Question

In a study of honeymoon vacations for newlyweds in the United States, it was determined that 79% take place outside of the country, 64% last longer than 7 days, and 50% are both outside the country and last longer than 7 days. Find the following probabilities: (a) What is the probability that a honeymoon vacation takes place outside of the country or lasts longer than 7 days? (b) What is the probability that a honeymoon vacation lasts longer than 7 days given that it takes place outside of the country? (c) What is the probability that a honeymoon vacation takes place outside of the country given that it does not last longer than 7 days?

Answer 1

Explanation:

Given that in a study of honeymoon vacations for newlyweds in the United States, it was determined that 79% take place outside of the country, 64% last longer than 7 days, and 50% are both outside the country and last longer than 7 days.

Let A - honey moon outside country

B - last longer than 7 days

Then
`P(A) = 0.79\\P(B) = 0.64\\P(AB) = 0.50`

a) the probability that a honeymoon vacation takes place outside of the country or lasts longer than 7 days

`=P(AUB) = P(A)+P(B)-P(AB)\\= 0.79+0.64-0.50\\=0.93`

b) the probability that a honeymoon vacation lasts longer than 7 days given that it takes place outside of the country

=
`P(B/A) = (P(AB))/(P(A)) \\=(0.50)/(0.79) =0.6329`

c) the probability that a honeymoon vacation takes place outside of the country given that it does not last longer than 7 days

=
`P(B/A') = (P(A'B))/(P(A')) =(P(B)-P(AB))/(1-P(A)) \\=(0.64-0.5)/(1-0.79) \\=0.6667`