(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(E_1\ "and"\ E_2) = P(E_1) × P(E_2|E_1)` ` = 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:
`P(R R) = 3/7 times 2/6 = 1/7`
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