(b) \(rac{7}{27}.\)Let S be the sample space of drawing a counter three times and replacing it each time. Then, n(S) = 3 × 3 × 3 = 27 Let A : Event of obtaining a total of 6 in the three draws of counters. Then, A = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), (2, 2, 2)} ⇒ n(A) = 7 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{7}{27}.\)