Problem

Part a

After the blue chip is drawn, there are 9 chips remaining in the bag. Since one blue chip has already been drawn, there are 2 remaining blue chips. Thus, the probability of the next trip drawn being blue is

\(P(a) = \frac{2}{9} \approx 0.222\)

Part b

Again, after one chip is drawn, there are nine chips remaining. In this case, no blue chip has been drawn, so the probability of the second chip being blue is

\(P(b) = \frac{3}{9} \approx 0.333\)

Part c

Assuming that no chips have been previously drawn, the probability of drawing two chips is the product of the probabilities of drawing a blue chip on each of the two draws: \(P(c) = P(c_1) \times P(c_2)\).

The probability of drawing a blue chip on the first draw is simply the number of blue chips (\(3\)) divided by the total number of chips (\(10\)). The probability of drawing a blue chip on the second draw, if a blue chip is selected on the first draw, is shown in Part a of this question. So the probability of drawing two blue chips in a row as the first two chips is

\(P(c) = P(c_1) \times P(c_2) = \frac{3}{10} \times \frac{2}{9} \approx 0.067\)

Part d

When drawing without replacement, the outcome of one draw affects the probability of subsequent draws, so the draws are not independent.