Seventy percent of the engines manufactured on an assembly line are defective. Suppose engines are randomly selected one at a time and tested. Given that the first two engines tested were defective, what is the probability that at least two more engines must be tested before the first nondefective is found?

Question

Mathematics

Kasen Walters

7 January, 04:05

Seventy percent of the engines manufactured on an assembly line are defective. Suppose engines are randomly selected one at a time and tested. Given that the first two engines tested were defective, what is the probability that at least two more engines must be tested before the first nondefective is found?

+4

Answers (1)

Know the Answer?

Dario Duarte · Answer 1

0.7

Step-by-step explanation:

Probability of defective = 0.7

P (non-defective) = 1 - 0.7 = 0.3

Let X be the trial at which we get the first non-defective

Then X follows a geometric distribution.

X ~ Geo (0.3)

What has happened already will not affect the probability.

P (atleast 2 more) = P (X > 1)

= 1 - P (X = 1)

= 1 - (0.3)

= 0.7