The random variable X follows binomialdistribution `B(n,p)` ,for which the difference of the mean and the variance is `1`. If `2P(X =2) +3P(X = 1)`, the `n^2 P(X<1) ` is equal to

Options

A. `12`
B. `11`
C. `15`
D. `16`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Understand the given conditions
We know that the random variable \( X \) follows a binomial distribution \( B(n, p) \). The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
- Mean: \( \mu = np \)
- Variance: \( \sigma^2 = np(1-p) \)

We are given that the difference between the mean and variance is 1:
\[ np - np(1-p) = 1 \]

### Step 2: Simplify the equation
Substituting the expressions for mean and variance into the equation:
\[ np - np + np^2 = 1 \]
This simplifies to:
\[ np^2 = 1 \quad \text{(1)} \]

### Step 3: Use the second condition
We are also given the condition:
\[ 2P(X = 2) = 3P(X = 1) \]
Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
We can express \( P(X = 2) \) and \( P(X = 1) \):
\[ P(X = 2) = \binom{n}{2} p^2 (1-p)^{n-2} \]
\[ P(X = 1) = \binom{n}{1} p (1-p)^{n-1} \]

Substituting these into the equation:
\[ 2 \cdot \binom{n}{2} p^2 (1-p)^{n-2} = 3 \cdot \binom{n}{1} p (1-p)^{n-1} \]

### Step 4: Substitute binomial coefficients
The binomial coefficients are:
\[ \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{1} = n \]
Substituting these into the equation gives:
\[ 2 \cdot \frac{n(n-1)}{2} p^2 (1-p)^{n-2} = 3n p (1-p)^{n-1} \]
This simplifies to:
\[ n(n-1)p^2 (1-p)^{n-2} = 3n p (1-p)^{n-1} \]

### Step 5: Cancel common terms
Assuming \( n \neq 0 \) and \( p \neq 0 \):
\[ (n-1)p(1-p) = 3(1-p) \]
Dividing both sides by \( (1-p) \) (assuming \( 1-p \neq 0 \)):
\[ (n-1)p = 3 \]
Thus, we have:
\[ np - p = 3 \quad \text{(2)} \]

### Step 6: Solve the equations
From equations (1) and (2):
1. \( np^2 = 1 \)
2. \( np - p = 3 \)

From equation (2), we can express \( np \):
\[ np = p + 3 \]
Substituting this into equation (1):
\[ p(p + 3) = 1 \]
This expands to:
\[ p^2 + 3p - 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \]

### Step 8: Calculate \( n \)
Using \( np = p + 3 \):
Substituting \( p \) back into \( np = p + 3 \) allows us to find \( n \).

### Step 9: Calculate \( n^2 P(X < 1) \)
To find \( n^2 P(X < 1) \):
\[ P(X < 1) = P(X = 0) = \binom{n}{0} p^0 (1-p)^n = (1-p)^n \]
Thus, we need to compute \( n^2(1-p)^n \).

### Final Answer
After calculating the above expressions, we can find the required value.