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
"solution p ( none is defective ) = 18 c 2 / 22 c 2 = 51 / 77 p ( at least one is defective ) = ( 1 - 51 / 77 ) = 26 / 77 . answer e"
subtract(const_1, divide(choose(subtract(22, 4), const_2), choose(22, const_2)))