Question

The Wildcats and the Leopards are evenly matched football teams. Whenthey play, there is a 0.5 probability that the Wildcats will win. If they play 9times, what is the probability that the Wildcats will win 6 of the games?Round your answer to the nearest tenth of a percent.A 24.6%B. 0.2%O C. 7.0%D. 16.4%

Answer 1

To answer this question, we need to use the probability using the Binomial Distribution. Because we are finding an exact probability, we can use the next formula:

$C(9,6)\cdot(\frac{1}{2})^6\cdot(\frac{1}{2})^{(6-3)}=0.1640$

Or the probability is about 16.40%.

C(9, 6) is the combination of 9 out of 6. They are going to play 9 games, but we are finding the probability that the Wildcats win 6. Then:

$C(n,k)=\frac{n!}{(n-k)!\cdot k!}\Rightarrow C(9,6)=\frac{9!}{(9-6)!\cdot6!}=\frac{9\cdot8\cdot7\cdot6!}{3!\cdot6!}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=84$

Then, the general formula for the Binomial Distribution is:

$C(n,k)\cdot(p)^k\cdot(q)^{n-k}$

In this case, the probability of p = q = 1/2, k = 6, n = 9. Then, applying the formula, we obtain a probability of 0.1640 or about 16.40%. The correct option is D.