Question

In a shipment of 10 electronic components, 2 are defective. Suppose that 5
components are selected at random for inspection. What is the probability that:
a. No defective components are included among the 5 selected?
b. Exactly one defective component being included?

Answer 1

There are 10 components, 2 are defective.

Therefore 10-2=8 components are not defective.

a) Number of ways to choose 5 non defective components is $C(8, 5)$

Number of ways to choose 5 components in total is $C(10, 5)$

Therefore

P(5 non defective components selected) = $\frac{C(8, 5)}{C(10,2)}=\frac{\frac{8!}{5!3!}}{\frac{10!}{5!5!}}=\frac{2}{9}$

b) Exactly one defective component being included: there is one defective and 4 non defective components

Number of ways to choose 4 non defective components is $C(8, 4)$

Number of ways to choose 1 defective component is $C(2, 1)=2$

Number of ways to choose 5 components in total is $C(10, 5)$

Therefore

P(exactly one defective component being included) = $\frac{2\cdot C(8,4)}{C(10,5)}=\frac{\frac{2\cdot8!}{4!4!}}{\frac{10!}{5!5!}}=\frac{5}{9}$

Answer:

a) $\frac{2}{9}$

b) $\frac{5}{9}$