The probability of rolling a 1 or 2 is \(\frac{2}{6}\), and the probability of rolling a 3, 4, 5, or 6 is \(\frac{4}{6}\).

For bag A, the probability of picking two white marbles is:

\(\frac{8}{15} \times \frac{7}{14}\)

For bag B, the probability of picking two white marbles is:

\(\frac{6}{15} \times \frac{5}{14}\)

The total probability that both marbles are white is:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} + \frac{4}{6} \times \frac{6}{15} \times \frac{5}{14}\)

Calculating each part:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} = \frac{56}{630}\)

\(\frac{4}{6} \times \frac{6}{15} \times \frac{5}{14} = \frac{60}{630}\)

Adding these probabilities gives:

\(\frac{56}{630} + \frac{60}{630} = \frac{116}{630} = \frac{58}{315}\)

Thus, the probability that both marbles are white is \(\frac{58}{315}\) or 0.184.

To find the probability that exactly 1 student is studying Music and exactly 2 are boys, we consider two scenarios:

1. 1 boy studying Music, 1 boy not studying Music, and 1 girl not studying Music.
2. 1 girl studying Music, 2 boys not studying Music.

For the first scenario:

\(P(B \text{ M}) \times P(B \text{ NM}) \times P(G \text{ NM}) = \frac{40}{160} \times \frac{56}{159} \times \frac{52}{158} \times 3! = 0.17387\)

For the second scenario:

\(P(G \text{ M}) \times P(B \text{ NM}) \times P(B \text{ NM}) = \frac{12}{160} \times \frac{56}{159} \times \frac{55}{158} \times \frac{3!}{2!} = 0.02759\)

Adding both probabilities gives:

\(P = 0.17387 + 0.02759 = 0.201\)

The possible outcomes for the sum of 9 when the die is thrown twice are: (3, 6), (6, 3), (4, 5), and (5, 4).

The probability of rolling a 3 is 0.1 and the probability of rolling a 6 is 0.3. Thus, the probability of (3, 6) is:

\(P(3, 6) = 0.1 \times 0.3 = 0.03\)

The probability of (6, 3) is the same as (3, 6) because the events are independent:

\(P(6, 3) = 0.3 \times 0.1 = 0.03\)

The probability of rolling a 4 is 0.2 and the probability of rolling a 5 is 0.1. Thus, the probability of (4, 5) is:

\(P(4, 5) = 0.2 \times 0.1 = 0.02\)

The probability of (5, 4) is the same as (4, 5):

\(P(5, 4) = 0.1 \times 0.2 = 0.02\)

Therefore, the total probability that the sum is 9 is:

\(P(\text{sum is 9}) = P(3, 6) + P(6, 3) + P(4, 5) + P(5, 4) = 0.03 + 0.03 + 0.02 + 0.02 = 0.1\)

To find \(r\), note that the only way to get a score of 6 is by landing on 3 on both spinners, so \(P(\text{score is } 6) = r^2 = \frac{1}{36}\). Solving for \(r\), we get \(r = \frac{1}{6}\).

For a score of 5, the possible outcomes are (2, 3) and (3, 2). Thus, \(P(\text{score is } 5) = P(2, 3) + P(3, 2) = qr + rq = 2qr = \frac{1}{9}\). Substituting \(r = \frac{1}{6}\), we have:

\(2q \times \frac{1}{6} = \frac{1}{9}\)

\(\frac{q}{3} = \frac{1}{9}\)

\(q = \frac{1}{3}\)

Since \(p + q + r = 1\), we have:

\(p + \frac{1}{3} + \frac{1}{6} = 1\)

\(p + \frac{1}{2} = 1\)

\(p = \frac{1}{2}\)

The total number of balloons is 40 (10 pink + 9 yellow + 12 green + 9 white).

We need to find the probability of selecting exactly 3 green balloons out of 7 selected balloons.

Using combinations, the number of ways to choose 3 green balloons from 12 is given by:

\(\binom{12}{3}\)

The number of ways to choose the remaining 4 balloons from the 28 non-green balloons is:

\(\binom{28}{4}\)

The total number of ways to choose 7 balloons from 40 is:

\(\binom{40}{7}\)

The probability is then given by:

\(\frac{\binom{12}{3} \times \binom{28}{4}}{\binom{40}{7}}\)

Calculating each part:

\(\binom{12}{3} = 220\)

\(\binom{28}{4} = 20475\)

\(\binom{40}{7} = 18643560\)

Thus, the probability is:

\(\frac{220 \times 20475}{18643560} = 0.242\)