Question

The probability that a part produced by a certain​ factory's assembly line will be defective is 0.036. Find the probabilities that in a run of 44 ​items, the following results are obtained.

​(a) Exactly 3 defective items

​(b) No defective items

​(c) At least 1 defective item

(Round to four decimal places as​ needed.)

Answer 1

Given that the probability that a part produced by a certain factory's assembly line will be defective is 0.036. 44 items are run.

Given the probability that it will be defective = 0.036 . Thus, the probability of success, p, i.e., the probability of not defective = 1 - 0.036 = 0.964 . Thus, p = 0.964 . 44 items are run. Thus, n = 44 .

From Binomial Probability, we have, , where, x is the number of success.

We also know, nr = n!/{r!(n-r)!}, where, n! = n(n-1)(n-2).........1

a) Here, we have to find the probability of exactly 3 defective items. Thus, there are 3 defective items. Thus, number of non-defective items = 44 - 3 = 41. Thus, x = 41.

Thus, x = 41, n = 44, p = 0.964 .

Thus, = 0.1374(rounded up to four decimal places)

Thus, P(X = 41) = 0.1374 .

Thus, the probability that in a run of 44 items, exactly 3 defective items are obtained = 0.1374 .

b) Here, we have to find the probability of no defective items. Thus, there are 0 defective items. Thus, the number of non-defective items = 44 - 0 = 44. Thus, x = 44.

Thus, x = 44, n = 44, p = 0.964 .

Thus, = 0.1992(rounded up to four decimal places).

Thus, P(X = 44) = 0.1992 .

Thus, the probability that in a run of 44 items, no defective items are obtained = 0.1992 .

c) Here, we have to find the probability of at least one defective item. Thus, we can find out this by subtracting the probability that none are defective from 1. Here, we have to find P(X < 44). We know, P(X < 44) = 1 - P(X = 44).

From part (b), we have, P(X = 44) = 0.1992 . Thus, P(X < 44) = 1 - 0.1992 = 0.8008 .

Thus, P(X < 44) = 0.8008 .

Thus, the probability that in a run of 44 items, at least one defective item is obtained = 0.8008 .