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This season, the probability that the Yankees will win a game is 0.51 and the probability that the Yankees will score 5 or more runs in a game is 0.52 . The probability that the Yankees lose and score fewer than 5 runs is 0.37 . What is the probability that the Yankees would score fewer than 5 runs when they win the game? Round your answer to the nearest thousandth.
Answer:

This season, the probability that the Yankees will win a game is $0$ . $51$ and the probability that the Yankees will score $5$ or more runs in a game is $0$ . $52$ . The probability that the Yankees lose and score fewer than $5$ runs is $0$ . $37$ . What is the probability that the Yankees would score fewer than $5$ runs when they win the game? Round your answer to the nearest thousandth. $\newline$ Answer:

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Find probabilities using the addition rule

Full solution

Q. This season, the probability that the Yankees will win a game is

0

.

51

and the probability that the Yankees will score

5

or more runs in a game is

0

.

52

. The probability that the Yankees lose and score fewer than

5

runs is

0

.

37

. What is the probability that the Yankees would score fewer than

5

runs when they win the game? Round your answer to the nearest thousandth.

\newline

Answer:

Events Denoted: Let's denote the events as follows: $\newline$ W: The Yankees win a game. $\newline$ S: The Yankees score $5$ or more runs in a game. $\newline$ L: The Yankees lose a game. $\newline$ F: The Yankees score fewer than $5$ runs in a game. $\newline$ We are given the following probabilities: $\newline$ $P(W) = 0.51$ (Yankees win) $\newline$ $P(S) = 0.52$ (Yankees score $5$ or more runs) $\newline$ $P(L \text{ and } F) = 0.37$ (Yankees lose and score fewer than $5$ runs) $\newline$ We want to find $P(F|W)$ , which is the probability that the Yankees score fewer than $5$ runs given that they win the game. $\newline$ First, we need to find $P(F)$ , the probability that the Yankees score fewer than $5$ runs in a game. We can use the complement rule since $P(S)$ is the probability of scoring $5$ or more runs: $\newline$ $P(F) = 1 - P(S)$ $\newline$ $P(F) = 1 - 0.52$ $\newline$ $P(F) = 0.48$
Find $P(F)$ : Next, we need to find $P(W \text{ and } F)$ , which is the probability that the Yankees win and score fewer than $5$ runs. We can find this by subtracting the probability of losing and scoring fewer than $5$ runs from the probability of scoring fewer than $5$ runs: $\newline$ $P(W \text{ and } F) = P(F) - P(L \text{ and } F)$ $\newline$ $P(W \text{ and } F) = 0.48 - 0.37$ $\newline$ $P(W \text{ and } F) = 0.11$
Find $P(W \text{ and } F)$ : Now, we can calculate $P(F|W)$ , the probability that the Yankees score fewer than $5$ runs given that they win, using the definition of conditional probability: $\newline$ $P(F|W) = \frac{P(W \text{ and } F)}{P(W)}$ $\newline$ $P(F|W) = \frac{0.11}{0.51}$ $\newline$ $P(F|W) \approx 0.2157$

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