Question

The employees of a firm that manufactures insulation are being tested for indications of asbestos in their lungs. The firm is requested to send three employees who have positive indications of asbestos to a medical center for further testing. If 40% of the employees have positive indications of asbestos in their lungs

a. What is the probability that the first positive test is the fourth employee?

b. If x is the number of employees tested until the first success occurs, what is the mean and standard deviation of x?

c. What is the probability that x will exceed 5?

d. What is the probability that ten employees must be tested to find three positives?

Answer 1

a) probability that the first positive test is the fourth employee =P(first 3 nt have and 4th one have positive test)

=(0.6)³*0.4 =0.0864

b)

mean =1/p =1/0.4 =2.5

standard deviation of x =((1-p)/p²)^1/2 =((1-0.4)/0.4²)^1/2 =1.9365

c)probability that x will exceed 5 =P(no success in first 5) =(0.6)⁵ =0.0778

d)

probability that ten employees must be tested to find three positives=P(till 9 employee 2 are positive and 3rd positive on 10th one) = $\binom{9}{2}(0.4)^{3}(0.6)^{7}$ =0.0645