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(a) Of the 1000 home runs, it was found that 80 were caught by fans. What is the probability that a randomly selected home run is caught by a fan? 0.08 (Type an integer or a decimal. Do not round.) (b) Of the 1000 home runs, it was found that 284 were dropped when a fan made a legitimate attempt to catch the ball. What is the probability that a randomly selected home run is droppe 0.284 (Type an integer or a decimal. Do not round.) (c) Of the 80 caught balls, it was determined that 32 were barehanded catches, 39 were caught with a glove, and 9 were caught in a hat. What is the probability a randomly selected caught a hat? Interpret this probability. Select the correct choice below and fill in the answer boxes within your choice. (Type an integer or decimal rounded to the nearest thousandth as needed.) A. The probability is approximately \( \square . S o, ~ f o r ~ e v e r y ~ \) 1000 home runs caught by fans, we expect about \( \square \) to have been caught in a hat. B. The probability is \( \square \).So, for every 1000 home runs caught by fans, we expect exactly \( \square \) to have been caught in a hat.

Ask by Barrett Allan. in the United States
Feb 27,2025

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Answer

- (a) Probability a home run is caught by a fan: 0.08 - (b) Probability a home run is dropped: 0.284 - (c) Probability a caught home run was caught in a hat: 0.113

Solution

Let's solve each part step by step. ### Part (a) We need to find the probability that a randomly selected home run is caught by a fan. **Known conditions:** - Total home runs = 1000 - Home runs caught by fans = 80 **Probability calculation:** The probability \( P \) that a home run is caught by a fan is given by the formula: \[ P = \frac{\text{Number of home runs caught by fans}}{\text{Total number of home runs}} = \frac{80}{1000} \] Calculating this gives: \[ P = 0.08 \] ### Part (b) Now, we need to find the probability that a randomly selected home run is dropped when a fan made a legitimate attempt to catch it. **Known conditions:** - Total home runs = 1000 - Home runs dropped by fans = 284 **Probability calculation:** The probability \( P \) that a home run is dropped is given by: \[ P = \frac{\text{Number of home runs dropped}}{\text{Total number of home runs}} = \frac{284}{1000} \] Calculating this gives: \[ P = 0.284 \] ### Part (c) Next, we need to find the probability that a randomly selected caught home run was caught in a hat. **Known conditions:** - Total caught home runs = 80 - Home runs caught in a hat = 9 **Probability calculation:** The probability \( P \) that a caught home run was caught in a hat is given by: \[ P = \frac{\text{Number of home runs caught in a hat}}{\text{Total caught home runs}} = \frac{9}{80} \] Calculating this gives: \[ P = 0.1125 \] Now, rounding to the nearest thousandth: \[ P \approx 0.113 \] ### Interpretation Now, we can fill in the answer boxes for the interpretation. **Choice A:** The probability is approximately \( 0.113 \). So, for every 1000 home runs caught by fans, we expect about \( 113 \) to have been caught in a hat. **Choice B:** The probability is \( 0.113 \). So, for every 1000 home runs caught by fans, we expect exactly \( 113 \) to have been caught in a hat. ### Final Answers - (a) Probability caught by a fan: **0.08** - (b) Probability dropped: **0.284** - (c) Probability caught in a hat: **0.113** (Choose either A or B with \( 113 \) filled in)

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To find the probability that a randomly selected caught home run was caught in a hat, we need to divide the number of hat catches by the total number of caught home runs. Here, 9 balls were caught in a hat and the total catches by fans is 80. Thus, the probability is: \( \text{Probability} = \frac{9}{80} = 0.1125 \) Now, rounding to the nearest thousandth gives us approximately 0.113. A. The probability is approximately \(0.113\). So, for every 1000 home runs caught by fans, we expect about \(113\) to have been caught in a hat. B. The probability is \(0.1125\). So, for every 1000 home runs caught by fans, we expect exactly \(9\) to have been caught in a hat.

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