a box contains 24 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
"solution p ( none is defective ) = 20 c 2 / 24 c 2 = 53 / 77 . p ( at least one is defective ) = ( 1 - 53 / 77 ) = 24 / 77 . answer d"
subtract(const_1, divide(choose(subtract(24, 4), const_2), choose(24, const_2)))