A pharmaceutical company has developed a test for a rare disease that is present in 0.5% of the population. The test is 98% accurate in determining a positive result, and the chance of a false positive is 7%. What is the probability that someone who tests positive actually has the disease? (Round your answer to two decimal places.)

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jchoubay1 jchoubay1 · Answer 1 · 2020-04-15T16:41:49+02:00

Answer:

The probability that someone who tests positive actually has the disease is P(S|T)=0.43

Step-by-step explanation:

A pharmaceutical company has developed a test for a rare disease that is present in 0.5% of the population.

The test is 98% a positive result and the chance of a false positive is 7%

Let S be the event that the rare disease present

P(S)=0.05 and

Let T be the event that the test is positive

P(T)=0.98

Then we have to find the Conditional probability [tex]P(S|T)[/tex] :

By using the formula for Conditional Probability is [tex]P(A|B)=\frac{P(A and B)}{P(B)}[/tex]

[tex]P(S|T)=\frac{P(S and T)}{P(S)}[/tex]

[tex]=\frac{(0.05)(0.98)}{(0.05)(0.98)+(0.93)(0.07)}[/tex]

[tex]=\frac{0.049}{0.049+0.0651}[/tex]

[tex]=\frac{0.049}{0.1141}[/tex]

=0.4294

=0.43