Question

The probability that a vehicle entering the Luray
Caverns has Canadian license plates is 0.12; the
probability that it is a camper is 0.28; and the probability
that it is a camper with Canadian license plates
is 0.09. What is the probability that
(a) a camper entering the Luray Caverns has Canadian
license plates?
(b) a vehicle with Canadian license plates entering the
Luray Caverns is a camper?
(c) a vehicle entering the Luray Caverns does not have
Canadian plates or is not a camper?

Answer 1

So you have
\[P(A)=P(\text{Canadian})=0.12\\
P(B)=P(\text{camper})=0.28\\
P(A\cap B)=P(\text{Canadian}\cap\text{camper})=0.09\]
The first question wants to find the probability of a Canadian driver given that he is a camper:
\[P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{0.09}{0.28}\]
The second question asks for the probability of a camper given that he is Canadian:
\[P(B|A)=\frac{P(B\cap A)}{P(A)}=\cdots\]
The third wants the probability of "not a camper" OR "not Canadian," which translates to
\[P(A^C\cup B^C)\]
where \(A^C\) is the complement of \(A\).

You can use deMorgan's laws to rewrite it:
\[A^C\cup B^C=(A\cap B)^C\]
And since this event and *its* complement (which is just \(A\cap B\)) are mutually exclusive, the probability of one occurring is given by the following relationships:
\[P(A\cap B)=1-P\left((A\cap B)^C\right)\\
\color{red}{P\left((A\cap B)^C\right)}=1-P(A\cap B)\]
Solve for the red part.

Answer 2

Thank you so much! =)

Answer 3

yw!