The following is the probability distribution of $X$: $\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3…

The probability distribution remains valid when all probabilities are non-negative. This requires $P(X=0) = \frac{1+p}{5} \geq 0$ implying $p \geq -1$, $P(X=1) = \frac{2-2p}{5} \geq 0$ implying $p \leq 1$, $P(X=2) = \frac{2-p}{5} \geq 0$ implying $p \leq 2$, and $P(X=3) = \frac{2p}{5} \geq 0$ implying $p \geq 0$.

Combining these inequalities yields the valid range $0 \leq p \leq 1$.

The minimum value of $p$ is therefore $0$. Substituting $p = 0$ gives the distribution:
$P(X=0) = \frac{1}{5}$, $P(X=1) = \frac{2}{5}$, $P(X=2) = \frac{2}{5}$, $P(X=3) = 0$.

The expected value is calculated as:
$E(X) = 0 \cdot \frac{1}{5} + 1 \cdot \frac{2}{5} + 2 \cdot \frac{2}{5} + 3 \cdot 0 = \frac{2}{5} + \frac{4}{5} = \frac{6}{5}$.

Multiplying by 5 gives $5E(X) = 5 \cdot \frac{6}{5} = 6$.