问题及解答

A certain sports car comes equipped with either an automatic or a manual transmission, and the car is available in one of four colors. Relevant probabilities for various combinations of transmission type and color are given in the accompanying table.

Let $A=\{\text{automatic transmission}\}$, $B=\{\text{black}\}$, and $C=\{\text{white}\}$.

1.   Calculate $P(A)$, $P(B)$ and $P(A\cap B)$.
2.   Calculate both $P(A|B)$ and $P(B|A)$, and explain in context what each of these probabilities represents.
3.   Calculate and interpret $P(A|C)$ and $P(A|C^c)$.
    

References:

Book P83, Exer 45.

(1) 

\[
\begin{split}
P(A)&=P(A\cap\{\text{White}\})+P(A\cap\{\text{Blue}\})+P(A\cap\{\text{Black}\})+P(A\cap\{\text{Red}\})\\
&=0.15+0.10+0.10+0.10\\
&=0.45
\end{split}
\]

\[
P(B)=P(\{\text{Black}\})=P(A\cap\{\text{Black}\})+P(M\cap\{\text{Black}\})=0.10+0.15=0.25
\]

\[
P(A\cap B)=P(A\cap\{\text{Black}\})=0.10
\]

(2)

\[
P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.10}{0.10+0.15}=\frac{2}{5}=0.4
\]

\[
P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{0.10}{0.45}=\frac{2}{9}\approx 0.222222
\]

(3)

\[
P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{0.15}{0.15+0.15}=\frac{1}{2}=0.5
\]

\[
P(A|C^c)=\frac{P(A\cap C^c)}{P(C^c)}=\frac{P(A\cap\{\text{Blue, Black, Red}\})}{1-P(C)}=\frac{0.10+0.10+0.10}{1-(0.15+0.15)}=\frac{3}{7}\approx 0.428571
\]