Let a random variable X take values 0, 1, 2, 3 with $$P(X = 0) = P(X = 1) = p$$, $$P(X = 2) = P(X = 3) = q$$ and $$E(X^2) = 2E(X)$$. Then the value of $$8p - 1$$ is :

The random variable $$X$$ can take four values, so we write its probability distribution:

$$P(X = 0) = p,\; P(X = 1) = p,\; P(X = 2) = q,\; P(X = 3) = q.$$

Step 1: Use the fact that total probability equals $$1$$.
$$2p + 2q = 1 \;$$ $$-(1)$$

Step 2: Compute the expected value $$E(X)$$.
Formula: $$E(X) = \sum x_i P(X = x_i).$$
$$E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q.$$ $$-(2)$$

Step 3: Compute the second moment $$E(X^2).$$
Formula: $$E(X^2) = \sum x_i^2 P(X = x_i).$$
$$E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q.$$ $$-(3)$$

Step 4: Apply the given condition $$E(X^2) = 2E(X).$$
Using $$(2)$$ and $$(3)$$:
$$p + 13q = 2(p + 5q).$$
Simplify:
$$p + 13q = 2p + 10q$$
$$0 = 2p + 10q - p - 13q = p - 3q$$
$$\Rightarrow\; p = 3q.$$ $$-(4)$$

Step 5: Solve for $$p$$ and $$q$$ with $$(1)$$ and $$(4).$$
Substitute $$p = 3q$$ into $$(1):$$
$$2(3q) + 2q = 1$$
$$6q + 2q = 1$$
$$8q = 1 \;\Rightarrow\; q = \frac{1}{8}.$$
Then from $$(4):$$
$$p = 3q = 3 \times \frac{1}{8} = \frac{3}{8}.$$

Step 6: Evaluate the required expression $$8p - 1.$$/> $$8p - 1 = 8 \times \frac{3}{8} - 1 = 3 - 1 = 2.$$

Hence $$8p - 1 = 2$$, which corresponds to Option B.