total no of bulbs=100.
total no of defective bulbs =20.
we have take out 10 bulbs, therefore total possible outcomes is .100C10
1. if all the 10 are defective means taken out from 20 defective bulbs i.e possible outcome is .20C10
therefore probability isp=
.20C10
.100C10
2.if all the 10 are good that is taken out from rest of the 80 which are good bulbs i.e possible outcome is .80C10
therefore robability is p=
.80C10
.100C10
4.atleast one is defective means we have to find all the probabilities when one is defective ,2 is defective and 3 is defective so on .so find probility when none is defective i.e all are good which is p=
.20C10
.100C10
therefore probability of atleast one defective is 1-p
= 1-
.20C10
.100C10