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:

$$P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)$$

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 independent of the first.

Tree Diagram Outcomes and Probabilities:

• First draw Green ($G_1$), Second draw Green ($G_2$): $P(G_1 \text{ and } G_2) = P(G_1) \times P(G_2) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$
• First draw Green ($G_1$), Second draw Blue ($B_2$): $P(G_1 \text{ and } B_2) = P(G_1) \times P(B_2) = \frac{5}{12} \times \frac{7}{12} = \frac{35}{144}$
• First draw Blue ($B_1$), Second draw Green ($G_2$): $P(B_1 \text{ and } G_2) = P(B_1) \times P(G_2) = \frac{7}{12} \times \frac{5}{12} = \frac{35}{144}$
• First draw Blue ($B_1$), Second draw Blue ($B_2$): $P(B_1 \text{ and } B_2) = P(B_1) \times P(B_2) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$

The probability of getting both blue balls is $P(B_1 \text{ and } B_2)$.

$$P(\text{both blue}) = P(B) \times P(B) = \frac{7}{12} \times \frac{7}{12} = \frac{49}{144}$$

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