(a) Tree Diagram

Let's represent the problem using a tree diagram. The first level of the diagram will represent the outcome of the first ball being picked, and the second level will represent the outcome of the second ball.

1. First Ball: It can either be brown (B) or green (G).

   • Probability of picking a brown ball (B) = $\frac{3}{12} = \frac{1}{4}$
   • Probability of picking a green ball (G) = $\frac{9}{12} = \frac{3}{4}$

2. Second Ball: The second ball can again be either brown or green, but now the total number of balls has decreased by one, and it depends on what color was chosen first:

   • If the first ball was brown, there are now 2 brown and 9 green left:

     • Probability of picking a second brown ball (B) = $\frac{2}{11}$
     • Probability of picking a green ball (G) = $\frac{9}{11}$

   • If the first ball was green, there are now 3 brown and 8 green left:

     • Probability of picking a brown ball (B) = $\frac{3}{11}$
     • Probability of picking a second green ball (G) = $\frac{8}{11}$

The tree diagram would look like this:

          First Ball
           /    \
        Brown  Green
      (B,1/4) (G,3/4)
        /  \     /  \
     B/2/11 G/9/11 B/3/11 G/8/11

(b) Probability that the balls picked are of different colors

There are two favorable outcomes for picking balls of different colors:

1. First ball is brown, second ball is green.
2. First ball is green, second ball is brown.

We can now calculate the probabilities for each of these outcomes:

1. First brown, then green:

   • Probability of first ball being brown: $\frac{1}{4}$
   • Probability of second ball being green: $\frac{9}{11}$
   • Total probability: $\frac{1}{4} \times \frac{9}{11} = \frac{9}{44}$

2. First green, then brown:

   • Probability of first ball being green: $\frac{3}{4}$
   • Probability of second ball being brown: $\frac{3}{11}$
   • Total probability: $\frac{3}{4} \times \frac{3}{11} = \frac{9}{44}$

Finally, the probability that the balls picked are of different colors is the sum of these two probabilities:

$P(\text{different colors}) = \frac{9}{44} + \frac{9}{44} = \frac{18}{44} = \frac{9}{22}$

Thus, the probability that the two balls picked are of different colors is $\frac{9}{22}$.

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Would you like further details on any part? Here are some related questions:

1. What is the probability of both balls being brown?
2. What is the probability of both balls being green?
3. How would the probability change if there were 4 brown and 8 green balls?
4. Can you explain why these probabilities add up to 1?
5. How do we handle "without replacement" in probability calculations?

Tip: When working with probabilities, always check that the sum of all possible outcomes equals 1 to verify correctness.