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

Answer:

Step-by-step 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

=$P(B/A) = \frac{P(AB)}{P(A)} \\=\frac{0.50}{0.79} =0.6329$

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

=$P(B/A') = \frac{P(A'B)}{P(A')} =\frac{P(B)-P(AB)}{1-P(A)} \\=\frac{0.64-0.5}{1-0.79} \\=0.6667$