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Let's calculate the probability step by step.

### Step 1: Determine the total number of balls in each prize type
- **Big Prize (B)**: 5 red (R) + 5 gold (G) = 10 balls
- **Medium Prize (M)**: 3 red (R) + 2 gold (G) = 5 balls
- **Little Prize (L)**: 1 red (R) + 1 gold (G) = 2 balls

### Step 2: Calculate the total number of balls in all prizes combined
Your friend has:
- 3 Big Prizes (B)
- 1 Medium Prize (M)
- 2 Little Prizes (L)

Total balls from each prize type:
- Big Prizes: \(3 \times 10 = 30\) balls
- Medium Prize: \(1 \times 5 = 5\) balls
- Little Prizes: \(2 \times 2 = 4\) balls

Total number of balls = 30 + 5 + 4 = 39 balls

### Step 3: Determine the total number of gold balls in all prizes
Gold balls from each prize type:
- Big Prizes: \(3 \times 5 = 15\) gold balls
- Medium Prize: \(1 \times 2 = 2\) gold balls
- Little Prizes: \(2 \times 1 = 2\) gold balls

Total number of gold balls = 15 + 2 + 2 = 19 gold balls

### Step 4: Determine the probability of choosing a gold ball from each type of prize
- Probability of choosing a gold ball from a Big Prize:
  \[ P(G|B) = \frac{5}{10} = \frac{1}{2} \]
- Probability of choosing a gold ball from a Medium Prize:
  \[ P(G|M) = \frac{2}{5} \]
- Probability of choosing a gold ball from a Little Prize:
  \[ P(G|L) = \frac{1}{2} \]

### Step 5: Determine the prior probabilities for each type of prize being chosen
Total prizes = 3 (B) + 1 (M) + 2 (L) = 6 prizes

- Probability of choosing a Big Prize:
  \[ P(B) = \frac{3}{6} = \frac{1}{2} \]
- Probability of choosing a Medium Prize:
  \[ P(M) = \frac{1}{6} \]
- Probability of choosing a Little Prize:
  \[ P(L) = \frac{2}{6} = \frac{1}{3} \]

### Step 6: Apply Bayes' Theorem to find the probability of having chosen a Big Prize given that a gold ball was selected
Bayes' Theorem:
\[ P(B|G) = \frac{P(G|B) \cdot P(B)}{P(G)} \]

Where \(P(G)\) is the total probability of picking a gold ball:
\[ P(G) = P(G|B) \cdot P(B) + P(G|M) \cdot P(M) + P(G|L) \cdot P(L) \]

\[ P(G) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{2}{5} \times \frac{1}{6}\right) + \left(\frac{1}{2} \times \frac{1}{3}\right) \]

\[ P(G) = \frac{1}{4} + \frac{2}{30} + \frac{1}{6} \]

\[ P(G) = \frac{1}{4} + \frac{1}{15} + \frac{1}{6} \]

Find a common denominator (60):
\[ P(G) = \frac{15}{60} + \frac{4}{60} + \frac{10}{60} \]

\[ P(G) = \frac{29}{60} \]

Now, apply Bayes' Theorem:
\[ P(B|G) = \frac{P(G|B) \cdot P(B)}{P(G)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)}{\frac{29}{60}} \]

\[ P(B|G) = \frac{\frac{1}{4}}{\frac{29}{60}} \]

\[ P(B|G) = \frac{1}{4} \times \frac{60}{29} \]

\[ P(B|G) = \frac{60}{116} \]

\[ P(B|G) = \frac{30}{58} \]

\[ P(B|G) = \frac{15}{29} \]

Thus, the probability that you were choosing from a big prize bag is \(\frac{15}{29}\).

Would you like more details or have any questions? Here are some other questions you might find interesting:

1. What is the probability of choosing from a medium prize bag given that a gold ball was selected?
2. What is the probability of choosing from a little prize bag given that a gold ball was selected?
3. How would the probabilities change if the number of prizes of each type were different?
4. What is the total number of red balls in all the prizes?
5. How would the probability calculation change if the distribution of balls in the prizes was different?

**Tip:** When dealing with probabilities, always ensure to consider all possible outcomes and their respective probabilities to get accurate results.

Learn how to calculate probabilities using Bayes' Theorem with a detailed example problem. This problem involves determining the probability of choosing a big prize given that a gold ball was selected, covering key concepts such as conditional probability and prior probabilities. Suitable for advanced high school or college-level students.

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