Problem: a box contains 25 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
Rationale: "solution p ( none is defective ) = 21 c 2 / 25 c 2 = 7 / 10 p ( at least one is defective ) = ( 1 - 7 / 10 ) = 3 / 10 . answer a"
annotated_formula: subtract(const_1, divide(choose(subtract(25, 4), const_2), choose(25, const_2)))