91Ó°ÊÓ

The event \({A_i}\)represents the awarded project i, for i=1, 2, 3.

The values are given as,

\(P\left( {{A_1}} \right) = .22,\;P\left( {{A_2}} \right) = .25,\;P\left( {{A_3}} \right) = .28,\;P\left( {{A_1} \cap {A_2}} \right) = .11,\;P\left( {{A_1} \cap {A_3}} \right) = .05,\)\(P\left( {{A_2} \cap {A_3}} \right) = .07\),\(P\left( {{A_1} \cap {A_2} \cap {A_3}} \right) = .01\).

a.

The event is represented as \({A_1} \cup {A_2}\) .

The event represents that the firm is awarded project 1 or project 2.

The probability is computed as,

\(\begin{aligned}P\left( {{A_1} \cup {A_2}} \right) &= P\left( {{A_1}} \right) + P\left( {{A_2}} \right) - P\left( {{A_1} \cap {A_2}} \right)\\ &= 0.22 + 0.25 - 0.11\\ &= 0.36\end{aligned}\)

Thus, the probability of the provided event is 0.36.

b.

The event is represented as \(A_1' \cap A_2'\) .

The event represents that the firm is not awarded project 1 and not awarded project 2.

The probability is computed as,

\(\begin{aligned}P\left( {A_1' \cap A_2'} \right) &= P\left( {{A_1} \cup {A_2}} \right)'\\ &= 1 - P\left( {{A_1} \cup {A_2}} \right)\\ &= 1 - 0.36\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{Using}}\;{\rm{part}}\;{\rm{a}}} \right)\\ &= 0.64\end{aligned}\)

Thus, the probability of the provided event is 0.64.

c.

The event is represented as \({A_1} \cup {A_2} \cup {A_3}\) .

The event represents that the firm awarded at least one of the three projects.

The probability is computed as,

\(\begin{aligned}P\left( {{A_1} \cup {A_2} \cup {A_3}} \right) &= P\left( {{A_1}} \right) + P\left( {{A_2}} \right) + P\left( {{A_3}} \right) - P\left( {{A_1} \cap {A_2}} \right) - P\left( {{A_2} \cap {A_3}} \right) - \left( {{A_1} \cap {A_3}} \right) + \left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\ &= 0.22 + 0.25 + 0.28 - 0.11 - 0.05 - 0.07 + 0.01\\ &= 0.53\end{aligned}\)

Thus, the probability of the provided event is 0.53.

d.

The event is represented as \(A_1' \cap A_2' \cap A_3'\).

The event represents that the firm does not awarded any of the three projects.

The probability is computed as,

\(\begin{aligned}P\left( {A_1' \cap A_2' \cap A_3'} \right) &= 1 - P\left( {{A_1} \cup {A_2} \cup {A_3}} \right)\\ &= 1 - 0.53\\ &= 0.47\end{aligned}\)

Thus, the probability of the provided event is 0.47.

e.

The event is represented as \(A_1' \cap A_2' \cap {A_3}\) .

The event represents that the firm does not awarded project 1 and project 2 but awarded project 3.

The probability is computed as,

\(\begin{aligned}P\left( {A_1' \cap A_2' \cap {A_3}} \right) &= P\left( {{A_3}} \right) - P\left( {{A_1} \cap {A_3}} \right) - P\left( {{A_2} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\ &= 0.28 - 0.05 - 0.07 + 0.01\\ &= 0.17\end{aligned}\)

Thus, the probability of the provided event is 0.17.

f.

The event is represented as \(\left( {A_1' \cap A_2'} \right) \cup {A_3}\) .

The event of not awarded both projects 1 and 2, or awarded project 3.

The probability is computed as,

\(\begin{aligned}P\left( {\left( {A_1' \cap A_2'} \right) \cup {A_3}} \right) &= P\left( {{A_1} \cup {A_2} \cup {A_3}} \right)' + P\left( {{A_3}} \right)\\ &= \left( {1 - P\left( {{A_1} \cup {A_2} \cup {A_3}} \right)} \right) + P\left( {{A_3}} \right)\\ &= 1 - 0.53 + 0.28\\ &= 0.75\end{aligned}\)

Thus, the probability of the provided event is 0.75.