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24.12.2019 •
Mathematics
Amachine 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?
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Ответ:
0.005714
Step-by-step 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](/tpl/images/0431/3849/ccc75.png)
=0.005714
Ответ: