**If the probability of success is** 𝟏/𝟏𝟎𝟎 **, how many trials are necessary in order that**

**probability of atleast one success is greater than half ?**

To solve the problem, we need to determine the number of trials necessary such that the probability of at least one success is greater than half (i.e., greater than \( \frac{1}{2} \)) when the probability of success in a single trial is \( \frac{1}{100} \).

### Step 1: Define the Probability of Success and Failure
Let:
- \( p \) be the probability of success in a single trial: \( p = \frac{1}{100} \).
- \( q \) be the probability of failure in a single trial: \( q = 1 - p = \frac{99}{100} \).

### Step 2: Probability of At Least One Success
The probability of at least one success in \( n \) trials is given by:
\[
P(\text{at least one success}) = 1 - P(\text{no success in } n \text{ trials})
\]
The probability of no success in \( n \) trials is:
\[
P(\text{no success in } n \text{ trials}) = q^n = \left(\frac{99}{100}\right)^n
\]
Thus, the probability of at least one success is:
\[
P(\text{at least one success}) = 1 - \left(\frac{99}{100}\right)^n
\]
We want this probability to be greater than \( \frac{1}{2} \), so:
\[
1 - \left(\frac{99}{100}\right)^n > \frac{1}{2}
\]
\[
\left(\frac{99}{100}\right)^n < \frac{1}{2}
\]

### Step 3: Solve the Inequality
Take the natural logarithm on both sides:
\[
\ln\left(\left(\frac{99}{100}\right)^n\right) < \ln\left(\frac{1}{2}\right)
\]
Simplify the logarithm:
\[
n \cdot \ln\left(\frac{99}{100}\right) < \ln\left(\frac{1}{2}\right)
\]
Since \( \ln\left(\frac{99}{100}\right) \) is negative, dividing both sides by it will reverse the inequality:
\[
n > \frac{\ln\left(\frac{1}{2}\right)}{\ln\left(\frac{99}{100}\right)}
\]
Substitute the logarithm values:
\[
n > \frac{\ln\left(0.5\right)}{\ln\left(0.99\right)}
\]

### Step 4: Calculate the Value of \( n \)
Using approximate values:
\[
\ln(0.5) \approx -0.6931, \quad \ln(0.99) \approx -0.01005
\]
So,
\[
n > \frac{-0.6931}{-0.01005} \approx 68.95
\]

### Conclusion
Since \( n \) must be an integer, \( n \) must be at least 69. Therefore, **69 trials** are necessary to ensure that the probability of at least one success is greater than half.

Would you like further details on this calculation or have any questions? Here are five related questions to consider:

1. What is the probability of getting exactly one success in 69 trials?
2. How does the probability of success change if the number of trials is increased to 100?
3. What is the minimum number of trials required if the probability of success is \( \frac{1}{200} \)?
4. How would you calculate the expected number of successes in 69 trials?
5. What is the complementary probability, and how is it used in probability calculations?

**Tip:** Understanding the complementary probability is key to solving problems involving "at least one" scenarios.

Learn how to determine the number of trials required so that the probability of at least one success is greater than half. This problem involves probability theory with a specific focus on calculating probabilities in repeated trials.

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