Question

The 1899 Cleveland Spiders were a historically awful baseball team, finishing with a win-loss record of 20-134. So, assume that team would win any given game with probability 13%. Find the probability that the Spiders would have won their first three games played that season.

Answer 1

18% because 20+3=23 and 134-3=131 and so you divide 23/131 and you get 17.55 which rounds up to 18%

Answer 2

Answer:

The probability that the Cleveland Spiders would have won their first three games is 0.00219.

Step-by-step explanation:

Let's define a random variable X as follows:

$X_i =$ "The Cleveland Spiders won the ith game"

$X_i =$ can be either 1 or 0, 1 meaning that they won and 0 that they lost the game. We want the to compute the probability that they've won on the first, second and third game. In other words:

$P(X_1 \cap X_2 \cap X_3)$

Let's assume the events are independent, then:

$P(X_1 = 1 \cap X_2 = 1 \cap X_3 = 1) = P(X_1 = 1) \times P(X_2 = 1) \times P(X_3 = 1) = 0.13 \times 0.13 \times 0.13 = (0.13)^3 \approx 0.00219$