a box contains 23 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 ) = 19 c 2 / 23 c 2 = 25 / 37 . p ( at least one is defective ) = ( 1 - 25 / 37 ) = 12 / 37 . answer c
subtract(const_1, divide(choose(subtract(23, 4), const_2), choose(23, const_2)))