Draw a tree diagram to illustrate all possible outcomes of the draws.

• $P(\text{First ball is Red}) = P(R_1) = \frac{3}{6} = \frac{1}{2}$
• $P(\text{First ball is White}) = P(W_1) = \frac{2}{6} = \frac{1}{3}$
• $P(\text{First ball is Blue}) = P(B_1) = \frac{1}{6}$

If the first ball was Red ($R_1$): (Remaining: 2R, 2W, 1B)

• $P(\text{Second ball is Red} | R_1) = P(R_2 | R_1) = \frac{2}{5}$
• $P(\text{Second ball is White} | R_1) = P(W_2 | R_1) = \frac{2}{5}$
• $P(\text{Second ball is Blue} | R_1) = P(B_2 | R_1) = \frac{1}{5}$

If the first ball was White ($W_1$): (Remaining: 3R, 1W, 1B)

• $P(\text{Second ball is Red} | W_1) = P(R_2 | W_1) = \frac{3}{5}$
• $P(\text{Second ball is White} | W_1) = P(W_2 | W_1) = \frac{1}{5}$
• $P(\text{Second ball is Blue} | W_1) = P(B_2 | W_1) = \frac{1}{5}$

If the first ball was Blue ($B_1$): (Remaining: 3R, 2W, 0B)

• $P(\text{Second ball is Red} | B_1) = P(R_2 | B_1) = \frac{3}{5}$
• $P(\text{Second ball is White} | B_1) = P(W_2 | B_1) = \frac{2}{5}$
• $P(\text{Second ball is Blue} | B_1) = P(B_2 | B_1) = \frac{0}{5} = 0$

Tree Diagram:

• First Draw: Red ($R_1$) ($P(R_1) = \frac{3}{6}$)
  • Second Draw: Red ($R_2$)
    • $P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) = \frac{3}{6} \times \frac{2}{5} = \frac{6}{30} = \frac{1}{5}$
  • Second Draw: White ($W_2$)
    • $P(R_1 \cap W_2) = P(R_1) \times P(W_2 | R_1) = \frac{3}{6} \times \frac{2}{5} = \frac{6}{30} = \frac{1}{5}$
  • Second Draw: Blue ($B_2$)
    • $P(R_1 \cap B_2) = P(R_1) \times P(B_2 | R_