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This season, the probability that the Yankees will win a game is 0.61 and the probability that the Yankees will score 5 or more runs in a game is 0.49 . The probability that the Yankees lose and score fewer than 5 runs is 0.3 . What is the probability that the Yankees will lose when they score 5 or more runs? Round your answer to the nearest thousandth.

This season, the probability that the Yankees will win a game is $0.61$ and the probability that the Yankees will score $5$ or more runs in a game is $0.49$ . The probability that the Yankees lose and score fewer than $5$ runs is $0.3$ . What is the probability that the Yankees will lose when they score $5$ or more runs? Round your answer to the nearest thousandth.

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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

0.61

and the probability that the Yankees will score

5

or more runs in a game is

0.49

. The probability that the Yankees lose and score fewer than

5

runs is

0.3

. What is the probability that the Yankees will lose when they score

5

or more runs? Round your answer to the nearest thousandth.

Define Events: Let's define the events: $\newline$ W: Yankees win a game. $\newline$ L: Yankees lose a game. $\newline$ S: Yankees score $5$ or more runs. $\newline$ We know: $\newline$ $P(W) = 0.61$ $\newline$ $P(S) = 0.49$ $\newline$ $P(L \text{ and not } S) = 0.3$ $\newline$ First, find $P(L)$ using the complement of $P(W)$ : $\newline$ $P(L) = 1 - P(W) = 1 - 0.61 = 0.39$
Calculate $P(L)$ : Next, calculate $P(L \text{ or not } S)$ using the complement of $P(S)$ : $P(L \text{ or not } S) = 1 - P(S) = 1 - 0.49 = 0.51$
Find $P(L \text{ or not } S)$ : Now, use the Inclusion-Exclusion Principle to find $P(L \text{ and } S)$ : $P(L \text{ and } S) = P(L) + P(S) - P(L \text{ or not } S) = 0.39 + 0.49 - 0.51 = 0.37$

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