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9 August, 23:21

The probability that a certain machine will produce a defective item is 0.20. if a random sample of 20 items is taken from the output of this machine, what is the probability that there will be 3 or more defectives in the sample?

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  1. 9 August, 23:35
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    Use binomial distribution, with p=0.20, n=20, x=3

    P (X=x) = C (n, x) p^x (1-p) ^ (n-x)

    P (X>=3)

    =1 - (P (X=0) + P (X=1) + P (X=2))

    =1 - (C (20,0) 0.2^0 (0.8) ^ (20-0) + C (20,1) 0.2^1 (0.8) ^ (20-1) + C (20,2) 0.2^2 (0.8) ^ (20-2))

    =1 - (0.0115292+0.057646+0.136909)

    =1-0.206085

    =0.793915
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