91Ó°ÊÓ

(a): The Tree Diagram:

The tree diagram which represents the experimental situation given in the exercise is given below.

The initial branches represent the first layer of events, and the adequate probabilities are given in the diagram. Second-generation branches represent the second layer of events, with the adequate conditional probabilities given. The third-generation branches represent the third layer of events, with adequate conditional probabilities given.

We are given relevant probabilities which we are going to use.

First layer probabilities:

\(\begin{aligned}{\rm{P(A) = 0}}{\rm{.75}}\\{\rm{P}}\left( {{{\rm{A}}^{\rm{\cent}}}} \right){\rm{ = 1 - P(A)}}\\{\rm{ = 0}}{\rm{.25}}\end{aligned}\)

Second layer probabilities are

\(\begin{aligned}{\rm{P(B}}\mid {\rm{A) = 0}}{\rm{.9}}\\{\rm{P}}\left( {{{\rm{B}}^{\rm{\cent}}}\mid {\rm{A}}} \right){\rm{ = 1 - P(B}}\mid {\rm{A)}}\\{\rm{ = 0}}{\rm{.1P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm{\cent}}}} \right)\\{\rm{ = 0}}{\rm{.8P}}\left( {{{\rm{B}}^{\rm{\cent}}}\mid {{\rm{A}}^{\rm{\cent}}}} \right)\\{\rm{ = 1 - P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm{\cent}}}} \right)\\{\rm{ = 0}}{\rm{.2}}\end{aligned}\)

Third layer probabilities are

\(\begin{aligned}{\rm{P(C}}\mid {\rm{A\c{C}B) = 0}}{\rm{.8}}\\{\rm{P}}\left( {{{\rm{C}}^{\rm{\cent}}}\mid {\rm{A\c{C}B}}} \right){\rm{ = 0}}{\rm{.2}}\\{\rm{P}}\left( {{\rm{C}}\mid {\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = 0}}{\rm{.6}}\\{\rm{P}}\left( {{{\rm{C}}^{\rm{\cent}}}\mid {\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = 0}}{\rm{.4}}\\{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B}}} \right){\rm{ = 0}}{\rm{.7}}\\{\rm{P}}\left( {{{\rm{C}}^{\rm{\cent}}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{{\rm{C}}^{\rm{\cent}}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = 0}}{\rm{.7}}\end{aligned}\)

At the event of the third-generation branches, there are the probability of the intersection of adequate three events that lead to that point, however, we are not supposed to do that now (see description of (a)).

(b):

The Multiplication Rule

\({\rm{P(A\c{C}B) = P(A}}\mid {\rm{B)*P(B)}}\)

Using the multiplication rule twice, we have

\(\begin{aligned}{\rm{P(A\c{C}B\c{C}C) = P(C}}\mid {\rm{A\c{C}B)*P(A\c{C}B)}}\\{\rm{ = P(C}}\mid {\rm{A\c{C}B)*P(B}}\mid {\rm{A)*P(A)}}\\{\rm{ = 0}}{\rm{.75*0}}{\rm{.9*0}}{\rm{.8}}\\{\rm{ = 0}}{\rm{.54}}\end{aligned}\)

(c):

For any events \({{\rm{D}}_{\rm{1}}}\), \({{\rm{D}}_{\rm{2}}}\), and \({{\rm{D}}_{\rm{3}}}\), we have that

\(\left. {{{\rm{D}}_{\rm{1}}}{\rm{\c{C}}}{{\rm{D}}_{\rm{2}}}{\rm{\c{C}}}{{\rm{D}}_{\rm{3}}}} \right){\rm{ + }}\left( {{{\rm{D}}_{\rm{1}}}{\rm{\c{C}}}{{\rm{D}}_{\rm{2}}}{\rm{\c{C}D}}_{\rm{3}}^{\rm{\cent}}} \right)\)

where " \({\rm{ + }}\) " stands for disjoint union of two events. Using this we have

1. \(\begin{aligned}{\rm{P(B\c{C}C) = P(B\c{C}C\c{C}A) + P}}\left( {{\rm{B\c{C}C\c{C}}}{{\rm{A}}^{\rm{\cent}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.54 + 0}}{\rm{.25*0}}{\rm{.8*0}}{\rm{.7}}\\{\rm{ = 0}}{\rm{.68,}}\end{aligned}\)
2. we compute the probabilities the same way as in (a), by using two times the multiplication rule (or a general theorem of the multiplication rule).

(d):

Similarly, as in (b) the following is true

\(\begin{aligned}{\rm{P(C) = P(A\c{C}B\c{C}C) + P}}\left( {{{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B\c{C}C}}} \right){\rm{ + P}}\left( {{\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}{\rm{\c{C}C}}} \right){\rm{ + P}}\left( {{{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}{\rm{\c{C}C}}} \right)\\{\rm{ = 0}}{\rm{.75*0}}{\rm{.9*0}}{\rm{.8 + 0}}{\rm{.75*0}}{\rm{.1*0}}{\rm{.6 + 0}}{\rm{.25*0}}{\rm{.8 \times 0}}{\rm{.7 + 0}}{\rm{.25*0}}{\rm{.2*0}}{\rm{.3}}\\{\rm{ = 0}}{\rm{.54 + 0}}{\rm{.45 + 0}}{\rm{.14 + 0}}{\rm{.015}}\\{\rm{ = 0}}{\rm{.74}}\end{aligned}\)

(e):

We are asked to find

\(\begin{aligned}{\rm{P(A}}\mid {\rm{B\c{C}C) = }}\frac{{{\rm{P(A\c{C}B\c{C}C)}}}}{{{\rm{P(B\c{C}C)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.54}}}}{{{\rm{0}}{\rm{.68}}}}\\{\rm{ = 0}}{\rm{.7941}}{\rm{.}}\end{aligned}\)

where we used definition of conditional probability:

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

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A\c{C}B)}}}}{{{\rm{P(B)}}}}\)

for any two event \({\rm{A}}\)and\({\rm{B}}\).