If E₁ denotes the event of getting the sum 6 in throwing two dice and E₂ the event of getting 2 in either of the throw, then P(E₂/E₁) is

2/5

Events favourable to E₁ are:

(1, 5), (2, 4), (3, 3), (4, 2) and (1, 5)

Events E₂ are (2, 4), (4, 2).

P(E₂ / E₁) = Probability of E₂ when E₁ has already occurred  = $\frac{2}{5}$

Hence (D) is the correct answer.