A doctor gives a patient a test for a particular virus. Before the results of th

Question

A doctor gives a patient a test for a particular virus. Before the results of the test, it is believed that .15% of the population has this virus. Based on experience, the doctor knows that, in 98% of the cases in which the virus is present, the test is positive; and in 93% of the cases in which the virus is not present, the test is negative. Use a hypothetical 100,000 table to find the following.

A). A patient was given the test and it came back positive. What is the chance that the virus is present? Please set up the table.

B). If the person was tested negative, what is the chance the patient is healthy (i.e., the virus is not present)?

C). Without doing any calculation, would you expect your answer in part (a) to increase or decrease if it is believed that 1 in every 1250 people has the virus? Explain your answer briefly.

Explanation / Answer

P(V) = 0.15 and P(V') = 0.85
P(P|V) = 0.98 and P(P'|V) = 0.02
P(P'|V') = 0.93 and P(P|V') = 0.07

A)
P(V|P) = P(P|V)*P(V) / (P(P|V)*P(V) + P(P|V')*P(V'))

P(V|P) = 0.98*0.15 / (0.98*0.15 + 0.07*0.85)

P(V|P) = 0.7119

B)
P(V'|P') = P(P'|V')*P(V') / (P(P'|V')*P(V') + P(P'|V)*P(V))

P(V'|P') = 0.93*0.85 / (0.93*0.85 + 0.02*0.15)

P(V'|P') = 0.9962

C)
1/1250 = 0.0008

This will decrease the probability in part A. This is because the denominator of the expression will increase and numerator will decrease.

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