A bag contains 5 green and 7 blue balls. Two balls are drawn randomly one after another with replacement. a) If A and B are two independent events, write the formula of the multiplication law of proba

For two independent events A and B, the multiplication law of probability states:

The probability of drawing a green ball is $P(G) = \frac{5}{12}$.
The probability of drawing a blue ball is $P(B) = \frac{7}{12}$.
Since the draws are with replacement, the probabilities for the second draw are the same as for the first draw.

Tree Diagram:

Start
  |
  |--- 1st Draw (G) --- P(G1) = 5/12 --- 2nd Draw (G) --- P(G2|G1) = 5/12 --- Outcome: GG, P(GG) = 5/12 * 5/12 = 25/144
  |                                   |
  |                                   |--- 2nd Draw (B) --- P(B2|G1) = 7/12 --- Outcome: GB, P(GB) = 5/12 * 7/12 = 35/144
  |
  |--- 1st Draw (B) --- P(B1) = 7/12 --- 2nd Draw (G) --- P(G2|B1) = 5/12 --- Outcome: BG, P(BG) = 7/12 * 5/12 = 35/144
                                    |
                                    |--- 2nd Draw (B) --- P(B2|B1) = 7/12 --- Outcome: BB, P(BB) = 7/12 * 7/12 = 49/144

Possible Outcomes and their Probabilities:

• $P(\text{GG}) = P(G) \times P(G) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$
• $P(\text{GB}) = P(G) \times P(B) = \frac{5}{12} \times \frac{7}{12} = \frac{35}{144}$
• $P(\text{BG}) = P(B) \times P(G) = \frac{7}{12} \times \frac{5}{12} = \frac{35}{144}$
• $P(\text{BB}) = P(B) \times P(B) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$

The probability of getting both blue balls is $P(\text{BB})$.

The probability of getting both blue balls is $\boxed{\frac{49}{144}}$.