Question

A machine produces 5% defective items. 12 items are inspected every hour. The machine has to be shut down if more than 2 defectives are found. What is the probability that machine has to be shut for the first time on third inspection?

Answer 1

0.005714

Explanation:

Given that a machine produces 5% defective items. 12 items are inspected every hour. The machine has to be shut down if more than 2 defectives are found.

The probability for any random item to be found defective = 0.05 a constant

since each item is independent of the other.

Also there are only two outcomes

Hence X no of defectives is binomial with p =0.05 and n =12

Probability that machine has to be shut for the first time on third inspection

=prob in the first two inspection no defects are found*prob that more than 2 defectives

= P(x=0) two times*P(X>2)

=
`0.54036*0.54036*0.01957`

=0.005714