Question

A box contains 6 defective bulbs and 10 functional bulbs. If 4 bulbs are to be chosen at random. what is the probability that there are 2 defective bulbs and 2 functional bulbs selected?

Answer 1

There are $16$ different bulbs. There are $C_{16}^4=\frac{16!}{4!12!}=\frac{13\cdot14\cdot15\cdot16}{2\cdot3\cdot4}=1820$ ways to choose $4$ bulbs. There are $C_6^2=15$ ways to make a set that contains $2$ defective bulbs and $C_{10}^2=45$ ways to make a set that contains $2$ functional bulbs. We get the following probability: $p=\frac{C_6^2C_{10}^2}{C_{16}^4}=\frac{15\cdot45}{1820}=\frac{135}{364}\approx0.3709$.

Thus, the answer is $0.3709$ (it is rounded to $4$ decimal places).