(a) The second event is dependent on the first.

P(E₁) = P(white) = 4/7

There are 6 balls left and out of those 6, three of them are red. So the probability that the second one is red is given by:

P(E₂ | E₁) = P(red) = 3/6 = 1/2

Dependent events, so

P(E₁ and E₂) = P(E₁) × P(E₂ | E₁) = 4/7 × 1/2 = 2/7

(b) Also dependent events. Using similar reasoning, but realising there will be 2 red balls on the second draw, we have: