Answer :
Answer:
a) There is a 6.69% probability that a randomly selected female student abstains from alcohol.
b) If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.
Step-by-step explanation:
This is a probability problem:
We have these following probabilities:
-20.7% of a woman attending an all-women college abstaining from alcohol.
-6% of a woman attending a coeducational college abstaining from alcohol.
-4.7% of a woman attending an all-women college
- 100%-4.7% = 95.3% of a woman attending a coeducational college.
(a) What is the probability that a randomly selected female student abstains from alcohol?
[tex]P = P_{1} + P_{2}[/tex]
[tex]P_{1}[/tex] is the probability of a woman attending an all-women college being chosen and abstaining from alcohol. There is a 0.047 probability of a woman attending an all-women college being chosen and a 0.207 probability that she abstain from alcohol. So:
[tex]P_{1} = 0.047*0.207 = 0.009729[/tex]
[tex]P_{2}[/tex] is the probability of a woman attending a coeducational college being chosen and abstaining from alcohol. There is a 0.953 probability of a woman attending a coeducational college being chosen and a 0.06 probability that she abstain from alcohol. So:
[tex]P_{2} = 0.953*0.06 = 0.05718[/tex]
So, the probability of a randomly selected female student abstaining from alcohol is:
[tex]P = P_{1} + P_{2} = 0.009729 + 0.05718 = 0.0669[/tex]
There is a 6.69% probability that a randomly selected female student abstains from alcohol.
(b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coedücational colege?
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened.
Here:
What is the probability of a woman attending a coeducational college, knowing that she abstains from alcohol.
It can be calculated by the following formula:
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
We have the following probabilities:
[tex]P(B)[/tex] is the probability of a woman from a coeducational college being chosen. So [tex]P(B) = 0.953[/tex]
[tex]P(A/B)[/tex] is the probability of a woman abstaining from alcohol, given that she attends a coeducational college. So [tex]P(A/B) = 0.06[/tex]
[tex]P(A)[/tex] is the probability of a woman abstaining from alcohol. From a), [tex]P(A) = 0.0669[/tex]
So, the probability that a randomly selected female student attends a coeducational college, given that she abstains from alcohol is:
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.953)*(0.06)}{(0.0669)} = 0.8287[/tex]
If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.