If `X` has a binomial distribution, `B (n,p)` with parameters `n` and `p` such that `P(X=2)=P(X=3)`, then `E(X)` , the mean of variable `X`, is

Options

A. `2 - p`
B. `3 - p `
C. `p/2`
D. `p/3`

Answer

To solve the problem, we need to find the mean \( E(X) \) of a binomial distribution \( B(n, p) \) given that \( P(X=2) = P(X=3) \).

### Step-by-Step Solution:

1. **Write the Probability Mass Function (PMF)**:
The PMF of a binomial distribution is given by:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

2. **Set Up the Equation**:
According to the problem, we have:
\[ P(X=2) = P(X=3) \]
Therefore, we can write:
\[ \binom{n}{2} p^2 (1-p)^{n-2} = \binom{n}{3} p^3 (1-p)^{n-3} \]

3. **Substituting the Binomial Coefficients**:
The binomial coefficients are:
\[ \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Substituting these into the equation gives:
\[ \frac{n(n-1)}{2} p^2 (1-p)^{n-2} = \frac{n(n-1)(n-2)}{6} p^3 (1-p)^{n-3} \]

4. **Cancel Common Terms**:
We can cancel \( n(n-1) \) from both sides (assuming \( n \geq 2 \)):
\[ \frac{1}{2} p^2 (1-p)^{n-2} = \frac{(n-2)}{6} p^3 (1-p)^{n-3} \]
Now, we can also cancel \( (1-p)^{n-3} \) from both sides (assuming \( p \neq 1 \)):
\[ \frac{1}{2} p^2 (1-p) = \frac{(n-2)}{6} p^3 \]

5. **Rearranging the Equation**:
Multiply both sides by 6 to eliminate the fraction:
\[ 3p^2(1-p) = (n-2)p^3 \]
Rearranging gives:
\[ 3p^2 - 3p^3 = (n-2)p^3 \]
Combining like terms:
\[ 3p^2 = (n-2 + 3)p^3 \quad \Rightarrow \quad 3p^2 = (n+1)p^3 \]

6. **Dividing by \( p^2 \) (assuming \( p \neq 0 \))**:
\[ 3 = (n+1)p \]
Thus, we can express \( p \) in terms of \( n \):
\[ p = \frac{3}{n+1} \]

7. **Finding the Mean**:
The mean \( E(X) \) of a binomial distribution is given by:
\[ E(X) = n \cdot p = n \cdot \frac{3}{n+1} \]
Therefore:
\[ E(X) = \frac{3n}{n+1} \]

### Final Answer:
The mean \( E(X) \) is \( \frac{3n}{n+1} \).