Let 'N' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.Required number = (LCM of 13 and 16) - (common difference of divisors and remainders) = (208) - (11) = 197.
Total balls = 5 + 4 + 4 = 13∴ n(S) = 13C3 = 2862 blue balls can be selected from 4 balls in 4C2 = 6 ways and the remaining one ball is to be selected from 9 balls in 9C1 = 9 ways ∴ n(E) = 6 x 9∴ P(E) = 6 x 9/286 = 27/143
EXAMIANS