Question

a coin flip determines who gets the ball first at the beginning of a football game, with the visiting team calling heads or tails. the captain of one particular team always calls heads. in the first four games as visitor of a season, find the probability that his team
(a) wins the toss one time.
(b) loses the toss four times.
(c) wins the toss more than three times.
(d) loses the toss no more than three times.
(e) loses the toss at least once.
write your answers in exact, simplified form.
part 1 of 5
(a) the probability that the team wins the toss one time is (\frac{1}{4}).
part 2 of 5
(b) the probability that the team loses the toss all four times is (\frac{1}{16}).
part 3 of 5
(c) the probability that the team wins the toss more than three times is

Explanation:

Step1: <Binomial Probability Formula>

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 4\) (number of coin - tosses), \(p=\frac{1}{2}\) (probability of winning a toss, i.e., getting heads), and \(1-p=\frac{1}{2}\).

Step2: <Case for losing more than three times (i.e., losing 4 times)>

Losing 4 times means \(k = 0\) (wins). Using the binomial formula:
\(C(4,0)=\frac{4!}{0!(4 - 0)!}=1\)
\(P(X = 0)=C(4,0)\times(\frac{1}{2})^{0}\times(\frac{1}{2})^{4}\)
Since \((\frac{1}{2})^{0}=1\), then \(P(X = 0)=1\times1\times\frac{1}{16}=\frac{1}{16}\)

Answer:

\(\frac{1}{16}\)