Question

This season, the probability that the Yankees will win a game is 0.48 and the probability that the Yankees will score 5 or more runs in a game is 0.52. The probability that the Yankees win and score 5 or more runs is 0.4. What is the probability that the Yankees lose and score fewer than 5 runs

Answer 1

Answer:

$P(A \cup B) = P(A) + P(B)- P(A \cap B)$

And we can solve for the probability of interest $P(A \cap B)$ and we got:

$P(A \cap B) = P(A) + P(B) -P(A \cup B)$

And if we replace the values given we got:

$P(A \cap B) = 0.48 + 0.4 -0.52 =0.36$

Step-by-step explanation:

For this case we define some notation:

A= The event that the Yankees will win a game

B= The event that the Yankees will score 5 or more runs in a game

We have defined the probabilities for these two events:

$P(A) = 0.48, P(B) =0.4$

We also know the probability that the Yankees win and score 5 or more runs given by:

$P(A \cup B) = 0.52$

And we are interested on the probability that the Yankees lose and score fewer than 5 runs.

For this case we can use the total probability rule given by:

$P(A \cup B) = P(A) + P(B)- P(A \cap B)$

And we can solve for the probability of interest $P(A \cap B)$ and we got:

$P(A \cap B) = P(A) + P(B) -P(A \cup B)$

And if we replace the values given we got:

$P(A \cap B) = 0.48 + 0.4 -0.52 =0.36$