# The probability that a vehicle entering the Luray Caverns has Canadian

The probability that a vehicle entering the Luray Caverns has Canadian license plates is 0.19; the probability that it is a camper is 0.38; and the probability that it is a camper with Canadian license plates is 0.13. What is the probability that
(b) a vehicle with Canadian license plates entering the Luray Caverns is not a camper?
(c) a vehicle entering the Luray Caverns does not have Canadian plates or is not a camper?

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Step 1
Let the event that a vehicle entering the Luray Caverns has Canadian license plates is "A".
Let the event that a vehicle entering the Luray Caverns is a camper is denoted by "B".
Let the event that a vehicle entering the Luray Caverns is a camper with Canadian license plates is denoted by $$\displaystyle{\left({A}\cap{B}\right)}$$.
So,
$$\displaystyle{P}{\left({A}\right)}={0.19}$$
$$\displaystyle{P}{\left({B}\right)}={0.38}$$
$$\displaystyle{P}{\left({A}\cap{B}\right)}={0.13}$$
Step 2
(a) We find the probability that a camper entering the Luray Caverns has Canadian license plates as,
$$\displaystyle{\left({A}{\mid}{B}\right)}={\frac{{{P}{\left({A}\cap{B}\right)}}}{{{P}{\left({B}\right)}}}}$$
$$\displaystyle={\frac{{{0.13}}}{{{0.38}}}}$$
$$\displaystyle={0.3421}$$
Hence, required probability is 0.3421.
(b) We find the probability that a vehicle with Canadian license plates entering the Luray Caverns is not a camper as,
$$\displaystyle{P}{\left({B}^{{{C}}}{\mid}{A}\right)}={\frac{{{P}{\left({B}^{{{C}}}\cap{A}\right)}}}{{{P}{\left({A}\right)}}}}$$
$$\displaystyle={\frac{{{P}{\left({A}\right)}-{P}{\left({A}\cap{B}\right)}}}{{{P}{\left({A}\right)}}}}$$ [By formula, $$\displaystyle{P}{\left({A}\right)}={P}{\left({A}\cap{B}\right)}+{P}{\left({B}^{{{C}}}\cap{A}\right)}$$]
$$\displaystyle={\frac{{{0.19}-{0.13}}}{{{0.19}}}}$$
$$\displaystyle={0.3158}$$
Hence, required probability is 0.3158.
(c) We find the probability that a vehicle entering the Luray Caverns does not have Canadian plates or is not a camper is,
$$\displaystyle{P}{\left({A}^{{{C}}}\cup{B}^{{{C}}}\right)}={P}{\left({A}\cap{B}\right)}^{{{C}}}$$ [By formula, $$\displaystyle{P}{\left({A}^{{{C}}}\cup{B}^{{{C}}}\right)}={P}{\left({A}\cap{B}\right)}^{{{C}}}$$]
$$\displaystyle={1}-{P}{\left({A}\cap{B}\right)}$$
$$\displaystyle={1}-{0.13}$$
$$\displaystyle={0.87}$$
Hence, required probability is 0.87.
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