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?

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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