Can you help me with this statistics problem?
A company requires all employees to take drug test. The company can afford only the inexpensive drug test- the one with a 5% false-positive rate and a 10% false-negative rate. (This means that 5% of those who aren't using drugs test positive incorrectly and 10% of those who are using drugs test negative). Suppose that 10% of the employees in the company are using drugs and answer the following questions.
a. If one employee is chosen at random what is the probability that the employee uses drugs and tests positive?
b. If one employee is chosen at random, what is the probability that the employee does not use drugs but tests positive anyway.
c. If one employee is chosen at random, what is the probability that the employee tests positive.
d. If we know that a randomly chosen employee has tested positive, what is the probability that he uses drugs.
Please explain how to determine the probabilities.
Sure! I can help you with this statistics problem. To determine the probabilities, we need to understand conditional probability and apply it to the given information.
Before we start, let's define some terms:
- A: The event that an employee uses drugs.
- B: The event that an employee tests positive.
Now, let's find the probabilities step by step:
a. To find the probability that an employee uses drugs and tests positive, we need to find P(A and B). According to the given information, the probability of testing positive given that an employee uses drugs is 1 - 10% = 90% (since 10% of drug users test negative). Also, we know that 10% of the employees use drugs (P(A) = 0.10). So, we can calculate P(A and B) as follows:
P(A and B) = P(B|A) * P(A)
= 0.90 * 0.10
= 0.09
= 9%
Therefore, the probability that a randomly chosen employee uses drugs and tests positive is 9%.
b. To find the probability that an employee does not use drugs but tests positive, we need to calculate P(not A and B). The probability of testing positive given that an employee does not use drugs is 5% (since 5% of non-drug users test positive). The probability of not using drugs is 1 - 10% = 90% (P(not A) = 0.90). So, we can calculate P(not A and B) as follows:
P(not A and B) = P(B|not A) * P(not A)
= 0.05 * 0.90
= 0.045
= 4.5%
Therefore, the probability that a randomly chosen employee does not use drugs but tests positive is 4.5%.
c. To find the probability that an employee tests positive, we need to find P(B). An employee can test positive either if they use drugs (P(A and B) = 9%) or if they do not use drugs (P(not A and B) = 4.5%). So, we can calculate P(B) as follows:
P(B) = P(A and B) + P(not A and B)
= 0.09 + 0.045
= 0.135
= 13.5%
Therefore, the probability that a randomly chosen employee tests positive is 13.5%.
d. To find the probability that a randomly chosen employee uses drugs given that they have tested positive, we need to calculate P(A|B). According to Bayes' theorem:
P(A|B) = P(B|A) * P(A) / P(B)
We already know that P(B|A) = 90% (as mentioned before), P(A) = 10%, and P(B) = 13.5% (as calculated in part c). So, we can calculate P(A|B) as follows:
P(A|B) = (P(B|A) * P(A)) / P(B)
= (0.90 * 0.10) / 0.135
= 0.0667 (rounded to 4 decimal places)
= 6.67%
Therefore, the probability that a randomly chosen employee uses drugs given that they have tested positive is approximately 6.67%.
I hope this explanation helps you understand how to determine the probabilities in this statistics problem. Let me know if you need any further assistance!