This is the question: Let's propose a hypothetical situation where you plan to apply your newly acquired analytical skills to the end of the baseball season. The Dodgers most recent win percentage is 0.569. Assume that the team that they are playing and home field advantage is completely random, which implies that their probability of winning is the same regardless of location. The Dodgers are going to play a three game series. The attached image provides you with the distribution of wins. Calculate the probability that they do not win a single game. Submit your answer to two decimal places and appropriate rounding.
Here is what I did: p(x=0)=[3!/0!(3-0)!]*(.569)^0*(4.31)^3=80.06 I just wanted to know whether I am right. Thanks in advance.
p(x=0) must be less than 1
To calculate the probability that the Dodgers do not win a single game in a three-game series, you need to consider the probability of losing each individual game and then multiply them together.
Given that the Dodgers' win percentage is 0.569, the probability of winning a single game would be 0.569, and the probability of losing a single game would be 1 - 0.569, which is 0.431.
To calculate the probability of not winning a single game, you need to calculate the probability of losing all three games. Since each game is independent, you can multiply the individual probabilities together.
p(x=0) = (0.431)^3 ≈ 0.083
So, the probability that the Dodgers do not win a single game in the three-game series is approximately 0.083 or 8.3%.
It seems there was a slight error in your calculation. You correctly raised the probability of winning (0.569) to the power of 0 (since we are looking for zero wins), but the rest of the exponent should be (0.431) (the probability of losing) raised to the power of 3 (the number of games in the series). The correct calculation would be:
p(x=0) = (0.569)^0 * (0.431)^3 ≈ 0.083
Therefore, your answer of 0.083 or 8.3% is correct.