Answer to Question #93236 in Statistics and Probability for Pramod .P.Holla

Question

On an average 20 in 100 blades manufactured by a firm are defective. If blades are supplied in packets of 25 each, what is the probability that a packet has
1. no defective blade.
2. defective blades

Answer 1

We can apply a binomial distribution in this problem.

Since we have 20 from 100 as defective details in average, we can assume that P(detail is bad) = 20 / 100 = 0.2, P(detail is good) = 1 - P(detail is bad) = 0.8

Then P(no defects in packet) = C(25, 0) * P(good)^25 * P(bad)^0, where C(n, k) is the binomial coefficient;

P(some defects in packet) = 1 - P(no defects in packet);

P(no defects in packet) = 0.0037;

P(some defects in packet) = 0.9962.