Answer:
p = 0.9942
Step-by-step explanation:
[tex]P(A|B)=p\Rightarrow \frac{P(A\cap B)}{P(B)}=p \Rightarrow P(A\cap B) =pP(B)[/tex]
[tex]P(A^c|B^c)=p\Rightarrow \frac{P(A^c\cap B^c)}{P(B^c)}=p \Rightarrow P(A^c\cap B^c) =pP(B^c)[/tex]
By De Morgan's Law and using the fact that P is a probability
[tex]P(A^c\cap B^c) =pP(B^c)\Rightarrow P((A\cup B)^c)=pP(B^c)\Rightarrow 1-P(A\cup B)=p(1-P(B))[/tex]
As on Saturday night, about 5% of all drivers are known to exceed the limit, P(B) = 0.05 and 1-P(B) = 0.95
We have then,
[tex]P(A\cap B)=0.05p[/tex]
and
[tex]P(A\cup B)=1-0.95p[/tex]
But
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]
hence
P(A)+0.05-0.05p = 1-0.95p
and
P(A) = 0.95-0.9p
If we want P(B|A) = 0.9, then
[tex]\frac{P(B\cap A)}{P(A)}=\frac{P(A\cap B)}{P(A)}=\frac{0.05p}{0.95-0.9p}=0.9[/tex]
and solving for p
0.05p = 0.855 - 081p
0.86p = 0.855
and finally
p = 0.9942
