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Question

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



Options:
A .  4/19
B .  7/19
C .  12/19
D .  21/95
Answer: Option B

P( None is defective)
= 16C2 / 20C2
= (16x15/2x1x2x1/20x19)
= 12/19.
P( at least one is defective)
= (1−12/19)
= 7/19.




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