Question 427435: Suppose a company wants to do random drug testing of its employees and that 2% of the employees use illegal drugs. If the test for detecting drug use is given to people who use illegal drugs, it returns a positive result 99% of the time. When the test is given to people who do not use drugs, it returns a positive result 6% of the time. What is the probability that an employee who receives a positive test result actually does NOT use drugs? (hint..use a tree using "user" and "non-user" instead of "ill" and "healthy") Round to four decimal places.
There are 20 grapefruit in a basket. 9 are sweet, and 11 are sour. If you select 2 at random,what is the probability that they will all be sweet? (Round the answer to four decimal places.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose a company wants to do random drug testing of its employees and that
2% of the employees use illegal drugs.
If the test for detecting drug use is given to people who use illegal drugs, it returns a positive result 99% of the time.
When the test is given to people who do not use drugs, it returns a positive result 6% of the time.
What is the probability that an employee who receives a positive test result actually does NOT use drugs?
(hint..use a tree using "user" and "non-user" instead of "ill" and "healthy") Round to four decimal places.
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P(nonuse|+) = P(non AND +)/[P(+ AND non)+P(+ AND use)
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= [P(+|non)*P(non)/[P(+|non)*P(non)+P(+|use)*P(use)
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= [0.06*0.98]/[same + 0.99*0.02]
= [0.0588]/[0.0588+0.0198]
= 0.7481
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There are 20 grapefruit in a basket. 9 are sweet, and 11 are sour. If you select 2 at random,what is the probability that they will all be sweet? (Round the answer to four decimal places.)
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P(2 sweet) = 9C2/20C2 = 0.1895
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Cheers,
Stan H.
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