Assume that 0.4% of the population has a condition that is not detectible by sim

Question

Assume that 0.4% of the population has a condition that is not detectible by simple external observation. A diagnostic test is available for this condition, but, like most tests, it is not perfect. The test correctly diagnoses, with a positive result, those with the condition 99.7% of the time. The test correctly identifies, with a negative result, those without the condition 98.5% of the time.

1. Let the event C 1 represent the presence of the condition and C 2 represent the absence of the condition, and let event T represent a positive test result (meaning the test indicated, either correctly or incorrectly, that a person has the condition) and T C represent a negative test result (meaning the test indicated, either correctly or incorrectly, that a person does not have the condition). List symbolically the information given in the statement above.

2.If a randomly selected member of the population tests positive for the condition, what is the probability that the person has the condition? (Report orround your answer to 3 decimal places.)

Find:

Formula(s):

Computations:

Answer:

3. If a randomly selected member of the population tests positive for the condition, what is the probability that the person does not have the condition?

4.If a randomly selected member of the population tests negative for the condition, what is the probability that the person has the condition? (Report or round your answer to 6 decimal

places.)

do question 3 and 4 please

Explanation / Answer

1)

P(C1) = 0.4% = 0.004

P(T|C1) = 0.997

P(TC|C2) = 0.985

2)

P(C1|T) = P(C1 and T)/P(T)

= P(C1) P(T|C1) /P(T)

= 0.004 * 0.997/(0.004 * 0.997 + 0.996 * (1- 0.985))

=0.210693153

3)

P(C2|T) = P(C2 and T)/P(T)

= 0.996 * (1- 0.985)/((0.004 * 0.997 + 0.996 * (1- 0.985))

=0.78930684699

4)

P(C1|Tc)

= P(C1 and Tc) /P(Tc)

= (0.004 * (1 - 0.997))/(0.004 * (1-0.997) + 0.996 * 0.985)

= 0.00001223151

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