If the sum of the mean and variance of a binomial distribution for 5 trials is `(75)/(16)`, find the binomial distribution.

Answer

To solve the problem, we need to find the binomial distribution given that the sum of the mean and variance for 5 trials is \( \frac{75}{16} \).

### Step-by-step Solution:

1. **Understand the Mean and Variance of a Binomial Distribution**:
- For a binomial distribution with \( n \) trials and probability \( p \) of success, the mean \( \mu \) is given by:
\[ \mu = n \cdot p \]
- The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
- Here, \( q = 1 - p \).

2. **Set up the Equation**:
- We know that the sum of the mean and variance is:
\[ \mu + \sigma^2 = \frac{75}{16} \]
- Substituting the formulas for mean and variance, we get:
\[ n \cdot p + n \cdot p \cdot q = \frac{75}{16} \]
- Given \( n = 5 \), we can rewrite this as:
\[ 5p + 5pq = \frac{75}{16} \]

3. **Simplify the Equation**:
- Factor out \( 5p \):
\[ 5p(1 + q) = \frac{75}{16} \]
- Since \( q = 1 - p \), we have:
\[ 1 + q = 1 + (1 - p) = 2 - p \]
- Substituting this back gives:
\[ 5p(2 - p) = \frac{75}{16} \]

4. **Multiply through by 16 to eliminate the fraction**:
- This results in:
\[ 80p(2 - p) = 75 \]
- Expanding this gives:
\[ 160p - 80p^2 = 75 \]
- Rearranging leads to:
\[ 80p^2 - 160p + 75 = 0 \]

5. **Solve the Quadratic Equation**:
- Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 80 \), \( b = -160 \), and \( c = 75 \).
\[ p = \frac{160 \pm \sqrt{(-160)^2 - 4 \cdot 80 \cdot 75}}{2 \cdot 80} \]
- Calculate the discriminant:
\[ 25600 - 24000 = 1600 \]
- Thus,
\[ p = \frac{160 \pm 40}{160} \]
- This gives us two potential solutions:
\[ p = \frac{200}{160} = \frac{5}{4} \quad \text{(not valid, as probability cannot exceed 1)} \]
\[ p = \frac{120}{160} = \frac{3}{4} \]

6. **Find \( q \)**:
- Since \( q = 1 - p \):
\[ q = 1 - \frac{3}{4} = \frac{1}{4} \]

7. **Construct the Binomial Distribution**:
- Now that we have \( n = 5 \), \( p = \frac{3}{4} \), and \( q = \frac{1}{4} \), we can find the probability distribution \( P(X = k) \) for \( k = 0, 1, 2, 3, 4, 5 \):
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- Calculate for each \( k \):
- \( P(X = 0) = \binom{5}{0} \left(\frac{3}{4}\right)^0 \left(\frac{1}{4}\right)^5 = 1 \cdot 1 \cdot \frac{1}{1024} = \frac{1}{1024} \)
- \( P(X = 1) = \binom{5}{1} \left(\frac{3}{4}\right)^1 \left(\frac{1}{4}\right)^4 = 5 \cdot \frac{3}{4} \cdot \frac{1}{256} = \frac{15}{1024} \)
- \( P(X = 2) = \binom{5}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^3 = 10 \cdot \frac{9}{16} \cdot \frac{1}{64} = \frac{90}{1024} \)
- \( P(X = 3) = \binom{5}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^2 = 10 \cdot \frac{27}{64} \cdot \frac{1}{16} = \frac{270}{1024} \)
- \( P(X = 4) = \binom{5}{4} \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^1 = 5 \cdot \frac{81}{256} \cdot \frac{1}{4} = \frac{405}{1024} \)
- \( P(X = 5) = \binom{5}{5} \left(\frac{3}{4}\right)^5 \left(\frac{1}{4}\right)^0 = 1 \cdot \frac{243}{1024} \cdot 1 = \frac{243}{1024} \)

### Final Probability Distribution Table:
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{1}{1024} \\ 1 & \frac{15}{1024} \\ 2 & \frac{90}{1024} \\ 3 & \frac{270}{1024} \\ 4 & \frac{405}{1024} \\ 5 & \frac{243}{1024} \\ \hline \end{array} \]