This season, the probability that the Yankees will win a game is 0.53 and the probability that the Yankees will score 5 or more runs in a game is 0.49 . The probability that the Yankees win and score 5 or more runs is 0.39 . What is the probability that the Yankees will win when they score fewer than 5 runs? Round your answer to the nearest thousandth. Answer:

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Events Denoted: Let's denote the events as follows: W: The Yankees win a game. S: The Yankees score 5 or more runs in a game. We are given the following probabilities: P(W) = 0.53 (Probability that the Yankees will win a game) P(S) = 0.49 (Probability that the Yankees will score 5 or more runs in a game) P(W and S) = 0.39 (Probability that the Yankees win and score 5 or more runs) We need to find the probability that the Yankees will win when they score fewer than 5 runs, which can be denoted as P(W and not S).Find Probability not S: First, we need to find the probability that the Yankees score fewer than 5 runs, which is the complement of the event S. The complement rule states that P(not S) = 1 - P(S). So, P(not S) = 1 - 0.49 = 0.51.Calculate P(W | not S): Now, we can use the formula for conditional probability to find P(W | not S), which is the probability that the Yankees win given that they score fewer than 5 runs. The formula is: P(W | not S) = P(W and not S) / P(not S). We already have P(not S), but we need to find P(W and not S). We can find this by subtracting P(W and S) from P(W), because P(W) includes all wins, both with 5 or more runs and with fewer than 5 runs. So, P(W and not S) = P(W) - P(W and S) = 0.53 - 0.39 = 0.14.Round to Nearest Thousandth: Now we can calculate P(W | not S) using the values we have: P(W | not S) = P(W and not S) / P(not S) = 0.14 / 0.51. Calculating this gives us P(W | not S) ≈ 0.2745.We round the answer to the nearest thousandth as requested: P(W | not S) ≈ 0.275.

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