a box contains 21 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 ) = 17 c 2 / 21 c 2 = 57 / 88 . p ( at least one is defective ) = ( 1 - 57 / 88 ) = 31 / 88 . answer e"
subtract(const_1, divide(choose(subtract(21, 4), const_2), choose(21, const_2)))