Binomial Wins 454Bc1

1. **State the problem.** The Falcons win each relay race with probability $p=0.8$, and they will race $n=6$ times. Let $W$ be the number of races won. 2. **Use the binomial model.** Since each race has only two outcomes, win or lose, the number of wins follows a binomial setting: $$W\sim \text{Binomial}(n=6,p=0.8)$$ The binomial probability formula is: $$P(W=k)=\binom{6}{k}(0.8)^k(0.2)^{6-k}$$ for $k=0,1,2,3,4,5,6$. 3. **Part a: possible values of $W$.** The number of wins can be any whole number from $0$ to $6$. So the possible values are: $$0,1,2,3,4,5,6$$ 4. **Part b: probability of exactly $5$ wins.** Substitute $k=5$ into the binomial formula: $$P(W=5)=\binom{6}{5}(0.8)^5(0.2)^1$$ Compute each part: $$\binom{6}{5}=6$$ $$P(W=5)=6(0.8)^5(0.2)$$ $$P(W=5)=6(0.32768)(0.2)$$ $$P(W=5)=0.393216$$ So the probability is $0.393216$. 5. **Part c: probability of losing $1$ or fewer races.** Losing $1$ or fewer races means the team loses $0$ races or $1$ race. Since there are $6$ total races, that means the team wins $6$ or $5$ races. So we add: $$P(\text{lose }1\text{ or fewer})=P(W=6)+P(W=5)$$ First find $P(W=6)$: $$P(W=6)=\binom{6}{6}(0.8)^6(0.2)^0$$ $$P(W=6)=1(0.8)^6(1)$$ $$P(W=6)=0.262144$$ From part b, $P(W=5)=0.393216$. Add them: $$P(\text{lose }1\text{ or fewer})=0.262144+0.393216$$ $$P(\text{lose }1\text{ or fewer})=0.65536$$ 6. **Part d: mean number of wins.** For a binomial random variable, the mean is: $$\mu=np$$ Substitute the values: $$\mu=6(0.8)$$ $$\mu=4.8$$ So the mean number of wins is $4.8$. 7. **Final answers.** a) $0,1,2,3,4,5,6$ b) $0.393216$ c) $0.65536$ d) $4.8$