Question

An employee suspected of having used an illegal drug is given two tests that operate independently of each other. Test A has probability 0.9 of being positive if the illegal drug has been used. Test B has probability 0.8 of being positive if the illegal drug has been used. What is the probability that neither test is positive if the illegal drug has been used?

Answer 1

Answer:

0.02 or 2%

Step-by-step explanation:

The probability that test A is negative if the illegal drug has been used (P(A|N)) is 0.1 while the probability that test B is negative if the illegal drug has been used (P(B|N)) is 0.2.

Therefore, the probability that neither test is positive if the illegal drug has been used is:

$P(A \cap B|N) = P(A|N)* P(B|N) = 0.1 *0.2\\P(A \cap B|N) = 0.02$

There is a 0.02 or 2% probability that neither test is positive if the illegal drug has been used.

Answer 2

Answer: 0.02

Step-by-step explanation:

For test A:

The probability of not being positive if the illegal drug has been used is = 1 - 0.9 = 0.1

For test B:

The probability of not being positive if the illegal drug has been used is= 1 - 0.8 = 0.2

Therefore the probability of neither A or B being positive if the illegal drug has been used is

= Pa×Pb = 0.1×0.2 = 0.02