Question

1. chs is playing the dallas cowboys (it was a really good year for chs). chs has to play 5 games, and has to win three to win the competition. if chs has a 7% chance of winning each game, what is the probability that chs wins the competition (3 wins of 5 games)?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is $P(X = k) = \binom{n}{k}p^{k}(1 - p)^{n - k}$, where $n = 5$ (number of trials), $k = 3$ (number of successes), and $p = 0.07$ (probability of success in a single trial). But we need the probability of winning at least 3 games (since winning 3, 4, or 5 games will result in winning the competition). So we need to calculate $P(X = 3)+P(X = 4)+P(X = 5)$.

Step2: Calculate $P(X = 3)$

First, calculate $\binom{5}{3}=\frac{5!}{3!(5 - 3)!}=\frac{5\times4\times3!}{3!\times2\times1}=10$. Then $P(X = 3)=\binom{5}{3}(0.07)^{3}(0.93)^{2}=10\times(0.07)^{3}\times(0.93)^{2}$. Calculate $(0.07)^{3}=0.000343$, $(0.93)^{2}=0.8649$, so $P(X = 3)=10\times0.000343\times0.8649\approx0.00296$.

Step3: Calculate $P(X = 4)$

$\binom{5}{4}=\frac{5!}{4!(5 - 4)!}=5$. $P(X = 4)=\binom{5}{4}(0.07)^{4}(0.93)^{1}=5\times(0.07)^{4}\times0.93$. $(0.07)^{4}=0.00002401$, so $P(X = 4)=5\times0.00002401\times0.93\approx0.000112$.

Step4: Calculate $P(X = 5)$

$\binom{5}{5}=1$. $P(X = 5)=\binom{5}{5}(0.07)^{5}(0.93)^{0}=1\times(0.07)^{5}\times1$. $(0.07)^{5}=0.0000016807$, so $P(X = 5)\approx0.00000168$.

Step5: Sum the probabilities

$P(X\geq3)=P(X = 3)+P(X = 4)+P(X = 5)\approx0.00296 + 0.000112+0.00000168\approx0.00307$.

Answer:

The probability that CHS wins the competition is approximately $0.0031$ (or $0.31\%$).