1. **Problem Statement:** We are given that 40% of all murders are domestic homicides, and a study selects 10 murder cases. We want to find the probability that between 1 and 3 of these cases are domestic homicides. 2. **Identify the distribution:** This is a binomial probability problem where: - Number of trials $n = 10$ - Probability of success (domestic homicide) $p = 0.4$ - We want $P(1 \leq X \leq 3)$ where $X$ is the number of domestic homicides. 3. **Binomial probability formula:** $$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$ where $\binom{n}{k}$ is the binomial coefficient. 4. **Using the binomial probability table:** We look up the probabilities for $X=1$, $X=2$, and $X=3$ and sum them: $$ P(1 \leq X \leq 3) = P(X=1) + P(X=2) + P(X=3) $$ 5. **Calculate each term:** - $P(X=1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 10 \times 0.4 \times 0.6^9$ - $P(X=2) = \binom{10}{2} (0.4)^2 (0.6)^8 = 45 \times 0.16 \times 0.6^8$ - $P(X=3) = \binom{10}{3} (0.4)^3 (0.6)^7 = 120 \times 0.064 \times 0.6^7$ 6. **Evaluate powers:** - $0.6^9 \approx 0.0101$ - $0.6^8 \approx 0.0168$ - $0.6^7 \approx 0.02799$ 7. **Calculate probabilities:** - $P(X=1) \approx 10 \times 0.4 \times 0.0101 = 0.0404$ - $P(X=2) \approx 45 \times 0.16 \times 0.0168 = 0.12096$ - $P(X=3) \approx 120 \times 0.064 \times 0.02799 = 0.2150$ 8. **Sum the probabilities:** $$ P(1 \leq X \leq 3) \approx 0.0404 + 0.12096 + 0.2150 = 0.37636 $$ **Final answer:** The probability that between 1 and 3 of the cases are domestic homicides is approximately **0.376**.