Question

the wildcats and the leopards are evenly matched football teams. when they play, there is a 0.5 probability that the wildcats will win. if they play 9 times, what is the probability that the wildcats will win 4 of the games? round your answer to the nearest tenth of a percent.

choice

a. 7.0%
b. 24.6%
c. 16.4%
d. 0.2%

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 $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial. Here, $n = 9$, $k = 4$, $p = 0.5$.

Step2: Calculate the combination

First, calculate the combination $\binom{9}{4}=\frac{9!}{4!(9 - 4)!}=\frac{9!}{4!5!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$.

Step3: Calculate the probability

Then, substitute into the formula: $P(X = 4)=\binom{9}{4}(0.5)^{4}(0.5)^{9 - 4}=126\times(0.5)^{4}\times(0.5)^{5}=126\times(0.5)^{9}$. Since $(0.5)^{9}=\frac{1}{512}$, then $126\times\frac{1}{512}=\frac{126}{512}\approx0.24609375$.

Step4: Convert to percentage

Convert to a percentage: $0.24609375\times100 = 24.609375\%$, which rounds to $24.6\%$.

Answer:

B. 24.6%