91Ó°ÊÓ

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

a)

Proposition: If \({\rm{A}}\) and \({\rm{B}}\) are independent, then

- \({A^\prime }\) and \({\rm{B}}\) are independent;

- \({A^\prime }\) and \({B^\prime }\) are independent;

- \({\rm{A}}\) and \({B^\prime }\)are independent.

We may deduce that \(A\) and \({\rm{B}}\) are independent (we are told that \(A\)and \({\rm{B}}\) are independent in the exercise).

We are requested to calculate the conditional probability of \({\rm{B}}\) (the European project succeeding) given the occurrence of \({A^\prime }\) (the Asian project failing).

If for two events \({\rm{A}}\) and \({\rm{B}}\) stands

\({\rm{P(A}}\mid {\rm{B) = P(A)}}\)we say that they are independent. They are dependent otherwise.

From the definition of independence, and the fact that \({A^\prime }\) and \({B^\prime }\) are independent we have

\(\begin{array}{c}P\left( {{B^\prime }\mid {A^\prime }} \right) &=& P\left( {{B^\prime }} \right)\\ &=& 1 - P(B)\\ &=& 1 - 0.7\\ &=& 0.3\end{array}\)

Therefore, the probability is \({\rm{0}}{\rm{.3}}\)

(b):

The likelihood of at least one of the two projects succeeding is equal to the sum of the two events.

\(\begin{array}{c}P(A \cup B) &=& P(A) + P(B) - P(A \cap B).......(1)\\ &=& 0.4 + 0.7 - 0.4 \cdot 0.7.......(2)\\ &=& 0.82,\end{array}\)

(1): here we use the proposition given below,

(2): from the definition and the fact that \({\rm{A}}\) and \({\rm{B}}\) are independent the following is true

\(\begin{array}{c}P(A \cap B) &=& P(A) \cdot P(B)\\ &=& 0.4 \cdot 0.7\\ &=& 0.28.\end{array}\)

Proposition: Two events \({\rm{A}}\) and \({\rm{B}}\) are independent if and only if

\(P(A \cap B) = P(A) \cdot P(B)\)

Proposition: For every two events \({\rm{A}}\)and \({\rm{B}}\)

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

Therefore, the probability is \({\rm{0}}{\rm{.82}}\)

Event that at least one of the two projects is successful is union of event \({\rm{A}}\)and \({\rm{B}}\)while event only the Asian project is successful is intersection of events \(A\)and \({B^\prime }.\)We are asked to find conditional probability of \(A \cup B\)given that the event \(A \cap {B^\prime }\)has occurred

The union of events \({\rm{A}}\) and \({\rm{B}}\) indicates that at least one of the two projects is successful, but the intersection of events \({\rm{A}}\) and \({B^\prime }\) indicates that only the Asian project is successful.

Given that the event\(A \cap {B^\prime }\)has happened, we must calculate the conditional probability of\(A \cap B\).

\(\begin{array}{c}P\left( {A \cap {B^\prime }\mid A \cup B} \right) &=& \frac{{P\left[ {\left( {A \cap {B^\prime }} \right) \cap (A \cup B)} \right]}}{{P(A \cup B)}}.........(1)\\ &=& \frac{{P\left( {A \cap {B^\prime }} \right)}}{{P(A \cup B)}}.........(2)\\ &=& \frac{{P(A) \cdot P\left( {{B^\prime }} \right)}}{{P(A \cup B)}}.........(3)\\ &=& \frac{{0.4 \cdot (1 - 0.7)}}{{0.82}}\\ &=& 0.146,\end{array}\)

(1): definition of conditional probability - given below,

(2): event \(A \cap {B^\prime }\) is subset of of event \(A \cup B\) (this is true for any two events),

(3): remember that \({\rm{A}}\) and \({\rm{B}}\) are independent, therefore \({\rm{A}}\) and \({B^\prime }\) are independent. The equality comes from the definition.

Conditional probability of A given that the event B has occurred, for which\({\rm{P(B) > 0}}\), is

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)for any two events \({\rm{A}}\) and \({\rm{B}}\).

Therefore, the probability is \({\rm{0}}{\rm{.146}}\)