Question

A motorist is driving eastbound at the posted maximum speed along a stretch of University Parkway in Baltimore. The probability that the motorist approaching a red light at the San Martin/39th St intersection is 0.4. The probability that the motorist encounters a red light again at the next intersection (Canterbury Rd) is 0.7. If the motorist does not encounter a red light at the first intersection (San Martin/39th), then the probability the motorist encounters a red light at the second intersection is 0.2. If in a certain instance the motorist is observed to have encountered a red light at the Canterbury Rd intersection, what is the probability the motorist encountered a red light at the San Martin/39th intersection

Answer 1

Answer:

0.7

Step-by-step explanation:

Let A and B be the events

A = the motorist encountered a red light at the 1st intersection

B = the motorist encountered a red light at the 2nd intersection

Then we have

$P(A) = 0.4$

$P(B|A) = 0.7 \; (P(B) \; given \; A \; occurred)$

$P(B|A^c) = 0.2 \; ( where\; A^c \; is \; the \; complement \; of \; A)$

We want to find P(A|B), the probability that A occurred given that B occurred.

Using Bayes' Theorem we have

$P(A|B)=\frac{P(B|A)P(B)}{P(B|A)P(B)+P(B|A^c)P(A^c)}$

So,  

$P(A|B)=\frac{0.7P(B)}{0.7P(B)+0.2(1-P(A))}=\frac{0.7P(B)}{0.7P(B)+0.12}$

and we just need to find P(B)

But

$0.7=P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B\cap A)}{0.4}\Rightarrow P(B\cap A)=0.28$

and  

$0.2=P(B|A^c)=\frac{P(B\cap A^c)}{P(A^c)}=\frac{P(B\cap A^c)}{0.6}\Rightarrow P(B\cap A^c)=0.12$

Since  

$(A\cap B)\cap (A\cap B^c)=\emptyset \;and\;(A\cap B)\cup (A\cap B^c)=B$

We have

$P(B)=P(A\cap B)+P(A\cap B^c)=0.12+0.28=0.4$

and finally,

$P(A|B)=\frac{0.7P(B)}{0.7P(B)+0.12}=\frac{0.7*0.4}{0.7*0.4+0.12}=\frac{0.28}{0.4}$

$\boxed{P(A|B)=0.7}$