If the probability of success in one Bernoulli trial is 0.01, how many independent trials are

necessary in order that the probability of at least one success is greater than 1

The probability of success in one Bernoulli trial is \( p = 0.01 \). If we conduct \( n \) independent trials, the probability of getting at least one success can be calculated as follows:

### Step 1: Probability of No Success in \( n \) Trials
The probability of no success in a single trial is \( 1 - p = 0.99 \).

The probability of no success in all \( n \) independent trials is:
\[
P(\text{no success in } n \text{ trials}) = (0.99)^n
\]

### Step 2: Probability of At Least One Success
The probability of getting at least one success in \( n \) trials is:
\[
P(\text{at least one success}) = 1 - P(\text{no success in } n \text{ trials}) = 1 - (0.99)^n
\]

### Step 3: Setting the Condition
We need to find the minimum \( n \) such that the probability of at least one success is greater than 1. Mathematically, this means:
\[
1 - (0.99)^n > 1
\]

However, the probability \( 1 - (0.99)^n \) will never exceed 1, as it is bounded between 0 and 1. It seems there might be a misunderstanding in the question. If the question intends to ask when the probability of at least one success is greater than some value close to 1 (e.g., 0.99), then we should set:
\[
1 - (0.99)^n > 0.99
\]

### Step 4: Solving for \( n \)
Now, let's solve for \( n \):
\[
1 - (0.99)^n > 0.99
\]
\[
(0.99)^n < 0.01
\]

Taking the natural logarithm on both sides:
\[
\ln(0.99^n) < \ln(0.01)
\]
\[
n \ln(0.99) < \ln(0.01)
\]
Since \( \ln(0.99) \) is negative, divide both sides by \( \ln(0.99) \) and reverse the inequality:
\[
n > \frac{\ln(0.01)}{\ln(0.99)}
\]

Now, calculate the value:
\[
n > \frac{\ln(0.01)}{\ln(0.99)} \approx \frac{-4.605}{-0.01005} \approx 458.2
\]

So, \( n \) must be greater than 458. Since \( n \) must be an integer, the minimum \( n \) is 459.

### Conclusion
You need **459** independent trials for the probability of at least one success to be greater than 0.99.

Would you like further clarification, or do you have any other questions?

---

1. What is the general formula for the probability of at least one success in \( n \) trials?
2. How does the required number of trials change if the probability of success per trial is higher?
3. Can this concept be applied to other distributions?
4. How does the number of trials affect the variance in the number of successes?
5. What is the significance of the natural logarithm in this context?

**Tip:** When solving problems involving probabilities close to 1, it's often helpful to use logarithms to simplify the calculation.

Understand how to calculate the probability of success in Bernoulli trials when the probability of success per trial is given. Explore detailed step-by-step solutions and applications of probability theory. Suitable for advanced high school students and beyond.

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