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 84%, indicating they have diabetes.

Question

 If the child does not have diabetes, the probability that the test will be positive is 10%. Suppose 5.6% 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(+/D) = 84%=.84

P(-/D)= 16%= .16 false negative probability

P(+/~ D)=.10 =10% false positive probability

= Type I and II errors

P(D)= 5.6%= 0.056

P(~D) = 94.4% = 0.944

P(a^b)=Pa(Pb/a) for dependent events, P(a^b)=PaPb for independent events

use Bayes' Theorem to find missing probabilities

P(a/b)Pb = P(b/a)Pa

a=+, b=D

P(+/D)PD = P(D/+)P+

a) P(D^+)= .86(.056)=0.04816= 4.816% chance both diabetic and tests positive

d) P(-^~D)= P(~D)P(-/~D)=.944(P(-/~D)=P(~D/-)P(-)= both not diabetic and tests negative