Mathematics, 13.09.2019 19:30 marieb39

Social sciences alcohol abstinence the harvard school of public health completed a study on alcohol consumption on college campuses. they concluded that 20.7% of women attending all-women colleges abstained from alcohol, compared to e6% of women attending coeducational colleges. approximately 4.7% of women college students attend all-women schools. source: harvard school of public health. (a) what is the 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?

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Answer from: zoeatlowapple

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?

$P = P_{1} + P_{2}$

$P_{1}$ 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:

$P_{1} = 0.047*0.207 = 0.009729$

$P_{2}$ 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:

$P_{2} = 0.953*0.06 = 0.05718$

So, the probability of a randomly selected female student abstaining from alcohol is:

$P = P_{1} + P_{2} = 0.009729 + 0.05718 = 0.0669$

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:

$P = \frac{P(B).P(A/B)}{P(A)}$

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:

P(B) is the probability of a woman from a coeducational college being chosen. So P(B) = 0.953

P(A/B) is the probability of a woman abstaining from alcohol, given that she attends a coeducational college. So P(A/B) = 0.06

P(A) is the probability of a woman abstaining from alcohol. From a), P(A) = 0.0669

So, the probability that a randomly selected female student attends a coeducational college, given that she abstains from alcohol is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.953)*(0.06)}{(0.0669)} = 0.8287$

If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

Answer from: Quest

4400

step-by-step explanation:

Answer from: Quest

i can't solve unless i know what the line is

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