Question

A cloth manufacturer finds that 8% of their production are defective. What is the probability that a batchof 10 willcontain (a) more than two defectives? (b) Less than seven defectives? (c) Exactly eight defectives.

Answer 1

p = 0.08

n = 10

P(X = x) = 10Cx * 0.08^x * (1 - 0.08)^10-x

a) P(X > 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

                  = 1 - (10C0 * 0.08⁰ * 0.92¹⁰ + 10C1 * 0.08¹ * 0.92⁹ + 10C2 * 0.08² * 0.92⁸)

                  = 1 - 0.9599

                  = 0.0401

b) P(X < 7) = 1 - (P(X = 8) + P(X = 9) + P(X = 10))

                  = 1 - (10C8 * 0.08⁸ * 0.92² + 10C9 * 0.08⁹ * 0.92¹ + 10C10 * 0.08¹⁰ * 0.92⁰)

                  = 1 - 0

                  = 1

c) P(X = 8) = 10C8 * 0.08⁸ * 0.92² = 0