Conditional Probability: 6. Employers often require their employees to be subjected to random drug testing. Suppose that 5% of employees use a certain type of drug. Drug testing accuracy varies. From years of data collection it is known that those who use this type of drug have a 94% chance of testing positive. It is also known that 3% of non- drug users (of the certain type of drug) still test positive. a.) Define all 4 of the marginal events b.) With proper notation write out all of the given probabilities c.) Make a tree diagram of all probabilities in proper notation. d.) Fill out a probability cross tabulation. e.) If the employee tests positive for a certain drug, what's the probability that the employee uses the drug? f.) If the employee tests positive for a certain drug, what's the probability that the employee does NOT use the drug?

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**Conditional Probability:**

6. Employers often require their employees to be subjected to random drug testing. Suppose that 5% of employees use a certain type of drug. Drug testing accuracy varies; from years of data collection, it is known that those who use this type of drug have a 92% chance of testing positive. It is also known that 3% of non-drug users (of the certain type of drug) still test positive.

   a) Define all 4 of the marginal events.

   b) With proper notation, write out all of the given probabilities.

   c) Make tree diagrams of all probabilities.

   d) Fill out the probability table.

   e) If an employee tests positive for a certain drug, what’s the probability that the employee uses the drug?

   f) If the employee tests positive for a certain drug, what’s the probability that the employee does NOT use the drug?

7. Each day, 10% of the email sent to my address is spam. If an email is actually spam, the spam filter will detect it with 99.5% accuracy. If an email is not spam, the spam filter will mistakenly identify it as spam with 0.5% probability. An email was marked as sp...

**Conditional Probability:**

a) The species is an amphibian.

b) The species is a reptile, endangered and not foreign.

c) The species is not endangered but is foreign.

d) The species is a reptile, endangered and foreign.

g) Assume the species is endangered, what’s the probability that it is a bird and that it’s from the U.S.?
Transcribed Image Text:**Conditional Probability:** 6. Employers often require their employees to be subjected to random drug testing. Suppose that 5% of employees use a certain type of drug. Drug testing accuracy varies; from years of data collection, it is known that those who use this type of drug have a 92% chance of testing positive. It is also known that 3% of non-drug users (of the certain type of drug) still test positive. a) Define all 4 of the marginal events. b) With proper notation, write out all of the given probabilities. c) Make tree diagrams of all probabilities. d) Fill out the probability table. e) If an employee tests positive for a certain drug, what’s the probability that the employee uses the drug? f) If the employee tests positive for a certain drug, what’s the probability that the employee does NOT use the drug? 7. Each day, 10% of the email sent to my address is spam. If an email is actually spam, the spam filter will detect it with 99.5% accuracy. If an email is not spam, the spam filter will mistakenly identify it as spam with 0.5% probability. An email was marked as sp... **Conditional Probability:** a) The species is an amphibian. b) The species is a reptile, endangered and not foreign. c) The species is not endangered but is foreign. d) The species is a reptile, endangered and foreign. g) Assume the species is endangered, what’s the probability that it is a bird and that it’s from the U.S.?
Expert Solution
Step 1: Basic probabilities

As per the guidelines of Bartleby, solution of only first 3 parts are given.

(a).

(i)  5% employees who take drugs are positive(+)

(ii). 5% employees who take drugs are not positive(-)

(iii) 3% employees who take drugs are positive(+)

(iv). 3% employees who take drugs are not positive(-)


(b). Let us make contingency table

Employees            +                   -

5%                      94%                06%

3%                        x %        100(1-x)%

P(5% employees who take drugs are positive) = 0.94

P(5% employees who take drugs are not positive) = 0.06

P(3% employees who take drugs are positive) = 0.x

P(3% employees who take drugs are not  positive) = 0.(1-x)







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