1. **Problem Statement:** Find the probability that at least 3 out of 5 randomly selected donors have Group O blood, given $p=0.45$ is the probability an individual has Group O blood. 2. **Formula and Explanation:** This is a binomial probability problem where the number of trials $n=5$, probability of success $p=0.45$, and we want $P(X \geq 3)$ where $X$ is the number of donors with Group O blood. The binomial probability formula is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ To find $P(X \geq 3)$, we use the complement rule: $$P(X \geq 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$ 3. **Calculate each term:** - $P(X=0) = \binom{5}{0} (0.45)^0 (0.55)^5 = 1 \times 1 \times 0.55^5 = 0.0503$ - $P(X=1) = \binom{5}{1} (0.45)^1 (0.55)^4 = 5 \times 0.45 \times 0.55^4 = 0.2061$ - $P(X=2) = \binom{5}{2} (0.45)^2 (0.55)^3 = 10 \times 0.2025 \times 0.1664 = 0.3361$ 4. **Sum and subtract from 1:** $$P(X \geq 3) = 1 - (0.0503 + 0.2061 + 0.3361) = 1 - 0.5925 = 0.4075$$ 5. **Interpretation:** The probability that at least 3 donors have Group O blood is approximately $0.4075$ or 40.75%.