Q.231. Conditional Probability.

Question 231.
To investigate the quality of a lot, a sample of 20 lamp is randomly selected. The retailer accepts a lot if there is less than 3 defective lamps. The proportion of defective item is 0.05. Given that the retailer accepts a lot, what is the probability that less than 2 defective lamps were observed?

Answer 231.
Let A = event that the retailer accepts the lot
and B = event that less than 2 defective bulbs are found
This is a conditional probability and
P(B/A) is required.

P(A)
= (0.95)^20 + 20C1 * (0.05) * (0.95)^19 + 20C2 * (0.05)^2 * (0.95)^18
= 0.35849 + 0.37735 + 0.18868
= 0.92452

P(B)
= (0.95)^20 + 20C1 * (0.05) * (0.95)^19
= 0.35849 + 0.37735
= 0.73584

P(B/A)
= P(B∩A) / P(A)
= P(B) / )(A) ... [because B is a subset of (A∩B)]
= (0.73584) / (0.92452)
= 0.7959.

Link to YA!