A random variable x follows binomial distribution with mean a and variance b then observe the following statements.
Statement-I : `a gt b gt 0`
Statement-II : `(a^(2))/(a-b)` is a positive integer
Statement-III : `a+b=1`
which of the above statments are true.

Options

A. only I, II
B. only II, III
C. only I, III
D. all the I, II, III

Answer

To solve the problem, we need to analyze the statements given the properties of a binomial distribution. A random variable \( X \) that follows a binomial distribution has the following characteristics:

1. **Mean** \( \mu = np \)
2. **Variance** \( \sigma^2 = npq \) where \( q = 1 - p \)

Given that the mean is \( a \) and the variance is \( b \), we can express these as:

- \( a = np \)
- \( b = npq = np(1-p) \)

### Step 1: Analyze Statement-I: \( a > b > 0 \)

From the definitions:
- Since \( a = np \) and \( b = np(1-p) \), we can see that \( b \) is positive if \( p \) is between 0 and 1 (i.e., \( 0 < p < 1 \)).
- We can also express \( b \) in terms of \( a \):
\[ b = np(1-p) = a(1-p) \]
Since \( 1-p \) is positive for \( 0 < p < 1 \), it follows that \( b < a \).

Thus, we have \( a > b > 0 \) is true.

### Step 2: Analyze Statement-II: \( \frac{a^2}{a-b} \) is a positive integer

Using the expressions for \( a \) and \( b \):
- We know \( b = a(1 - \frac{b}{a}) \) implies \( a - b = a - np(1-p) = np + np^2 = np(1 + p) \).

Now substituting:
\[ \frac{a^2}{a-b} = \frac{(np)^2}{np(1 + p)} = \frac{np}{1 + p} \]
Since \( np \) is positive (as discussed), \( \frac{np}{1 + p} \) is also positive.

To check if it is an integer, we note that since \( n \) is the number of trials (an integer), and \( p \) is a probability (which can be rational), \( \frac{np}{1+p} \) can be a positive integer depending on the values of \( n \) and \( p \).

Thus, Statement-II is also true.

### Step 3: Analyze Statement-III: \( a + b = 1 \)

From our definitions:
\[ a + b = np + np(1-p) = np + np - np^2 = np(2 - p) \]
This does not equal 1 in general, as it depends on the values of \( n \) and \( p \). Therefore, Statement-III is false.

### Conclusion

- **Statement-I**: True
- **Statement-II**: True
- **Statement-III**: False

Thus, the correct statements are I and II.

### Final Answer
The true statements are I and II.

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