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.

1 answer

p(x=0) must be less than 1