91Ó°ÊÓ

If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

Given:

\(\begin{array}{r}P\left( {{A_1}} \right) = 0.55\\P\left( {{A_2}} \right) = 0.65\\P\left( {{A_3}} \right) = 0.70\\P\left( {{A_1} \cup {A_2}} \right) = 0.80\\P\left( {{A_2} \cap {A_3}} \right) = 0.40\\P\left( {{A_1} \cup {A_2} \cup {A_3}} \right) = 0.88\end{array}\)

(a) General addition rule for any two events:

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

Solve the rule to \(P(A \cap B)\):

\(P(A \cap {\rm{B}}){\rm{ = P(A) + P(B) - P(A}} \cup {\rm{B)}}\)

Use this equation for \({\rm{A = }}{{\rm{A}}_{\rm{1}}}\)and \({\rm{B = }}{{\rm{A}}_{\rm{2}}}\):

\(\begin{array}{c}P\left( {{A_1} \cap {A_2}} \right) \rm &=& P\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{1}}} \cup {{\rm{A}}_2}} \right)\\ \rm &=& 0{\rm{.55 + 0}}{\rm{.65 - 0}}{\rm{.80}}\\ \rm &=& 0{\rm{.40}}\\ \rm &=& 40\% \end{array}\)

Therefore, The probability that the individual likes both vehicle #1 and vehicle #2 is

\(\begin{array}{l}P\left( {{A_1} \cap {A_2}} \right) &=& 0.40\\ &=& 40\% \end{array}\)

(b) Conditional probability is a term used to describe the chance of something happening:

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

Use the conditional probability definition with \({\rm{B = }}{{\rm{A}}_{\rm{2}}}\)and \({\rm{A = }}{{\rm{A}}_{\rm{3}}}\):

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {{\rm{A}}_{\rm{3}}}} \right) &=& \frac{{P\left( {{A_2} \cap {A_3}} \right)}}{{P\left( {{A_3}} \right)}}\\ \rm &=& \frac{{{\rm{0}}{\rm{.40}}}}{{{\rm{0}}{\rm{.70}}}}\\ &=& \frac{4}{7} \\ &\approx& 0.5714\\ &=& {\rm{57}}{\rm{.14\% }}\end{array}\)

Therefore, Interpretation is \(\begin{array}{l}P\left( {{A_2}\mid {A_3}} \right) &=& \frac{4}{7} \approx 0.5714\\ &=& 57.14\% \end{array}\)

(c) We know that the probability \({\rm{P(B|A)}}\) is equal to the probability \({\rm{P(B)}}\) if \({\rm{A}}\)and \({\rm{B}}\)are independent events (property independent events).

Because \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}\mid {{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}\frac{{\rm{4}}}{{\rm{7}}}\)is not equivalent to \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.65 = }}\frac{{{\rm{13}}}}{{{\rm{20}}}}\), \({{\rm{A}}_{\rm{2}}}\)and \({{\rm{A}}_3}\) are no longer independent.

Note: The fact that \({\rm{P}}\left( {{A_2} \cap {A_3}} \right){\rm{ = 0}}{\rm{.40}}\) is not the product of \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.65}}\) and \({\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.70}}\) also indicates that \({{\rm{A}}_{\rm{2}}}\)and \({{\rm{A}}_{\rm{3}}}\) are not independent (Multiplication rule for independent events).

Therefore, Not independent.

(d) Complement rule:

\(P\left( {{A^c}} \right) = 1 - P(A)\)

Use the complement rule for \({\rm{A = }}{{\rm{A}}_{\rm{1}}}\):

\(P\left( {A_1^c} \right) = 1 - P\left( {{A_1}} \right) = 1 - 0.55 = 0.45\)

In the Venn diagram below, we note that \(\left( {{A_2} \cup {A_3}} \right) \cap A_1^c\)is the union of the three events \({A_1} \cup {A_2} \cup {A_3}\) without \({{\rm{A}}_{\rm{1}}}\).

\(\begin{array}{c}{\rm{P}}\left( {\left( {{A_2} \cup {A_3}} \right) \cap A_1^c} \right)\rm &=& P\left( {{A_1} \cup {A_2} \cup {A_3}} \right) - P\left( {{A_1}} \right)\\ \rm &=& 0{\rm{.88 - 0}}{\rm{.55}}\\ \rm &=& 0{\rm{.33}}\end{array}\)

Definition Conditional probability:

\(P(B\mid A) = \frac{{P(A \cap B)}}{{P(A)}}\)

Use the definition of conditional probability with \(B = {A_2} \cup {A_3}\) and \(A = A_1^c\).

\(\begin{array}{c}{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}} \cup {A_3}\mid A_1^c} \right) &=& \frac{{P\left( {\left( {{A_2} \cup {A_3}} \right) \cap A_1^c} \right)}}{{P\left( {A_1^c} \right)}}\\ \rm &=& \frac{{{\rm{0}}{\rm{.33}}}}{{{\rm{0}}{\rm{.45}}}} \rm &=& \frac{{{\rm{33}}}}{{{\rm{45}}}}\\ \rm &=& \frac{{{\rm{11}}}}{{{\rm{15}}}} &\approx& {\rm{0}}{\rm{.7333}}\\ \rm &=& 73{\rm{.33\% }}\end{array}\)

Therefore, The probability that he/she liked at least one of the other two vehicles is \(\begin{array}{l}P\left( {{A_2} \cup {A_3}\mid A_1^c} \right) &=& \frac{{11}}{{15}} \approx 0.7333\\ &=& 73.33\% \end{array}\)