Question #271906

It is known that from a lot of 100 computers there are 20 defective ones. 5 of these computers are selected at random and tested. Find the probability that only 2 of the 5 are defective.

1
Expert's answer
2021-12-01T16:28:18-0500

Let X=X= the number of defective computers out of randomly selected 5 computers:


P(X=2)=(202)(803)(1005)P(X=2)=\dfrac{\dbinom{20}{2}\dbinom{80}{3}}{\dbinom{100}{5}}

=20!2!(202)!80!3!(803)!100!5!(1005)!=\dfrac{\dfrac{20!}{2!(20-2)!}\cdot \dfrac{80!}{3!(80-3)!}}{\dfrac{100!}{5!(100-5)!}}

=19082160752875200.207344=\dfrac{190\cdot82160}{75287520}\approx0.207344

The probability that only 2 of the 5 are defective is 0.207344.0.207344.



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