Question

A certain disease has an incidence rate of 0.8%. If the false negative rate is 5% and the false positive rate is

2%, compute the probability that a person who tests positive actually has the disease.
Give your answer accurate to at least 3 decimal places
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Answer 1

The probability that a person who tests positive actually has the disease is approximately 0.277.

Step-by-step explanation:

To find the probability that a person who tests positive actually has the disease, we can use Bayes' theorem.

Let D be the event that a person has the disease and P be the event that a person tests positive.

Given:

• Incidence rate of the disease = 0.8% = 0.008
• False negative rate = 5% = 0.05
• False positive rate = 2% = 0.02

The probability that a person has the disease and tests positive is given by:

P(D|P) = P(P|D) * P(D) / P(P)

To find P(P), we can use the law of total probability:

P(P) = P(P|D) * P(D) + P(P|~D) * P(~D)

Since the false negative rate is given, we can calculate P(P|D) as 1 - 0.05 = 0.95.

Also, P(D) = 0.008 (given incidence rate) and P(~D) = 1 - P(D) = 0.992.

Now, substituting the values:

P(P) = 0.95 * 0.008 + 0.02 * 0.992

P(P) = 0.0076 + 0.01984 = 0.02744

Finally, substituting the values in Bayes' theorem:

P(D|P) = (0.95 * 0.008) / 0.02744

P(D|P) = 0.0076 / 0.02744

= 0.27741935

Therefore, the probability that a person who tests positive actually has the disease is approximately 0.277 (rounded to 3 decimal places).