A box contains 22 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	Choices:
A) 4/19
B) 7/19
C) 12/19
D) 21/99
E) 26/77
Solution   P( None is defective)   = 18C2 / 22C2   = 51/77   P( at least one is defective)   = (1- 51/77)   = 26/77.   Answer 26/77
<result>E</result>
A box contains 22 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