Answer
The outcomes in the sample space are:
\[S=\left\{ \begin{align}
& ({{p}_{1}},{{i}_{1}}),\text{ }({{p}_{1}},{{i}_{2}}),\text{ }({{p}_{1}},{{i}_{3}}),\text{ }({{p}_{2}},{{i}_{1}}),\text{ }({{p}_{2}},{{i}_{2}}), \\
& ({{p}_{2}},{{i}_{3}}),\text{ }({{p}_{1}},{{p}_{2}}),\text{ }({{i}_{1}},{{i}_{2}}),\text{ }({{i}_{1}},{{i}_{3}}),\text{ }({{i}_{2}},{{i}_{3}}) \\
\end{align} \right\}\]
Work Step by Step
The outcomes of the event A are:
\[A=\left\{ \begin{align}
& ({{p}_{1}},{{i}_{1}}),\text{ }({{p}_{1}},{{i}_{2}}),\text{ }({{p}_{1}},{{i}_{3}}),\text{ }({{p}_{2}},{{i}_{1}}),\text{ }({{p}_{2}},{{i}_{2}}), \\
& ({{p}_{2}},{{i}_{3}}),\text{ }({{i}_{1}},{{i}_{2}}),\text{ }({{i}_{1}},{{i}_{3}}),\text{ }({{i}_{2}},{{i}_{3}}) \\
\end{align} \right\}\]
Since the woman has her purse snatched by two teenagers and she suspects five persons.
Out of five, 2 are perpetrators and 3 are innocent.
Let us define two perpetrators by${{p}_{1}}$and${{p}_{2}}$ and three 3 innocent people by ${{i}_{1}}$, ${{i}_{2}}$ and ${{i}_{3}}$.
We can choose 2 out of 5 suspects in ${}^{5}{{C}_{2}}=10$ ways.
Therefore, the sample space S can be defined as:
\[S=\left\{ \begin{align}
& ({{p}_{1}},{{i}_{1}}),\text{ }({{p}_{1}},{{i}_{2}}),\text{ }({{p}_{1}},{{i}_{3}}),\text{ }({{p}_{2}},{{i}_{1}}),\text{ }({{p}_{2}},{{i}_{2}}), \\
& ({{p}_{2}},{{i}_{3}}),\text{ }({{p}_{1}},{{p}_{2}}),\text{ }({{i}_{1}},{{i}_{2}}),\text{ }({{i}_{1}},{{i}_{3}}),\text{ }({{i}_{2}},{{i}_{3}}) \\
\end{align} \right\}\]
Let event A denote that she makes at least one incorrect identification. This means the event consists of at least one innocent. Hence, event A contains all outcomes in the sample space except $({{p}_{1,}}{{p}_{2}})$.
The outcomes of event A are:
\[A=\left\{ \begin{align}
& ({{p}_{1}},{{i}_{1}}),\text{ }({{p}_{1}},{{i}_{2}}),\text{ }({{p}_{1}},{{i}_{3}}),\text{ }({{p}_{2}},{{i}_{1}}),\text{ }({{p}_{2}},{{i}_{2}}), \\
& ({{p}_{2}},{{i}_{3}}),\text{ }({{i}_{1}},{{i}_{2}}),\text{ }({{i}_{1}},{{i}_{3}}),\text{ }({{i}_{2}},{{i}_{3}}) \\
\end{align} \right\}\]
