1. **Problem Statement:** We have a group of 7 spectators randomly selected from a crowd where 70% are males. We want to find: i. The probability exactly 4 are males. ii. The probability that at most 5 are females. 2. **Relevant Formula:** This is a binomial probability problem. The binomial probability formula is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where: - $n$ is the number of trials (7 people), - $k$ is the number of successes (males or females), - $p$ is the probability of success (for males, $p=0.7$), - $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$. 3. **Part i: Probability exactly 4 are males** - Here, $n=7$, $k=4$, $p=0.7$. - Calculate the binomial coefficient: $$\binom{7}{4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$ - Calculate the probability: $$P(X=4) = 35 \times (0.7)^4 \times (0.3)^3$$ - Calculate powers: $$0.7^4 = 0.2401, \quad 0.3^3 = 0.027$$ - Multiply all: $$P(X=4) = 35 \times 0.2401 \times 0.027 = 35 \times 0.0064827 = 0.2269$$ 4. **Part ii: Probability at most 5 are females** - Probability of female $= 1 - 0.7 = 0.3$. - "At most 5 females" means number of females $\leq 5$. - Since total is 7, number of males $\geq 2$ (because females $\leq 5$). - We want $P(\text{females} \leq 5) = P(\text{males} \geq 2) = 1 - P(\text{males} < 2) = 1 - P(X=0) - P(X=1)$. Calculate $P(X=0)$: $$P(X=0) = \binom{7}{0} (0.7)^0 (0.3)^7 = 1 \times 1 \times 0.3^7 = 0.0002187$$ Calculate $P(X=1)$: $$P(X=1) = \binom{7}{1} (0.7)^1 (0.3)^6 = 7 \times 0.7 \times 0.000729 = 7 \times 0.7 \times 0.000729 = 0.00357$$ Sum these: $$P(X<2) = 0.0002187 + 0.00357 = 0.00379$$ Therefore: $$P(\text{females} \leq 5) = 1 - 0.00379 = 0.9962$$ **Final answers:** - i. $P(4 \text{ males}) = 0.2269$ - ii. $P(\text{at most 5 females}) = 0.9962$