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:

We are given that $$X$$ follows a binomial distribution $$B(n, p)$$, so the probability $$P(X = k)$$ is given by the formula:

$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \quad \text{for} \quad k = 0, 1, 2, \ldots, n $$

We are told that $$P(X = 2) = P(X = 3)$$. Substituting the binomial probability formula, we get:

$$ \binom{n}{2} p^2 (1-p)^{n-2} = \binom{n}{3} p^3 (1-p)^{n-3} $$

Expressing the binomial coefficients, $$\binom{n}{2} = \frac{n!}{2!(n-2)!}$$ and $$\binom{n}{3} = \frac{n!}{3!(n-3)!}$$, the equation becomes:

$$ \frac{n!}{2!(n-2)!} p^2 (1-p)^{n-2} = \frac{n!}{3!(n-3)!} p^3 (1-p)^{n-3} $$

Since $$n!$$ is common and non-zero (assuming $$n \geq 3$$ for the probabilities to be non-trivial), we can divide both sides by $$n!$$:

$$ \frac{1}{2!(n-2)!} p^2 (1-p)^{n-2} = \frac{1}{3!(n-3)!} p^3 (1-p)^{n-3} $$

We know $$2! = 2$$ and $$3! = 6$$, so:

$$ \frac{1}{2(n-2)!} p^2 (1-p)^{n-2} = \frac{1}{6(n-3)!} p^3 (1-p)^{n-3} $$

Rearranging terms, we move everything to one side:

$$ \frac{p^2 (1-p)^{n-2}}{2(n-2)!} - \frac{p^3 (1-p)^{n-3}}{6(n-3)!} = 0 $$

Factoring out common terms $$p^2 (1-p)^{n-3}$$ (assuming $$0 < p < 1$$ and $$n \geq 3$$ so that this factor is non-zero):

$$ p^2 (1-p)^{n-3} \left[ \frac{1-p}{2(n-2)!} - \frac{p}{6(n-3)!} \right] = 0 $$

This implies:

$$ \frac{1-p}{2(n-2)!} - \frac{p}{6(n-3)!} = 0 $$

So:

$$ \frac{1-p}{2(n-2)!} = \frac{p}{6(n-3)!} $$

Noting that $$(n-2)! = (n-2)(n-3)!$$, substitute:

$$ \frac{1-p}{2(n-2)(n-3)!} = \frac{p}{6(n-3)!} $$

Since $$(n-3)!$$ is common and non-zero, multiply both sides by $$(n-3)!$$:

$$ \frac{1-p}{2(n-2)} = \frac{p}{6} $$

Cross-multiplying:

$$ 6(1-p) = 2(n-2)p $$

Expanding both sides:

$$ 6 - 6p = 2p(n-2) $$

Bringing all terms to one side:

$$ 6 - 6p - 2p(n-2) = 0 $$

Factoring $$p$$:

$$ 6 = 6p + 2p(n-2) $$

Simplifying the right side:

$$ 6 = p \left[ 6 + 2(n-2) \right] $$

Expanding inside the brackets:

$$ 6 + 2(n-2) = 6 + 2n - 4 = 2n + 2 $$

So:

$$ 6 = p(2n + 2) $$

Factor out 2:

$$ 6 = 2p(n + 1) $$

Dividing both sides by 2:

$$ 3 = p(n + 1) $$

Therefore:

$$ p(n + 1) = 3 \quad \text{(Equation 1)} $$

The mean of $$X$$, denoted $$E(X)$$, for a binomial distribution is $$E(X) = np$$. From Equation 1, we solve for $$n$$:

$$ n + 1 = \frac{3}{p} $$

So:

$$ n = \frac{3}{p} - 1 = \frac{3 - p}{p} $$

Now substitute into $$E(X) = np$$:

$$ E(X) = \left( \frac{3 - p}{p} \right) p = 3 - p $$

Thus, the mean $$E(X)$$ is $$3 - p$$. Comparing with the options:

A. $$2 - p$$

B. $$3 - p$$

C. $$\frac{p}{2}$$

D. $$\frac{p}{3}$$

Hence, the correct answer is Option B.