Question

Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability that the first defect is caused by the seventh componenttested. Find the expected number and variance of the number of components tested before adefective component is found.

Answer 1

Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

$p = (0.98)^6*0.02 = 0.0177$

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with $p = 0.02$

Expected number before 1 defective(n = 1). So

$E = \frac{n}{p} = \frac{1}{0.02} = 50$

Variance is:

$V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208$

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.