Question 1205136
To solve the problem, we need to analyze the number of possible combinations of bottlecap codes and then determine how many bottlecaps Student B needs to buy to have a 1% chance of winning. We will also calculate the probability of winning for Student C if he buys 24 bottlecaps.

### Step 1: Determine the Total Number of Combinations

Each bottlecap has 3 characters, and we need to determine how many different characters can be used. Let's assume the characters can be any of the 26 letters of the alphabet (A-Z). 

1. **Total Combinations**:
   Each character can be one of 26 letters, and since there are 3 characters, the total number of combinations is:
   $$
   \text{Total Combinations} = 26^3 = 26 \times 26 \times 26 = 17576
   $$

### Step 2: Determine Winning Combinations

The problem states that there are 5 winning combinations. Therefore, the probability of selecting a winning combination with one bottlecap is:
$$
P(\text{winning}) = \frac{\text{Number of Winning Combinations}}{\text{Total Combinations}} = \frac{5}{17576}
$$

### Step 3: Calculate the Number of Bottlecaps for Student B

To find out how many bottlecaps Student B needs to buy to have at least a 1% chance of winning, we can use the complement probability.

1. **Probability of Not Winning with One Bottlecap**:
   $$
   P(\text{not winning}) = 1 - P(\text{winning}) = 1 - \frac{5}{17576} = \frac{17571}{17576}
   $$

2. **Probability of Not Winning with $ n $ Bottlecaps**:
   The probability of not winning with $ n $ bottlecaps is:
   $$
   P(\text{not winning with } n) = \left( \frac{17571}{17576} \right)^n
   $$

3. **Setting Up the Equation for 1% Chance**:
   We want the probability of winning to be at least 1%, which means:
   $$
   1 - P(\text{not winning with } n) \geq 0.01
   $$
   This simplifies to:
   $$
   P(\text{not winning with } n) \leq 0.99
   $$
   Therefore:
   $$
   \left( \frac{17571}{17576} \right)^n \leq 0.99
   $$

4. **Taking the Natural Logarithm**:
   Taking the natural logarithm of both sides:
   $$
   n \cdot \ln\left( \frac{17571}{17576} \right) \leq \ln(0.99)
   $$
   Since $ \ln\left( \frac{17571}{17576} \right) $ is negative, we can rearrange:
   $$
   n \geq \frac{\ln(0.99)}{\ln\left( \frac{17571}{17576} \right)}
   $$

5. **Calculating the Values**:
   $$
   \ln(0.99) \approx -0.01005
   $$
   $$
   \ln\left( \frac{17571}{17576} \right) \approx \ln(1 - \frac{5}{17576}) \approx -\frac{5}{17576} \approx -0.000284
   $$
   Therefore:
   $$
   n \geq \frac{-0.01005}{-0.000284} \approx 35.4
   $$

Thus, Student B needs to buy at least **36 bottlecaps** to have a 1% chance of winning.

### Step 4: Calculate the Probability for Student C with 24 Bottlecaps

Now, we need to calculate the probability that Student C wins with 24 bottlecaps.

1. **Probability of Not Winning with 24 Bottlecaps**:
   $$
   P(\text{not winning with 24}) = \left( \frac{17571}{17576} \right)^{24}
   $$

2. **Calculating the Probability**:
   $$
   P(\text{not winning with 24}) \approx \left( 1 - \frac{5}{17576} \right)^{24} \approx \left( 1 - 0.000284 \right)^{24}
   $$
   Using the approximation $ (1 - x)^n \approx e^{-nx} $ for small $ x $:
   $$
   P(\text{not winning with 24}) \approx e^{-24 \cdot 0.000284} \approx e^{-0.006816} \approx 0.9932
   $$

3. **Probability of Winning with 24 Bottlecaps**:
   $$
   P(\text{winning with 24}) = 1 - P(\text{not winning with 24}) \approx 1 - 0.9932 \approx 0.0068
   $$

Thus, the probability that Student C wins with 24 bottlecaps is approximately **0.68%**.

### Summary of Results

- **Student B needs to buy at least 36 bottlecaps** to have a 1% chance of winning.
- **Student C has approximately a 0.68% chance of winning** with 24 bottlecaps.