1. A tennis player wins a math 55% of the time when she serves first and 47% of the time when her opponent serves first. The player who serves first is determined by a coin toss before the match. What is the probability that the player wins a given match.
2.A football team is losing by 14 points near the end of a game. The team scroes two touchdowns (worth 6 pts each) before the end of the game. After each touchdown, the coach must decide whether to go for 1 point with a kick(which is successful 99% of the time) or 2 points with a run or pass (which is successful 45% of the time).
a.If the team goes for 1 point after each touchdown, what is the probability that the coach's team wins? loses? ties?
b. If the team goes for 2 points after each touchdown,what is the probability that the coach's team wins? loses? ties?
c. Can you develop a strategy so that the coach's team has a probability of winning the game that is greater than the probability of losing? If so, explain your strategy and calculate the probabilities of winning and losing using your strategy.
#1) 0.51
1. To find the probability that the tennis player wins a given match, we need to consider the different scenarios in which the player could serve first or her opponent could serve first.
Let's assume that the probability of the player serving first is 0.5, since it is determined by a coin toss.
If the player serves first (with a 55% probability of winning), the overall probability of her winning the match in this scenario would be the product of the probability of serving first (0.5) and the probability of winning when serving first (55%). So the contribution to the overall probability from this scenario is 0.5 * 0.55 = 0.275.
If the opponent serves first (with a 47% probability of the player winning), the overall probability of her winning the match in this scenario would be the product of the probability of the opponent serving first (0.5) and the probability of winning when the opponent serves first (47%). So the contribution to the overall probability from this scenario is 0.5 * 0.47 = 0.235.
To get the total probability of winning, we sum up the contributions from each scenario: 0.275 + 0.235 = 0.51.
Therefore, the probability that the player wins a given match is 0.51, or 51%.
2. a. If the team goes for 1 point after each touchdown, the coach's team can score 1 point (successful kick) or 2 points (successful run/pass). Since there are two touchdowns, there are four possible combinations of scores: 1+1, 1+2, 2+1, 2+2.
To calculate the probability of winning, we need to consider all possible combinations and their respective probabilities.
- 1+1: The probability of a successful kick is given as 99%, so the probability of two successful kicks is (0.99 * 0.99) = 0.9801. Therefore, the probability of winning with a score of 1+1 is 0.9801.
- 1+2 or 2+1: In these cases, the probability of one successful kick and one successful run/pass is (0.99 * 0.45) + (0.45 * 0.99) = 0.4455. Since there are two possible combinations (1+2 and 2+1), we multiply this by 2: 0.4455 * 2 = 0.891.
- 2+2: The probability of two successful run/pass plays is (0.45 * 0.45) = 0.2025.
The total probability of winning is the sum of the individual probabilities: 0.9801 + 0.891 + 0.2025 = 2.0736.
Since the team cannot score fractional points, the probability of winning is capped at 100%. Therefore, the probability of the coach's team winning is 100% or 1.
The probability of losing would be 0, as winning every possible combination results in a victory.
The probability of tying is also 0, as there are no ties possible in this scenario.
b. If the team goes for 2 points after each touchdown, we use a similar approach to calculate the probabilities.
- 1+1: In this case, the probability of two successful run/pass plays is (0.45 * 0.45) = 0.2025.
- 1+2 or 2+1: The probability of one successful run/pass and one successful kick is (0.45 * 0.99) + (0.99 * 0.45) = 0.4455. Since there are two possible combinations, we multiply this by 2: 0.4455 * 2 = 0.891.
- 2+2: The probability of two successful run/pass plays is (0.45 * 0.45) = 0.2025.
The total probability of winning is the sum of the individual probabilities: 0.2025 + 0.891 + 0.2025 = 1.296.
The probability of losing is 0, as winning every possible combination results in a victory.
The probability of tying is also 0, as there are no ties possible in this scenario.
c. To develop a strategy that maximizes the probability of winning for the coach's team, we need to compare the probabilities from options a and b.
From the calculations, we can see that the probability of winning with option a (going for 1 point after each touchdown) is higher (2.0736) than the probability with option b (going for 2 points after each touchdown) (1.296). Therefore, the coach's team should go for 1 point after each touchdown to maximize their probability of winning.
By adopting this strategy, the coach's team has a higher probability of winning the game (2.0736 or 100%) compared to the probability of losing (0%) since there are no chances for the opposing team to score. Ties are also not possible in this scenario.