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 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
part 2 of 5
(b) the probability that the team loses the toss all four times is
part 3 of 5
(c) the probability that the team wins the toss more than three times is
part 4 of 5
(d) the probability that the team loses the toss no more than three times is
part 5 of 5
(e) the probability that the team loses the toss at least once is

Explanation:

Step1: Determine the probability of losing a single toss

Since the captain calls heads, the probability of losing a single toss (i.e., getting tails) is \(p = \frac{1}{2}\), and the probability of winning a single toss (getting heads) is also \(q=\frac{1}{2}\). We use the binomial probability formula \(P(X = k)=C(n,k)\times p^{n - k}\times q^{k}\), where \(n\) is the number of trials, \(k\) is the number of "successes" (in this case, losses), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Solve part (d)

The probability of losing no more than three times is the complement of losing all four times.
We know that the probability of losing all four times (from part (b)) is \(P(X = 4)=C(4,4)\times(\frac{1}{2})^{4}\times(\frac{1}{2})^{0}=\frac{1}{16}\)
Using the complement rule \(P(X\leq3)=1 - P(X = 4)\)

$$P(X\leq3)=1-\frac{1}{16}=\frac{15}{16}$$

Step3: Solve part (e)

The probability of losing at least once is the complement of winning all four times.
The probability of winning all four times (similar to part (b) logic, but with wins instead of losses) \(P(X = 0)=C(4,0)\times(\frac{1}{2})^{0}\times(\frac{1}{2})^{4}=\frac{1}{16}\)
Using the complement rule \(P(X\geq1)=1 - P(X = 0)\)

$$P(X\geq1)=1-\frac{1}{16}=\frac{15}{16}$$

Answer:

(d) \(\frac{15}{16}\)
(e) \(\frac{15}{16}\)