Question

There are 10 defective light bulbs in a bin containing 20 light bulbs. In how many ways can you select at most 2 defective light bulbs when you choose 10 light bulbs at random?

Answer 1

Let $X=$ the number of defective light bulbs in the sample.

$$X=0: \dbinom{10}{0}\dbinom{10}{10-0}=1(1)=1$$

$$X=1: \dbinom{10}{1}\dbinom{10}{10-1}=10(10)=100$$

$$X=2: \dbinom{10}{2}\dbinom{10}{10-2}=45(45)=2025$$

$$1+100+2025=2126$$

We can select at most 2 defective light bulbs in 2126 ways.