im having trouble with this math question

Question

Suppose a lab is developing a robust new test to look for diabetes in children. If a child actually has diabetes, the probability that the test will be positive is 82%, indicating they have diabetes. If the child does not have diabetes, the probability that the test will be positive is 10%. Suppose 5.3% of all children actually have diabetes. (Round your answer to 3 decimal places, if needed.)

(a) What is the probability that a randomly chosen child has diabetes and tested positive?

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(b) What is the probability that a randomly chosen child tested positive?

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(c) Given that a test comes up positive, what is the probability the child has diabetes?

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(d) What is the probability that a randomly chosen child does not have diabetes and tested negative?

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(e) What is the probability that a randomly chosen child tested negative?

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(f) Given that a test comes up negative, what is the probability the child does not have diabetes?

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Answer 1

P(P/D) = .82= 82% probability of testing positive if the child has the disease

P(P/~D) = .10 = probability of testing positive if the child does not have the disease

P(D)= .053= 5.3% probability of having the disease

P(~D) =1-P(D) =1-.053 = .947= 94.7% probability of not having the disease

P(P) = .529 = 52.9% probability of testing positive

P(~P) = .471 = 47.1% probability of testing negative

Use Bayes' Theorem

P(P/D)P(D) = P(D/P)P(P)

P(D/P) = P(P/D)P(D)/P(P)

P(D/P) = P(P/D)P(D)/[P(P/D)P(D) + P(P/~D)P(~D)]

=.82(.053)/[.82(.053)+(.1)(.947)]

= .4346/.529

=.8213

= 82.13% probability you have the disease if you tested positive