Step 1: Recall the variance formula for a binomial distribution.
For $X \sim B(n, p)$, the variance is $\operatorname{Var}(X) = npq$ where $q = 1 - p$. This is because each trial contributes $pq$ to the variance, and there are $n$ independent trials.
Step 2: Set up the equation.
Here $n = 15$ and $\operatorname{Var}(X) = 3.15$, so \[ 15 \cdot p(1-p) = 3.15 \]
Step 3: Simplify.
Divide both sides by 15: \[ p(1-p) = \frac{3.15}{15} = 0.21 \]
Step 4: Form a quadratic equation.
Expanding: $p - p^2 = 0.21$, which rearranges to \[ p^2 - p + 0.21 = 0 \]
Step 5: Solve the quadratic by factorisation.
We look for two numbers whose product is $0.21$ and whose sum is $1$. Those numbers are $0.3$ and $0.7$ (since $0.3 \times 0.7 = 0.21$ and $0.3 + 0.7 = 1$). So \[ (p - 0.3)(p - 0.7) = 0 \] giving $p = 0.3$ or $p = 0.7$.
Step 6: Verify both solutions are valid probabilities.
Both $0.3$ and $0.7$ lie in $[0,1]$. Also, $15 \times 0.3 \times 0.7 = 15 \times 0.21 = 3.15$. Confirmed.
Step 7: State the final answer.
The two possible values of $p$ are \[ \boxed{0.3 \text{ and } 0.7} \]