问题及解答

For customers purchasing a full set of tires at a particular tire store, consider the events
\[
A=\{\text{tires purchased were made in the United States}\}
\]
\[
B=\{\text{purchaser has tires balanced immediately}\}
\]
\[
C=\{\text{purchaser requests front-end alignment}\}
\]
along with $A^c$, $B^c$, and $C^c$. Assume the following unconditional and conditional probabilities:
\[
P(A)=.75\qquad P(B|A)=.9\qquad P(B|A^c)=.8
\]
\[
P(C|A\cap B)=.8\qquad P(C|A\cap B^c)=.6
\]
\[
P(C|A^c\cap B)=.7\qquad P(C|A^c\cap B^c)=.3
\]

1.  Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.
2.  Compute $P(A\cap B\cap C)$.
3.  Compute $P(B\cap C)$.
4.  Compute $P(C)$.
5.  Compute $P(A|B\cap C)$, the probability of a purchase of U.S. tires given that both balancing and an alignment were requested.
    

 

1. 

 

---

2.

\[
P(B\cap A)=P(B|A)\cdot P(A)=0.9\times 0.75=0.675
\]

Since

\[
P(C|A\cap B)=\frac{P(C\cap A\cap B)}{P(A\cap B)},
\]

we have

\[
P(A\cap B\cap C)=P(C|A\cap B)\cdot P(A\cap B)=0.8\times 0.675=0.54
\]

---

3.

\[
B\cap C=(A\cap (B\cap C))\cup (A^c\cap(B\cap C))
\]

Hence, 

\[
P(B\cap C)=P(A\cap B\cap C)+P(A^c\cap B\cap C)
\]

On the other hand, 

\[
P(C|A^c\cap B)=\frac{P(C\cap A^c\cap B)}{P(A^c\cap B)},
\]

we have

\[
\begin{split}
P(A^c\cap B\cap C)&=P(C|A^c\cap B)\cdot P(A^c\cap B)\\
&=0.7\times P(B\cap A^c)=0.7\times P(A^c)\cdot P(B|A^c)\\
&=0.7\times (1-0.75)\times 0.8\\
&=0.14
\end{split}
\]

Hence, $P(B\cap C)=0.54+0.14=0.68$.

---

4.

\[
0.8=P(C|A\cap B)=\frac{P(C\cap A\cap B)}{P(A\cap B)}=\frac{0.54}{P(A\cap B)}
\]

It infers that $P(A\cap B)=\frac{0.54}{0.8}=0.675$.

 

\[
P(C|A^c\cap B^c)=\frac{P(C\cap A^c\cap B^c)}{P(A^c\cap B^c)}
\]

It infers that

\[
\begin{split}
P(C\cap A^c\cap B^c)&=P(C|A^c\cap B)\cdot P(A^c\cap B^c)=0.3\times P((A\cup B)^c)\\
&=0.3\times (1-P(A\cup B))
\end{split}
\]

From the condition $P(C|A^c\cap B)=0.7$, we have

\[
0.7=P(C|A^c\cap B)=\frac{C\cap A^c\cap B}{P(A^c\cap B)}=\frac{0.14}{P(A^c\cap B)},
\]

then, $P(A^c\cap B)=\frac{0.14}{0.7}=0.2$. Thus, $P(A\cup B)=P(A)+P(B\cap A^c)=0.75+0.2=0.95$.

Therefore,

\[
P(C\cap A^c\cap B^c)=0.3\times (1-P(A\cup B))=0.3\times(1-0.95)=0.015.
\]

At the last,

\[
\begin{split}
P(C)&=P(B\cap C)+P(B^c\cap C)\\
&=\bigl[P(B\cap C\cap A)+P(B\cap C\cap A^c)\bigr]+\bigl[P(A\cap B^c\cap C)+P(A^c\cap B^c\cap C)\bigr]\\
&=\bigl[0.54+0.14\bigr]+\bigl[0.045+0.015\bigr]\\
&=0.74
\end{split}
\]

 

---

5.

\[
P(A|B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}=\frac{0.54}{0.54+0.14}\approx 0.794118
\]

Another method by using Venn diagram to solve this question. It will convert the question to solve a linear system.

 

Suppose the probabilities of the events devided by the three regions $A,B,C$ are $x,y,z,u,v,w,t$ as above.

For example let $t=P(A\cap B\cap C)$.

Then, by assumption, we have

\[
\begin{aligned}
P(A)=0.75 &\Rightarrow x+v+u+t=0.75\\
P(B|A)=0.9&\Rightarrow\frac{u+t}{P(A)}=0.9\Rightarrow u+t=0.75\times 0.9=0.675\\
P(B|A^c)=0.8&\Rightarrow\frac{P(B\cap A^c)}{P(A^c)}=0.8\Rightarrow\frac{y+w}{1-P(A)}=0.8\Rightarrow y+w=(1-0.75)\times 0.8=0.2\\
P(C|A\cap B)=0.8&\Rightarrow\frac{P(C\cap A\cap B)}{P(A\cap B)}=0.8\Rightarrow\frac{t}{u+t}=0.8\Rightarrow t=4u\\
P(C|A\cap B^c)=0.6&\Rightarrow\frac{P(C\cap A\cap B^c)}{P(A\cap B^c)}=0.6\Rightarrow\frac{v}{x+v}=0.6\Rightarrow v=\frac{3}{2}x\\
P(C|A^c\cap B)=0.7&\Rightarrow\frac{P(C\cap A^c\cap B)}{P(A^c\cap B)}=0.7\Rightarrow\frac{w}{y+w}=0.7\Rightarrow w=\frac{7}{3}y\\
P(C|A^c\cap B^c)=0.3&\Rightarrow\frac{P(C\cap A^c\cap B^c)}{P(A^c\cap B^c)}=0.3\Rightarrow\frac{z}{1-((x+v+u+t)+(y+w))}=0.3\\
&\Rightarrow\frac{z}{1-(0.75+0.2)}=0.3\Rightarrow z=0.015
\end{aligned}
\]

We take a summary,

\[
\begin{cases}
x+v+u+t&=0.75,\\
u+t&=0.675,\\
y+w&=0.2,\\
t&=4u,\\
v&=\frac{3}{2}x,\\
w&=\frac{7}{3}y,\\
z&=0.015.
\end{cases}
\]

Simplify it,

\[
\begin{cases}
x+v=0.075\Rightarrow x+\frac{3}{2}x=0.075\Rightarrow x=0.03\\
u+t=0.675\Rightarrow u+4u=0.675\Rightarrow u=0.135\\
y+\frac{7}{3}y=0.2\Rightarrow y=0.06
\end{cases}
\]

Thus, we get 

\[
\begin{aligned}
t&=4u=4\times 0.135=0.54,\\
v&=\frac{3}{2}x=\frac{3}{2}\times 0.03=0.045,\\
w&=\frac{7}{3}y=\frac{7}{3}\times 0.06=0.14.
\end{aligned}
\]

Back to the computation of the probabilities,

\[
\begin{aligned}
P(A\cap B\cap C)&=t=0.54,\\
P(B\cap C)&=t+w=0.54+0.14=0.68,\\
P(C)&=v+t+w+z=0.045+0.54+0.14+0.015=0.74,\\
P(A|B\cap C)&=\frac{P(A\cap B\cap C)}{P(B\cap C)}=\frac{t}{t+w}=\frac{0.54}{0.54+0.14}\approx 0.794118
\end{aligned}
\]