Question

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A) \({\rm{5 }}{\rm{.4}}\) and P(B) \({\rm{5 }}{\rm{.7}}\).

a. If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.

b. What is the probability that at least one of the two projects will be successful?

c. Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

Short answer

The probabilities are

\(\begin{array}{l}{\rm{a}}{\rm{. }}P\left( {{B^\prime }\mid {A^\prime }} \right) &=& 0.3{\rm{; }}\\{\rm{b}}{\rm{. }}P(A \cup B) &=& 0.82;{\rm{ }}\\{\rm{c}}{\rm{. }}P\left( {A \cap {B^\prime }\mid A \cup B} \right) &=& 0.146.{\rm{ }}\end{array}\)

Step-by-step solution

Definition

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

Finding 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 \({\rm{0}}{\rm{.3}}\)

Finding probability

(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 \({\rm{0}}{\rm{.82}}\)

Finding probability

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\(A \cap B\).

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

Explanation

(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 \({\rm{0}}{\rm{.146}}\)