1. **Problem Statement:** We are given a binomial probability scenario where a surgical technique has a 70% chance of success per patient, and it is performed on 7 patients. We want to find: a. Probability exactly 5 patients succeed. b. Probability fewer than 5 patients succeed. c. Probability at least 3 patients succeed. 2. **Formula and Explanation:** The binomial probability formula is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where: - $n$ = number of trials (patients), here 7 - $k$ = number of successes - $p$ = probability of success, here 0.7 - $\binom{n}{k}$ = combination of $n$ choose $k$ 3. **Calculations:** **a. Exactly 5 successes:** $$P(X=5) = \binom{7}{5} (0.7)^5 (0.3)^2$$ Calculate $\binom{7}{5} = \frac{7!}{5!2!} = 21$ $$P(X=5) = 21 \times (0.7)^5 \times (0.3)^2$$ **b. Less than 5 successes:** $$P(X \leq 4) = \sum_{k=0}^4 P(X=k)$$ Calculate each term using the binomial formula and sum. **c. At least 3 successes:** $$P(X \geq 3) = 1 - P(X \leq 2) = 1 - \sum_{k=0}^2 P(X=k)$$ 4. **Second Problem:** A manufacturer sells fix-a-flat with 78% success rate. There are 18 tires. Find: **a. Probability all 18 tires are repaired:** $$P(X=18) = \binom{18}{18} (0.78)^{18} (0.22)^0 = (0.78)^{18}$$ **b. Probability exactly 10 tires are repaired:** $$P(X=10) = \binom{18}{10} (0.78)^{10} (0.22)^8$$ **c. Probability at least 12 tires are repaired:** $$P(X \geq 12) = 1 - P(X \leq 11) = 1 - \sum_{k=0}^{11} P(X=k)$$ 5. **Explanation:** - Use binomial coefficients to find combinations. - Raise success and failure probabilities to appropriate powers. - Sum probabilities for cumulative cases. - Use complement rule for "at least" probabilities. 6. **Final answers:** Exact numeric values require calculation or software.