1. **Problem statement:** We want to find the probability that a binomial random variable $X$ with parameters $n=10$ and $p=0.3$ takes the value 3, i.e., $P(X=3)$. 2. **Formula:** The probability mass function of a binomial distribution is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. 3. **Apply the formula:** $$P(X=3) = \binom{10}{3} (0.3)^3 (0.7)^7$$ 4. **Calculate the binomial coefficient:** $$\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$ 5. **Calculate powers:** $$(0.3)^3 = 0.027$$ $$(0.7)^7 \approx 0.0823543$$ 6. **Multiply all parts:** $$P(X=3) = 120 \times 0.027 \times 0.0823543 \approx 0.2679$$ 7. **Interpretation:** The probability $P(X=3)$ is approximately 0.27. **Final answer:** D) 0.27