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Question 2: Suppose a new Internet company Mumble.com was to require all employees to take a drug test. Mumble.com can a ord only the inexpensive drug test { the one with a 5% false positive and a 10% false negative rate. (That means that ve percent of those who are not using drugs will incorrectly test positive, and ten percent of those who are actually using drugs will test negative.) Suppose that 10% of those who work for Mumble.com are using the drugs for which Mumble.com is checking. An employee is chosen at random. (a) (3 marks). What is the probability the employee both uses drugs and tests positive? (b) (2 marks). What is the probability the employee does not use drugs but tests positive anyway? (c) (2 marks). What is the probability the employee tests positive? (d) (2 marks). If the employee has tested positive, what is the probability he or she uses drugs?

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(a) The probability she uses drugs is $0.1$. Given she uses drugs, the probability she tests positive is $0.9$. Multiply.

(b) Same idea, a little simpler.

(c) Add the answers to (a) and (b).

(d) Let $D$ be the event she uses drugs, and $P$ be the event she tests positive. We want $\Pr(D|P)$. By the defining formula for conditional probability, we have $$\Pr(D|P)=\frac{\Pr(D\cap P)}{\Pr(P)}.$$ The numerator has been calculated in (a), and the denominator in (c).

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Hint: This question uses both conditional probability and Bayes' Rule.

Also, please show us what work you have done in your attempt to solve this problem. We are not here to answer your homework questions without any effort on your part. Thanks.

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