1. **Problem statement:** We have a binomial distribution with number of trials $n=10$ and probability of success $p=0.3$ (a passenger carries overweight luggage in a flight). We want to find: a. The probability exactly 5 flights have overweight luggage. b. The probability fewer than 3 flights have overweight luggage. 2. **Formula:** The binomial probability mass function is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $X$ is the number of successes, $k$ is the exact number of successes, $n$ is the number of trials, and $p$ is the probability of success. 3. **Calculate part (a):** Probability exactly 5 flights have overweight luggage: $$P(X=5) = \binom{10}{5} (0.3)^5 (0.7)^5$$ Calculate the binomial coefficient: $$\binom{10}{5} = \frac{10!}{5!5!} = 252$$ Calculate powers: $$(0.3)^5 = 0.00243$$ $$(0.7)^5 = 0.16807$$ Multiply all: $$P(X=5) = 252 \times 0.00243 \times 0.16807 \approx 0.103$$ 4. **Calculate part (b):** Probability fewer than 3 flights have overweight luggage means $X < 3$, i.e., $X=0,1,2$. $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$ Calculate each term: - $$P(X=0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 1 \times 1 \times 0.02825 = 0.02825$$ - $$P(X=1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 10 \times 0.3 \times 0.04035 = 0.12105$$ - $$P(X=2) = \binom{10}{2} (0.3)^2 (0.7)^8 = 45 \times 0.09 \times 0.05765 = 0.23347$$ Sum these probabilities: $$P(X<3) = 0.02825 + 0.12105 + 0.23347 = 0.38277$$ 5. **Final answers:** a. Probability exactly 5 flights have overweight luggage is approximately $0.103$. b. Probability fewer than 3 flights have overweight luggage is approximately $0.383$. These calculations use the binomial distribution formula and basic probability rules to find the required probabilities.