \\+\{7,10\}+\{7,14\}+\{10,14\}) \\&=\frac{6}{10}, \\\end{aligned}\)

where the same explanations as before are used.