Question

A sample of 10 is taken from a day's output of a machine that produces parts of which 5% are defective. A full inspection of a day's production is conducted whenever the sample of 10 gives 2 or more defective parts. Find the probability of a full inspection? What assumptions do you need to make?

Answer 1

here this is is binomial distribution for which parameter n=10 and p=0.05

we are assuming that sample is random ; and probability of one product being defective is independent of other product.

P(full inspection)=P(2 or more defective)=P(X>=2)=1-P(X<=1)=1-(P(X=0)+P(X=1))

=1-¹⁰C₀(0.05)⁰(0.95)¹⁰-¹⁰C₁(0.05)¹(0.95)⁹ =0.0861