(c) \(rac{4}{1155}\)Let n(S) = Number of ways of selecting 3 numbers from 100 numbers = 100C3 Let E : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100 = Event of selecting three multiples of 6 from 1 to 100 Selecting 3 numbers from = {6,12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96} ⇒ n(E) = 16C3 ∴ P (Selecting 3 numbers exactly divisible by 6 between 1 and 100)= \(rac{n(E)}{n(S)}\) = \(rac{^{16}C_3}{^{100}C_3}\) = \(rac{16 imes15 imes14}{100 imes99 imes98}\) = \(rac{4}{1155}\).