A coin flip determines who gets the ball first at the beginning of a football game, with the visiting team calling heads or talls. 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 two times. (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 three times. Write your answers in exact, simplified form. Part 1 of 5 (a)The probability that the team wins the toss two times is (b)The probability that the team loses the toss all four times is (c)The probability that the team wins the toss more than three times is (d)The probability that the team loses the toss no more than three times is (e)The probability that the team loses the toss at least three times is

Step 1 ：The problem is asking for the probability of different outcomes of a coin toss. A coin toss is a binomial experiment because it has exactly two outcomes (heads or tails), which are mutually exclusive, and the probability of each outcome is equal (0.5 for heads and 0.5 for tails).

Step 2 ：The probability of k successes (heads) in n trials (tosses) is given by the binomial probability formula: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where: \(P(X=k)\) is the probability of k successes in n trials, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, and \((1-p)\) is the probability of failure on a single trial.

Step 3 ：In this case, n=4 (the number of games), p=0.5 (the probability of winning a toss), and k varies depending on the question.

Step 4 ：For part (a), the probability that the team wins the toss two times is calculated as \(P(X=2) = C(4, 2) * (0.5^2) * ((1-0.5)^(4-2))\), which simplifies to \(\boxed{0.375}\).

Step 5 ：For part (b), the probability that the team loses the toss all four times is calculated as \(P(X=0) = C(4, 0) * (0.5^0) * ((1-0.5)^(4-0))\), which simplifies to \(\boxed{0.0625}\).

Step 6 ：For part (c), the probability that the team wins the toss more than three times is calculated as \(P(X=4) = C(4, 4) * (0.5^4) * ((1-0.5)^(4-4))\), which simplifies to \(\boxed{0.0625}\).

Step 7 ：For part (d), the probability that the team loses the toss no more than three times is the sum of the probabilities of losing 0, 1, 2, or 3 times. This is calculated as \(P(X=0) + P(X=1) + P(X=2) + P(X=3)\), which simplifies to \(\boxed{0.9375}\).

Step 8 ：For part (e), the probability that the team loses the toss at least three times is the sum of the probabilities of losing 3 or 4 times. This is calculated as \(P(X=3) + P(X=4)\), which simplifies to \(\boxed{0.3125}\).