In a binomial distribution `B(n , p=1/4)` , if the probability of at least one success is greater than or equal to `9/(10)` , then n is greater than (1) `1/((log)_(10)^4-(log)_(10)^3)`
(2) `1/((log)_(10)^4+(log)_(10)^3)`
(3) `9/((log)_(10)^4-(log)_(10)^3)`
(4) `4/((log)_(10)^4-(log)_(10)^3)`

Options

A. `(1)/("log"_(10)4-"log"_(10)3)`
B. `(1)/("log"_(10)4+"log"_(10)3)`
C. `(9)/("log"_(10)4-"log"_(10)3)`
D. `(4)/("log"_(10)4-"log"_(10)3)`

Explanation

To solve the problem, we need to find the value of \( n \) such that the probability of at least one success in a binomial distribution \( B(n, p = \frac{1}{4}) \) is greater than or equal to \( \frac{9}{10} \).

### Step-by-step Solution:

1. **Understanding the Probability of At Least One Success**:
The probability of at least one success can be expressed as:
\[ P(X \geq 1) = 1 - P(X = 0) \]
where \( P(X = 0) \) is the probability of zero successes.

2. **Calculating \( P(X = 0) \)**:
For a binomial distribution, the probability of zero successes is given by:
\[ P(X = 0) = \binom{n}{0} p^0 (1-p)^n = (1-p)^n \]
Here, \( p = \frac{1}{4} \) and \( 1 - p = \frac{3}{4} \). Thus:
\[ P(X = 0) = \left(\frac{3}{4}\right)^n \]

3. **Setting Up the Inequality**:
We want:
\[ 1 - P(X = 0) \geq \frac{9}{10} \]
Substituting \( P(X = 0) \):
\[ 1 - \left(\frac{3}{4}\right)^n \geq \frac{9}{10} \]

4. **Rearranging the Inequality**:
Rearranging gives:
\[ \left(\frac{3}{4}\right)^n \leq \frac{1}{10} \]

5. **Taking Logarithms**:
Taking logarithms on both sides:
\[ \log\left(\left(\frac{3}{4}\right)^n\right) \leq \log\left(\frac{1}{10}\right) \]
This simplifies to:
\[ n \cdot \log\left(\frac{3}{4}\right) \leq \log\left(\frac{1}{10}\right) \]

6. **Solving for \( n \)**:
Since \( \log\left(\frac{3}{4}\right) \) is negative, we can divide by it, which reverses the inequality:
\[ n \geq \frac{\log\left(\frac{1}{10}\right)}{\log\left(\frac{3}{4}\right)} \]

7. **Simplifying the Logs**:
We know:
\[ \log\left(\frac{1}{10}\right) = -1 \quad \text{(since } 10^1 = 10\text{)} \]
and using the property of logs:
\[ \log\left(\frac{3}{4}\right) = \log(3) - \log(4) \]
Thus:
\[ n \geq \frac{-1}{\log(3) - \log(4)} = \frac{1}{\log(4) - \log(3)} \]

8. **Final Result**:
This can be expressed in terms of base 10 logarithms:
\[ n \geq \frac{1}{\log_{10}(4) - \log_{10}(3)} \]

### Conclusion:
The value of \( n \) must be greater than \( \frac{1}{\log_{10}(4) - \log_{10}(3)} \).