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Question \( 20,5.1 \times 1 \) HW Score: \( 40 \%, 20 \) of 50 points Savo mework 10 (5.1, 5.2) Part 1 of 3 Points: 0 of 1 A baseball player hit 61 home runs in a season. Of the 61 home runs, 21 went to right field, 17 went to right center field, 11 went to center field, 10 went to left center field, and 2 went to left field. (a) What is the probability that a randomly selected home run was hit to right field? (b) What is the probability that a randomly selected home run was hit to leff field? (c) Was it unusual for this player to hit a home run to left field? Explain. (a) The probability that a randomly selected home run was hit to right field is \( \square \) (Round to three decimal places as needed.)

Ask by Harmon Daniel. in the United States
Mar 19,2025

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The probability that a randomly selected home run was hit to right field is 0.344.

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To find the probability of randomly selecting a home run hit to a specific field, you would use the formula: \[ \text{Probability} = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} \] (a) For right field: There were 21 home runs hit to right field. \[ \text{Probability (right field)} = \frac{21}{61} \approx 0.344 \] So, rounding to three decimal places, the answer is \( 0.344 \). (b) For left field: There were 2 home runs hit to left field. \[ \text{Probability (left field)} = \frac{2}{61} \approx 0.033 \] So, rounding to three decimal places, the answer is \( 0.033 \). (c) To determine if it was unusual for the player to hit a home run to left field, we can look at the probability calculated. Typically, an event is considered unusual if its probability is less than 0.05 (or 5%). Since \( 0.033 < 0.05 \), it can be concluded that hitting a home run to left field was indeed unusual for this player.

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