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?
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?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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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