Health, 21.09.2019 05:30 moniquejackson2

Suppose that 3% of all adults suffer from diabetes and that 28% of all adults are obese. suppose also that 2% of all adults both are obese and suffer from diabetes.
(a) find the probability that a randomly chosen adult who is obese also suffers from diabetes. round your answer to decimal places.
(b) find the probability that a randomly chosen adult is obese, given that he or she suffers from diabetes. round your answer to decimal places.

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Answer from: tommyaberman

a) There is a 7.14% p(a) probability that a randomly chosen adult who is obese also suffers from diabetes.

b) There is a 66.64% probability of the adult being obese, given that he suffers from diabetes.

Explanation:

The problem states that:

28% of all adults are obese. This means that there is a 28% probability of a randomly chosen adult being obese.

3% of all adults suffer from diabetes. This means that there is a 3% probability of a randomly chosen adult suffering from diabetes.

2% of all adults both are obese and suffer from diabetes. This means that there is a 2% probability of a randomly chosen adult both being obese and suffering from diabeters.

(a) Find the probability that a randomly chosen adult who is obese also suffers from diabetes. Round your answer to decimal places:

28% of all adults are obese and 2% of all adults are both obese and suffer from diabetes.

So the probability of a randomly chosen adult who is obese suffering from diabetes is:

$P = \frac{2}{28} = 0.0714$

There is a 7.14% probability that a randomly chosen adult who is obese also suffers from diabetes.

(b) Find the probability that a randomly chosen adult is obese, given that he or she suffers from diabetes.

What is the probability of B happening, knowing that A has happened.

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.

So:

What is the probability of the adult being obese, given that he suffers from diabetes.

P(B) is the probability of the adult being obese. So P(B) = 0.28

P(A/B) is the probability of the adult suffering from diabetes, given that he is obese. We found in a) that 7.14% of the adults who are obese also suffer from diabetes. So:

P(A/B) = 0.0714

P(A) is the probability of the adult suffering from diabetes. So:

P(A) = 0.03

Finally

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.28*0.0714}{0.03} = 0.6664$

There is a 66.64% probability of the adult being obese, given that he suffers from diabetes.

Answer from: Quest

hmmm

true, i think.

i checked and it seems the answer is, true.

Answer from: Quest

well say a good goal is to do 150 sit ups in a week! or 90 pull ups a day. even 60 push ups in a month these are all fair goals!

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