1. **State the problem:** We are given a binomial distribution with parameters $n=10$ (number of trials) and $p=0.10$ (probability of a voter being Independent). 2. **Part a: Probability exactly 5 voters are Independent** The binomial probability formula is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ For $k=5$: $$P(X=5) = \binom{10}{5} (0.10)^5 (0.90)^5$$ Calculate $\binom{10}{5} = \frac{10!}{5!5!} = 252$. Compute the probability: $$P(X=5) = 252 \times (0.10)^5 \times (0.90)^5 = 252 \times 0.00001 \times 0.59049 = 252 \times 0.0000059049 = 0.001488$$ 3. **Part b: Probability less than 3 voters are Independent** This means $P(X<3) = P(X=0) + P(X=1) + P(X=2)$. Calculate each: $$P(X=0) = \binom{10}{0} (0.10)^0 (0.90)^{10} = 1 \times 1 \times 0.34868 = 0.34868$$ $$P(X=1) = \binom{10}{1} (0.10)^1 (0.90)^9 = 10 \times 0.10 \times 0.38742 = 0.38742$$ $$P(X=2) = \binom{10}{2} (0.10)^2 (0.90)^8 = 45 \times 0.01 \times 0.43047 = 0.19371$$ Sum these probabilities: $$P(X<3) = 0.34868 + 0.38742 + 0.19371 = 0.92981$$ 4. **Part c: Probability at least 8 voters are Independent** This means $P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)$. Calculate each: $$P(X=8) = \binom{10}{8} (0.10)^8 (0.90)^2 = 45 \times 10^{-8} \times 0.81 = 3.645 \times 10^{-6}$$ $$P(X=9) = \binom{10}{9} (0.10)^9 (0.90)^1 = 10 \times 10^{-9} \times 0.9 = 9 \times 10^{-9}$$ $$P(X=10) = \binom{10}{10} (0.10)^{10} (0.90)^0 = 1 \times 10^{-10} \times 1 = 10^{-10}$$ Sum these probabilities: $$P(X \geq 8) \approx 3.645 \times 10^{-6} + 9 \times 10^{-9} + 10^{-10} = 3.654 \times 10^{-6}$$ **Final answers:** - a) $P(X=5) \approx 0.00149$ - b) $P(X<3) \approx 0.92981$ - c) $P(X \geq 8) \approx 0.00000365$