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:0.7Step-by-step explanation:Let A and B be the eventsA = the motorist encountered a red light at the 1st intersectionB = the motorist encountered a red light at the 2nd intersectionThen we have[tex] P(A) = 0.4 [/tex][tex] P(B|A) = 0.7 \; (P(B) \; given \; A \; occurred) [/tex][tex] P(B|A^c) = 0.2 \; ( where\; A^c \; is \; the \; complement \; of \; A)[/tex]We want to find P(A|B), the probability that A occurred given that B occurred.Using Bayes' Theorem we have[tex]P(A|B)=\frac{P(B|A)P(B)}{P(B|A)P(B)+P(B|A^c)P(A^c)}[/tex]So,  [tex]P(A|B)=\frac{0.7P(B)}{0.7P(B)+0.2(1-P(A))}=\frac{0.7P(B)}{0.7P(B)+0.12}[/tex]and we just need to find P(B)But[tex]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[/tex]and  [tex]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[/tex]Since  [tex](A\cap B)\cap (A\cap B^c)=\emptyset \;and\;(A\cap B)\cup (A\cap B^c)=B[/tex]We have[tex]P(B)=P(A\cap B)+P(A\cap B^c)=0.12+0.28=0.4[/tex]and finally,[tex]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}[/tex][tex]\boxed{P(A|B)=0.7}[/tex]