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.)

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 :

By using the formula for Conditional Probability is

=0.4294

=0.43

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)