91Ó°ÊÓ

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

The tree diagram below depicts the experimental circumstance presented in the exercise.

The first layer of occurrences is represented by the starting branches, and the appropriate probabilities are shown in the diagram. The second layer of events is represented by the second-generation and third-generation branches, which have the appropriate conditional probabilities.

We display the product of individual probabilities that lead to that location, which is really the probabilities of the crossings, to the right of the third-generation branches, using the multiplication formula described below (twice) (the ones that we need for the exercise).

The Multiplication Rule

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

The tree diagram sums up every information we have from the exercise \({\rm{59}}\) and exercise \({\rm{69}}{\rm{.}}\)

From exercise \({\rm{59}}\) we have the following

\(\begin{array}{l}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.4;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.35;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.25;}}\\{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.3;}}\\{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.6;}}\\{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.5}}\end{array}\)

From the exercise \({\rm{69}}\) we have

\(\begin{array}{l}P\left( {C\mid {A_1} \cap B} \right) = 0.7;\\P\left( {C\mid {A_1} \cap {B^\prime }} \right) = 0.5\\P\left( {C\mid {A_2} \cap B} \right) = 0.6;\\P\left( {C\mid {A_2} \cap B} \right) = 0.5;\\P\left( {C\mid {A_3} \cap B} \right) = 0.5;\\P\left( {C\mid {A_3} \cap B} \right) = 0.4.\end{array}\)

As previously stated, we may compute the crossings we require using the multiplication method.

\(\begin{array}{l}P\left( {{A_1} \cap B \cap C} \right) &=& P\left( {{A_1}} \right) \cdot P\left( {B\mid {A_1}} \right) \cdot P\left( {C\mid B \cap {A_1}} \right)\\ &=& 0.4 \cdot 0.3 \cdot 0.7\\ &=& 0.084P\\\left( {{A_1} \cap {B^\prime } \cap C} \right) &=& P\left( {{A_1}} \right) \cdot P\left( {{B^\prime }\mid {A_1}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_1}} \right)\\ &=& 0.4 \cdot 0.7 \cdot 0.5\\ &=& 0.14P\\\left( {{A_2} \cap B \cap C} \right) &=& P\left( {{A_2}} \right) \cdot P\left( {B\mid {A_2}} \right) \cdot P\left( {C\mid B \cap {A_2}} \right)\\ &=& 0.35 \cdot 0.6 \cdot 0.6\\ &=& 0.126\\P\left( {{A_2} \cap {B^\prime } \cap C} \right) &=& P\left( {{A_2}} \right) \cdot P\left( {{B^\prime }\mid {A_2}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_2}} \right)\\ &=& 0.35 \cdot 0.4 \cdot 0.5\\ &=& 0.7P\\\left( {{A_3} \cap B \cap C} \right) &=& P\left( {{A_3}} \right) \cdot P\left( {B\mid {A_3}} \right) \cdot P\left( {C\mid B \cap {A_3}} \right)\\ &=& 0.25 \cdot 0.5 \cdot 0.5\\ &=& 0.0625P\\\left( {{A_3} \cap {B^\prime } \cap C} \right) &=& P\left( {{A_3}} \right) \cdot P\left( {{B^\prime }\mid {A_3}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_3}} \right)\\ &=& 0.25 \cdot 0.5 \cdot 0.4\\ &=& 0.05\end{array}\)

We now have everything we require to construct the tree diagram.

(a):

Event plus and fill-up and credit card is an intersection of those event

\(\begin{array}{l}P\left( {{A_2} \cap B \cap C} \right) &=& P\left( {{A_2}} \right) \cdot P\left( {B\mid {A_2}} \right) \cdot P\left( {C\mid B \cap {A_2}} \right)\\ &=& 0.35 \cdot 0.6 \cdot 0.6\\ &=& 0.126.\end{array}\)

(b):

Event premium and non-fill-up and credit card is intersection

\(\begin{array}{l}P\left( {{A_3} \cap {B^\prime } \cap C} \right) &=& P\left( {{A_3}} \right) \cdot P\left( {{B^\prime }\mid {A_3}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_3}} \right)\\ &=& 0.25 \cdot 0.5 \cdot 0.4\\ &=& 0.05.\end{array}\)

(c):

The intersection of events premium and credit card is

\(\begin{array}{l}P\left( {{A_3} \cap C} \right)\mathop = \limits^{(1)} P\left( {{A_3} \cap B \cap C} \right) + P\left( {{A_3} \cap {B^\prime } \cap C} \right)\\ = P\left( {{A_3}} \right) \cdot P\left( {B\mid {A_3}} \right) \cdot P\left( {C\mid B \cap {A_3}} \right) + P\left( {{A_3}} \right) \cdot P\left( {{B^\prime }\mid {A_3}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_3}} \right)\\ = 0.0625 + 0.05\\ = 0.1125\end{array}\)

(1): this is true for any three events (because \({B^\prime } \cup B\)is almost certain event).

(d):

The intersection of events fill-up and credit card is

\(\begin{array}{l}P(B \cap C)\mathop = \limits^{(1)} P\left( {{A_1} \cap B \cap C} \right) + P\left( {{A_2} \cap B \cap C} \right) + P\left( {{A_3} \cap B \cap C} \right)\\ = P\left( {{A_1}} \right) \cdot P\left( {B\mid {A_1}} \right) \cdot P\left( {C\mid B \cap {A_1}} \right) + P\left( {{A_2}} \right) \cdot P\left( {B\mid {A_2}} \right) \cdot P\left( {C\mid B \cap {A_2}} \right) + P\left( {{A_3}} \right) \cdot P\left( {B\mid {A_3}} \right) \cdot P\left( {C\mid B \cap {A_3}} \right)\\ = 0.084 + 0.126 + 0.0625\\ = 0.2725\end{array}\)

(1): because \({A_1} \cup {A_2} \cup {A_3}\)is almost certain event.

(e):

The probability of only credit card can be calculated using The Law of Total Probability.

The Law of Total Probability

If \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\) are mutually exclusive, and\( \cup _{i = 1}^n{A_i} = \Omega \), then for any event \({\rm{B}}\) the following is true

\(\begin{array}{l}{\rm{P(B) = P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + \ldots + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{n}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{n}}}} \right)\\{\rm{ = }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)\end{array}\)

We have the following

\(\begin{array}{c}P(C) &=& P\left( {{A_1} \cap B \cap C} \right) + P\left( {{A_1} \cap {B^\prime } \cap C} \right) + P\left( {{A_2} \cap B \cap C} \right) + P\left( {{A_2} \cap {B^\prime } \cap C} \right) + P\left( {{A_3} \cap B \cap C} \right) + P\left( {{A_3} \cap {B^\prime } \cap C} \right)\\ &=& P\left( {{A_1}} \right) \cdot P\left( {B\mid {A_1}} \right) \cdot P\left( {C\mid B \cap {A_1}} \right) + P\left( {{A_1}} \right) \cdot P\left( {{B^\prime }\mid {A_1}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_1}} \right) + P\left( {{A_2}} \right) \cdot P\left( {B\mid {A_2}} \right) \cdot P\left( {C\mid B \cap {A_2}} \right) + P\left( {{A_2}} \right) \cdot P\left( {{B^\prime }\mid {A_2}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_2}} \right) + P\left( {{A_3}} \right) \cdot P\left( {B\mid {A_3}} \right) \cdot P\left( {C\mid B \cap {A_3}} \right) + P\left( {{A_3}} \right) \cdot P\left( {{B^\prime }\mid {A_3}} \right) \cdot P\left( {C\mid {B^\prime } \cap {A_3}} \right)\\ &=& 0.084 + 0.14 + 0.126 + 0.07 + 0.0625 + 0.05\\ &=& 0.5325.\end{array}\)

(f):

If the next customer uses a credit card, the probability that premium was requested is conditional probability of \({{\rm{A}}_{\rm{3}}}\) given that the event \({\rm{C}}\) has occurred

\(\begin{array}{l}P\left( {{A_3}\mid C} \right) &=& \frac{{P\left( {{A_3} \cap C} \right)}}{{P(C)}}\\ &=& \frac{{0.1125}}{{0.5325}}\\ &=& 0.2113\end{array}\)

where we used the following 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 probabilities are

\(\begin{array}{l}{\rm{ a}}{\rm{. }}P\left( {{A_2} \cap B \cap C} \right) &=& 0.126;\\{\rm{ b}}{\rm{. }}P\left( {{A_3} \cap {B^\prime } \cap C} \right) &=& 0.05;\\{\rm{ c}}{\rm{. }}P\left( {{A_3} \cap C} \right) &=& 0.1125;{\rm{ }}\\{\rm{ d}}{\rm{. }}P(B \cap C) &=& 0.2725;\\{\rm{ e}}{\rm{. }}P(C) &=& 0.5325;\\{\rm{ f}}{\rm{. }}P\left( {{A_3}\mid C} \right) &=& 0.2113.{\rm{ }}\end{array}\)