Use the following scenario to answer the next 5 questions. Millions of Americans are subjected to drug testing in the workplace. The tests are not always reliable. A false positive rate of 5% is considered to be a conservative estimate. Suppose that there are 20 employees in your company, which tests its workers weekly.

11) What is the probability of no false-positives for this week’s test?

12) What is the probability of at least 1 false-positive for this week’s tests?



13) What is the likelihood that the test will give 5 to 10 false positives?


14) What is the chance of at most 3 false positives?

15) How many false positives should the company expect?

11) To find the probability of no false-positives for this week's test, we need to find the probability of not having a false positive for each employee and then multiply those probabilities together. Since the false positive rate is 5%, the probability of not having a false positive is 1 - 0.05 = 0.95.

Probability of no false positive = (0.95)^20 ≈ 0.3585

So, the probability of no false-positives for this week's test is approximately 0.3585.

12) To find the probability of at least 1 false-positive for this week's tests, we can subtract the probability of no false-positives (which we found in question 11) from 1.

Probability of at least 1 false-positive = 1 - 0.3585 = 0.6415

So, the probability of at least 1 false-positive for this week's tests is approximately 0.6415.

13) To find the likelihood that the test will give 5 to 10 false positives, we need to calculate the probabilities for each possible number of false positives within this range (i.e., 5, 6, 7, 8, 9, and 10) and then sum them.

Let's calculate the probability for each individual case using the binomial distribution formula:

Probability of k false positives = (number of ways to choose k false positives out of 20) * (probability of false positive)^(k) * (probability of not having a false positive)^(20 - k)

For k = 5:
Probability of 5 false positives = (20 choose 5) * (0.05)^(5) * (0.95)^(20 - 5)

For k = 6:
Probability of 6 false positives = (20 choose 6) * (0.05)^(6) * (0.95)^(20 - 6)

Repeat the same steps for k = 7, 8, 9, and 10.

Finally, sum up the probabilities for all these cases to get the likelihood.

14) To find the chance of at most 3 false positives, we need to calculate the probabilities for 0, 1, 2, and 3 false positives and then sum them.

Probability of at most 3 false positives = Probability of 0 false positives + Probability of 1 false positive + Probability of 2 false positives + Probability of 3 false positives

For each individual case, use the same binomial distribution formula as in question 13 to calculate the probabilities. Sum up these probabilities to find the chance of at most 3 false positives.

15) To find the expected number of false positives, we need to multiply the total number of tests (20) by the probability of having a false positive (0.05).

Expected number of false positives = Total number of tests * Probability of false positive

Expected number of false positives = 20 * 0.05

So, the company should expect 1 false positive.