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 expected value \( E(X) \) of a binomial distribution \( B(n, p) \) given that \( P(X = 2) = P(X = 3) \).

### Step-by-Step Solution:

1. **Understand the Binomial Probability Formula**:
The probability mass function for 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**:
Since we know that \( P(X = 2) = P(X = 3) \), we can write:
\[ \binom{n}{2} p^2 (1-p)^{n-2} = \binom{n}{3} p^3 (1-p)^{n-3} \]

3. **Substitute the Binomial Coefficients**:
The binomial coefficients can be expressed as:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \quad \text{and} \quad \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
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. **Simplify the Equation**:
Cancel \( n(n-1) \) from both sides (assuming \( n(n-1) \neq 0 \)):
\[ \frac{1}{2} p^2 (1-p) = \frac{n-2}{6} p^3 \]
Rearranging gives:
\[ 3(1-p) = (n-2)p \]

5. **Rearranging to Find n**:
Rearranging the equation:
\[ 3 - 3p = np - 2p \]
This simplifies to:
\[ 3 = np - 2p + 3p \implies 3 = np + p \implies 3 = p(n + 1) \]
Thus, we can express \( n \) in terms of \( p \):
\[ n = \frac{3}{p} - 1 \]

6. **Calculate the Expected Value**:
The expected value \( E(X) \) of a binomial distribution is given by:
\[ E(X) = np \]
Substituting our expression for \( n \):
\[ E(X) = p\left(\frac{3}{p} - 1\right) = 3 - p \]

### Final Answer:
Thus, the expected value \( E(X) \) is:
\[ \boxed{3 - p} \]