A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur independently of each other. (a) find the probability of exactly 3 defective computers in a shipment of twenty. (b) find the probability that the engineer has to test at least 5 computers in order to find 2 defective ones.

(a) 0.059582148 probability of exactly 3 defective out of 20

(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

0.05^3 * (1 - 0.05) ^ (20-3) * 20! / (3! (20-3) !)

= 0.05^3 * 0.95^17 * 20! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

= 0.05^3 * 0.95^17 * 20*19 * (2*3*3) / (2*3)

= 0.05^3 * 0.95^17 * 20*19*3

= 0.000125 * 0.418120335 * 1140

= 0.059582148

(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05) ^4 = 0.95^4 = 0.81450625.

The probability of getting exactly 1 defective computer out of 4 is 0.05 * (1-0.05) ^3*4! / (1! (4-1) !)

= 0.05*0.95^3*24 / (1!3!)

= 0.05*0.857375*24/6

= 0.171475

So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.