To solve the problem, we use the properties of the Binomial distribution \( B(n, p) \):
- Mean (\( \mu \)): \( \mu = np \),
- Variance (\( \sigma^2 \)): \( \sigma^2 = npq \), where \( q = 1
- p \).
Given:
1. The sum of the mean and variance is 5:
\[
\mu + \sigma^2 = 5 \quad \Rightarrow \quad np + npq = 5.
\]
2. The product of the mean and variance is 6:
\[
\mu \cdot \sigma^2 = 6 \quad \Rightarrow \quad np \cdot npq = 6.
\]
Step 1: Simplify the equations
From the first equation:
\[
np + npq = 5 \quad \Rightarrow \quad np(1 + q) = 5.
\]
From the second equation:
\[
np \cdot npq = 6 \quad \Rightarrow \quad n^2 p^2 q = 6.
\]
Step 2: Express \( q \) in terms of \( p \)
Since \( q = 1
- p \), substitute into the first equation:
\[
np(1 + 1
- p) = 5 \quad \Rightarrow \quad np(2
- p) = 5.
\]
Step 3: Solve for \( n \) and \( p \)
From the second equation:
\[
n^2 p^2 q = 6.
\]
Substitute \( q = 1
- p \):
\[
n^2 p^2 (1
- p) = 6.
\]
From the first equation, solve for \( np \):
\[
np = \frac{5}{2
- p}.
\]
Substitute \( np = \frac{5}{2
- p} \) into the second equation:
\[
\left( \frac{5}{2
- p} \right)^2 p^2 (1
- p) = 6.
\]
Simplify:
\[
\frac{25 p^2 (1
- p)}{(2
- p)^2} = 6.
\]
Multiply through by \( (2
- p)^2 \):
\[
25 p^2 (1
- p) = 6 (2
- p)^2.
\]
Expand \( (2
- p)^2 \):
\[
25 p^2 (1
- p) = 6 (4
- 4p + p^2).
\]
Simplify:
\[
25 p^2
- 25 p^3 = 24
- 24 p + 6 p^2.
\]
Rearrange:
\[
-25 p^3 + 19 p^2 + 24 p
- 24 = 0.
\]
Step 4: Solve the cubic equation
The cubic equation \(
-25 p^3 + 19 p^2 + 24 p
- 24 = 0 \) can be solved numerically or by inspection. One solution is \( p = 1 \), but this is not valid since \( p \) must be between 0 and 1. Another solution is \( p = \frac{3}{5} \).
Substitute \( p = \frac{3}{5} \) into \( np = \frac{5}{2
- p} \):
\[
n \cdot \frac{3}{5} = \frac{5}{2
- \frac{3}{5}} = \frac{5}{\frac{7}{5}} = \frac{25}{7}.
\]
Thus:
\[
n = \frac{25}{7} \cdot \frac{5}{3} = \frac{125}{21}.
\]
Step 5: Compute \( 6(n + p
- q) \)
Substitute \( n = \frac{125}{21} \), \( p = \frac{3}{5} \), and \( q = 1
- p = \frac{2}{5} \):
\[
n + p
- q = \frac{125}{21} + \frac{3}{5}
- \frac{2}{5} = \frac{125}{21} + \frac{1}{5}.
\]
Find a common denominator:
\[
n + p
- q = \frac{125 \cdot 5 + 1 \cdot 21}{105} = \frac{625 + 21}{105} = \frac{646}{105}.
\]
Multiply by 6:
\[
6(n + p
- q) = 6 \cdot \frac{646}{105} = \frac{3876}{105} = 36.914.
\]
This does not match any of the options. Let's re
-examine the problem.
Step 6: Re
-examining the problem
The correct approach is to use the given equations:
1. \( np + npq = 5 \),
2. \( np \cdot npq = 6 \).
Substitute \( q = 1
- p \):
\[
np + np(1
- p) = 5 \quad \Rightarrow \quad np(2
- p) = 5,
\]
\[
np \cdot np(1
- p) = 6 \quad \Rightarrow \quad n^2 p^2 (1
- p) = 6.
\]
Let \( x = np \). Then:
\[
x(2
- p) = 5,
\]
\[
x^2 (1
- p) = 6.
\]
From the first equation:
\[
x = \frac{5}{2
- p}.
\]
Substitute into the second equation:
\[
\left( \frac{5}{2
- p} \right)^2 (1
- p) = 6.
\]
Simplify:
\[
\frac{25 (1
- p)}{(2
- p)^2} = 6.
\]
Multiply through by \( (2
- p)^2 \):
\[
25 (1
- p) = 6 (2
- p)^2.
\]
Expand \( (2
- p)^2 \):
\[
25 (1
- p) = 6 (4
- 4p + p^2).
\]
Simplify:
\[
25
- 25 p = 24
- 24 p + 6 p^2.
\]
Rearrange:
\[
6 p^2 + p
- 1 = 0.
\]
Solve the quadratic equation:
\[
p = \frac{
-1 \pm \sqrt{1 + 24}}{12} = \frac{
-1 \pm 5}{12}.
\]
Thus:
\[
p = \frac{4}{12} = \frac{1}{3} \quad \text{or} \quad p = \frac{
-6}{12} =
-\frac{1}{2}.
\]
Since \( p \) must be between 0 and 1, \( p = \frac{1}{3} \).
Substitute \( p = \frac{1}{3} \) into \( x = \frac{5}{2
- p} \):
\[
x = \frac{5}{2
- \frac{1}{3}} = \frac{5}{\frac{5}{3}} = 3.
\]
Thus:
\[
np = 3 \quad \Rightarrow \quad n = \frac{3}{p} = \frac{3}{\frac{1}{3}} = 9.
\]
Now, compute \( 6(n + p
- q) \):
\[
n + p
- q = 9 + \frac{1}{3}
- \frac{2}{3} = 9
- \frac{1}{3} = \frac{26}{3}.
\]
Multiply by 6:
\[
6(n + p
- q) = 6 \cdot \frac{26}{3} = 52.
\]
Final Answer:
\[
\boxed{52}
\]