Question

The ABC Manufacturing Corp. has been having trouble with one of its machines. Out of all of the parts the machine​ produces, 15 ​% are defective. An inspector selects a group of 19 parts at random from the​ machine's output. Express your answers to four decimal places. Express the probabilities as decimal fractions​ -- not as percentages. a. What is the probability that exactly two of these parts will be​ defective? b. What is the probability that none of these parts will be ​defective? c. What is the mean of the number of defective parts among the group of parts ​selected? d. What is the variance of the number of defective parts among the ​group?

Please show step by step. I don't know where to start

Answer 1

a.

The probability that a randomly selected part is defective = 0.15

Let X be the number of defective parts in sample of 19. Then X ~ Binomial(n = 19, p = 0.15)

Probability that exactly two of these parts will be​ defective = P(X = 2) =  ¹⁹C₂ * 0.15² * (1 - 0.15)^19-2

= 171 * 0.15² * 0.85¹⁷

= 0.2428

b.

Probability that none of these parts will be ​defective = P(X = 0) = ¹⁹C₀ * 0.15⁰ * (1 - 0.15)^19-0

=  0.85¹⁹

= 0.0456

c.

Mean of the number of defective parts = np = 19 * 0.15 = 2.85

d.

Variance of the number of defective parts = np(1-p) = 19 * 0.15 * (1 - 0.85) = 0.4275