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

0.7 i had the same question on edge

Explanation:

Answer 2

0.7

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)=(P(B|A)P(B))/(P(B|A)P(B)+P(B|A^c)P(A^c))`

So,

`P(A|B)=(0.7P(B))/(0.7P(B)+0.2(1-P(A)))=(0.7P(B))/(0.7P(B)+0.12)`

and we just need to find P(B)

But

`0.7=P(B|A)=(P(B\cap A))/(P(A))=(P(B\cap A))/(0.4)\Rightarrow P(B\cap A)=0.28`

and

`0.2=P(B|A^c)=(P(B\cap A^c))/(P(A^c))=(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)=(0.7P(B))/(0.7P(B)+0.12)=(0.7*0.4)/(0.7*0.4+0.12)=(0.28)/(0.4)`

`\boxedP(A`