1. **State the problem:** 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 (PMF) of a binomial distribution is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k}$ is the binomial coefficient representing the number of ways to choose $k$ successes out of $n$ trials. 3. **Apply the formula:** Substitute $n=10$, $k=3$, and $p=0.3$: $$P(X=3) = \binom{10}{3} (0.3)^3 (0.7)^7$$ 4. **Calculate the binomial coefficient:** $$\binom{10}{3} = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$ 5. **Calculate powers:** $$(0.3)^3 = 0.027$$ $$(0.7)^7 \approx 0.0823543$$ 6. **Multiply all parts:** $$P(X=3) = 120 \times 0.027 \times 0.0823543 \approx 0.2673$$ **Final answer:** $$P(X=3) \approx 0.2673$$