If the sum and the product of the mean and variance of a Binomial Distribution are `1-8` and `0.8` respectively, find the probability distribution and the probability of at least one success

Answer

To solve the problem, we need to find the probability distribution and the probability of at least one success in a binomial distribution given the sum and product of the mean and variance.

### Step-by-Step Solution:

1. **Understand the Binomial Distribution**:
- The mean (μ) of a binomial distribution is given by \( μ = np \).
- The variance (σ²) is given by \( σ² = npq \), where \( q = 1 - p \).

2. **Set Up the Given Information**:
- We are given that the sum of the mean and variance is \( 1.8 \):
\[ np + npq = 1.8 \]
- We are also given that the product of the mean and variance is \( 0.8 \):
\[ np \cdot npq = 0.8 \]

3. **Rewrite the Equations**:
- From the first equation, factor out \( np \):
\[ np(1 + q) = 1.8 \]
- From the second equation, rewrite it as:
\[ n^2 p^2 q = 0.8 \]

4. **Express \( q \) in Terms of \( p \)**:
- Since \( q = 1 - p \), substitute \( q \) into the equations:
\[ np(1 + (1 - p)) = 1.8 \implies np(2 - p) = 1.8 \]
- The second equation becomes:
\[ n^2 p^2 (1 - p) = 0.8 \]

5. **Solve for \( n \) and \( p \)**:
- From \( np(2 - p) = 1.8 \):
\[ np = \frac{1.8}{2 - p} \]
- Substitute \( np \) into the second equation:
\[ \left(\frac{1.8}{2 - p}\right)^2 (1 - p) = 0.8 \]

6. **Cross Multiply and Rearrange**:
- This leads to a quadratic equation in terms of \( p \):
\[ 1.8^2(1 - p) = 0.8(2 - p)^2 \]
- Expand and simplify to find the values of \( p \).

7. **Find the Roots of the Quadratic**:
- After simplification, you will find two potential values for \( q \):
\[ q = \frac{4}{5} \quad \text{and} \quad q = \frac{5}{4} \]
- Since \( q \) must be less than 1, we take \( q = \frac{4}{5} \) and thus \( p = \frac{1}{5} \).

8. **Calculate \( n \)**:
- Substitute \( p \) back into the first equation to find \( n \):
\[ n \cdot \frac{1}{5} \cdot \left(1 + \frac{4}{5}\right) = 1.8 \implies n \cdot \frac{1}{5} \cdot \frac{9}{5} = 1.8 \]
- Solve for \( n \):
\[ n = 5 \]

9. **Determine the Probability Distribution**:
- The parameters of the binomial distribution are \( n = 5 \) and \( p = \frac{1}{5} \).
- The probability distribution is \( X \sim B(5, \frac{1}{5}) \).

10. **Calculate the Probability of At Least One Success**:
- The probability of at least one success is given by:
\[ P(X \geq 1) = 1 - P(X = 0) \]
- Calculate \( P(X = 0) \):
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^5 = \left(\frac{4}{5}\right)^5 = \frac{1024}{3125} \]
- Thus,
\[ P(X \geq 1) = 1 - \frac{1024}{3125} = \frac{2101}{3125} \approx 0.672 \]

### Final Answer:
- The probability distribution is \( X \sim B(5, \frac{1}{5}) \).
- The probability of at least one success is approximately \( 0.672 \).