91Ó°ÊÓ

The Multiplication Rule

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

First layer of event is

\(\begin{aligned}{{\rm{A}}_{\rm{1}}}\rm &= \{ short book \} \\{{\rm{A}}_{\rm{2}}}\rm &= \{ medium book \} \\{{\rm{A}}_{\rm{3}}}\rm &= \{ long book \} \end{aligned}\)

and the second layer events are

\(\begin{aligned}\rm W &= \{ submited in word \} \\\rm L &= \{ submited in LaTeX \} \end{aligned}\)

The adequate probabilities for the first layer are

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.6}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.3}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.1}}\end{aligned}\)

The adequate conditional probabilities are

\(\begin{aligned}{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm &= 0{\rm{.8}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.2}}\\{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm &= 0{\rm{.5}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.5}}\\{\rm{P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm &= 0{\rm{.3}}\\{\rm{P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm &= 1 - P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.7}}\end{aligned}\)

The tree diagram also contains the probabilities of the intersections. We use the multiplication rule to calculate every of the four intersections displayed in the tree diagram

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.6*0}}\rm.8 = 0{\rm{.48}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{1}}}} \right)\rm = 0{\rm{.6*0}}\rm.2 = 0{\rm{.12}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.3*0}}\rm.5 = 0{\rm{.15}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{2}}}} \right)\rm = 0{\rm{.3*0}}\rm.5 = 0{\rm{.15}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}W}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.1*0}}\rm.3 = 0{\rm{.03}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}L}}} \right)\rm &= P\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{*P}}\left( {{\rm{L}}\mid {{\rm{A}}_{\rm{3}}}} \right)\rm = 0{\rm{.1*0}}\rm.7 = 0{\rm{.07}}\end{aligned}\)

From the tree diagram given below, we will calculate everything easier.

(a):

We need to find the probability that the selected review was submitted in Word format

\({\rm{P(W)}}\)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 \({\rm{E }}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\), then for any event \({\rm{B}}\) the following is true\(\begin{aligned}\rm P(B) &= P\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left({{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + \ldots + P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{n}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{n}}}} \right)\\\rm &=\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left({{{\rm{A}}_{\rm{i}}}} \right)\end{aligned}\)

For \({\rm{B = W}}\) we have

\(\begin{aligned}\rm P(W) &= P\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{1}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{2}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ + P}}\left( {{\rm{W}}\mid {{\rm{A}}_{\rm{3}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right)\\\rm &= 0{\rm{.48 + 0}}{\rm{.15 + 0}}{\rm{.03}}\\\rm &= 0{\rm{.66}}\end{aligned}\)

(b):

Bayes' Theorem

For \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\) mutually exclusive events, and \({\rm{E }}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\) the prior probabilities are\(\left. {{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{,i = 1,2, \ldots ,n}}} \right)\). If \({\rm{B}}\)is event for which\({\rm{P(B) > 0}}\), then the posterior probability of \({{\rm{A}}_{\rm{i}}}\)given that \({\rm{B}}\) has occurred

\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}\mid {\rm{B}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}{\rm{{C}B}}} \right)}}{{{\rm{P(B)}}}}{\rm{ = }}\frac{{{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{j}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}} \right)}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{\rm{,}}\;\;\;{\rm{j = 1,2, \ldots ,n}}{\rm{.}}\)

From the theorem, the posterior probabilities are

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.48}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.727;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.15}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.227;}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}\mid {\rm{W}}} \right)\rm &= \frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}{\rm{{C}W}}} \right)}}{{{\rm{P(W)}}}}\\\rm &= \frac{{{\rm{0}}{\rm{.03}}}}{{{\rm{0}}{\rm{.66}}}}\\\rm &= 0{\rm{.046}}\end{aligned}\)