Question

According to Exercise 16, the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to bith countries is 0.04. What'

Probability
ANSWERED
asked 2020-11-08
According to Exercise 16, the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to bith countries is 0.04.
What's the probability that someone who has traveled to Mexico has visited Canada too?
Are traveling to Mexico and to Canada disjoint events?
Are traveling to Mexico and to Canada independent events?
Explain.

Answers (1)

2020-11-09
Let C be the event that someone has visited Canada, and M be the event that someone has visited Mexico.
We need to find \(\displaystyle{P}{\left({C}{\mid}{M}\right)}\). By definition, \(\displaystyle{P}{\left({C}{\mid}{M}\right)}=\frac{{{P}{\left({C}⋂{M}\right)}}}{{P}}{\left({M}\right)}=\frac{{0.04}}{{0.09}}=\frac{{4}}{{9}}\sim{44.44}\%\)
If they were disjoint, we would have \(\displaystyle{P}{\left({C}⋂{M}\right)}={0}\frac{=}{{0.04}}\)
Therefore, they are not disjoint.
They are independent if and only if \(\displaystyle{P}{\left({C}⋂{M}\right)}={P}{\left({C}\right)}{P}{\left({M}\right)}\)
However, \(\displaystyle{P}{\left({C}⋂{M}\right)}={0.04}\)
\(\displaystyle{P}{\left({C}\right)}{P}{\left({M}\right)}={0.18}\cdot{0.09}={0.0162}\)
Therefore, \(\displaystyle{P}{\left({C}⋂{M}\right)}\frac{=}{{P}}{\left({C}\right)}{P}{\left({M}\right)}\)
so C and M are not independent.
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