Answer to Question #271906 in Statistics and Probability for Tony

Question

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.

Answer 1

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

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

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

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

The probability that only 2 of the 5 are defective is $0.207344.$