A bag contains 6 white balls and 4 black balls. Two balls are drawn one after another with replacement. a) If A and B are mutually exclusive events, then write the principle of P(A U B). b) Present th

Since the drawing is with replacement, the probabilities for the second draw are the same as the first.
If the first ball is white:

If the first ball is black:

• Outcome (White, White): $P(W,W) = \frac{6}{10} \times \frac{6}{10} = \frac{36}{100}$
• Outcome (White, Black): $P(W,B) = \frac{6}{10} \times \frac{4}{10} = \frac{24}{100}$
• Outcome (Black, White): $P(B,W) = \frac{4}{10} \times \frac{6}{10} = \frac{24}{100}$
• Outcome (Black, Black): $P(B,B) = \frac{4}{10} \times \frac{4}{10} = \frac{16}{100}$

The tree diagram would visually represent these steps:
Start with a single point.
Branch 1: First Draw

• Branch for White (probability 6/10)
• Branch for Black (probability 4/10) From each of these branches, create a second set of branches for the Second Draw:
• From First White branch:
  • Branch for Second White (probability 6/10) -> Outcome (W,W) with probability 36/100
  • Branch for Second Black (probability 4/10) -> Outcome (W,B) with probability 24/100
• From First Black branch:
  • Branch for Second White (probability 6/10) -> Outcome (B,W) with probability 24/100
  • Branch for Second Black (probability 4/10) -> Outcome (B,B) with probability 16/100

The outcomes for getting same colored balls are (White, White) and (Black, Black).

The final answer is $\boxed{\frac{13}{25}}$.

The outcomes for getting different colored balls are (White, Black) and (Black, White).

Divide both sides by their greatest common divisor, which is 4.

The final answer is $\boxed{13:12}$.