A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is
Choose the correct answer:	A) 7/19	B) 6/19	C) 5/19	D) 4/19	E) None of these
Explanation:   Please remember that Maximum portability is 1.   So we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs.   So here we go,   Total cases of non defective bulbs   16C2=16∗15/2∗1=120   total cases = 20C2=20∗19/2∗1=190   probability = 120/190=12/19   P of at least one defective = 1−12/19=7/19   Option 7/19
<result>A</result>
A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is