Question 1205136: Student A was tasked to program an algorithm for a number generator. The generator will be used by A Soda Company for their bottlecaps.
Soda Company will be holding a bottlecap lottery where 5 select bottlecaps will be winning combinations. There would be three characters printed on the bottlecaps with each character having the chance to be repeated. Those that are not winning codes are duplicated 10 times.
If Student B, was to join, how many bottles of Soda will he need to buy to have a 1% chance at winning.
And if Student C would join too, what is the chance that 24 bottlecaps have at winning?
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! 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.
|
|
|
