a 4 - digit code can consist of the digits 0 ~ 9 except that it does not contain the digits 1 , 2 , 4 , and 6 at all . if repeated digits are allowed , what is the probability that it has at least one even digit ?
p ( no even digits ) = 2 / 3 * 2 / 3 * 2 / 3 * 2 / 3 = 16 / 81 p ( at least one even digit ) = 1 - 16 / 81 = 65 / 81 the answer is d .
divide(subtract(power(subtract(const_10, const_4), 4), power(const_4, 4)), power(subtract(const_10, const_4), 4))