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Batting averages: Tony Gwynn (San Diego Padres) had a lifetime batting average of 0.347 ranking him as one of the greatest hitters of all time. Suppose he came to bat five times in any given game. a. What is the probability that he will get exactly three hits? b. What is the probability that he will get at least three hits?

Short Answer

Expert verified
a. 0.1777, b. 0.1965

Step by step solution

01

Define the Problem

The problem involves finding the probabilities in a binomial distribution context, where Tony Gwynn has a probability of success (hitting) equal to 0.347 in each at-bat.
02

Define Variables for Binomial Distribution

Let \( p = 0.347 \) be the probability of a successful hit, and \( q = 1 - p = 0.653 \) be the probability of not hitting. Also, \( n = 5 \) (number of trials or at-bats), and we are interested in finding the probability of exactly \( k = 3 \) hits.
03

Calculate Probability of Exactly Three Hits

The probability of exactly \( k \) hits in a binomial distribution can be found using the formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]Substitute the values: \[ P(X = 3) = \binom{5}{3} (0.347)^3 (0.653)^2 \] Calculate each component:\( \binom{5}{3} = 10 \); \( (0.347)^3 \approx 0.0417 \); \( (0.653)^2 \approx 0.4264 \).Multiply them together: \[ P(X = 3) \approx 10 \times 0.0417 \times 0.4264 \approx 0.1777 \]
04

Calculate Probability of Getting At Least Three Hits

To find the probability of getting at least three hits, sum the probabilities of getting 3, 4, and 5 hits: \[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \] Using the binomial formula for \( k = 4 \) and \( k = 5 \):For \( k = 4 \):\[ P(X = 4) = \binom{5}{4} (0.347)^4 (0.653)^1 \approx 5 \times 0.0052 \times 0.653 \approx 0.017 \]For \( k = 5 \): \[ P(X = 5) = \binom{5}{5} (0.347)^5 (0.653)^0 \approx 1 \times 0.0018 \times 1 \approx 0.0018 \]Finally, sum the probabilities:\[ P(X \geq 3) \approx 0.1777 + 0.017 + 0.0018 \approx 0.1965 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation in a binomial distribution involves determining the likelihood of a certain number of successes in a series of independent trials. For Tony Gwynn's batting scenario, we're dealing with repeated chances at success (hits during at-bats), making it ripe for a binomial distribution. The formula used here is:- \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]- Where: - \( n \) is the number of trials (at-bats). - \( k \) is the number of successful trials (hits). - \( p \) is the probability of success on a single trial (Gwynn's batting average of 0.347).- The term \( \binom{n}{k} \) is a binomial coefficient, indicating the number of ways to choose \( k \) successes from \( n \) trials.We've broken down the problem by calculating the probability of exactly 3 hits (using the values \( n = 5 \), \( k = 3 \), and \( p = 0.347 \)). Multiplying probabilities for successful and unsuccessful outcomes gives us a probability of about 0.1777. Such calculations provide a systematic way to predict outcomes based on given probabilities.
Batting Average
A batting average is a scalar statistic used to measure a batter's ability to hit the ball. It's calculated by dividing the number of hits by the number of "at-bats." In this context, Tony Gwynn’s remarkable career average of 0.347 means he successfully hit the ball 34.7% of the time he came to bat. Understanding batting averages provides insight not just into player performance but also into the likelihood of a specific outcome occurring, such as getting a hit. The higher the batting average, the higher the probability of a successful at-bat. For Gwynn, his above-average batting rate heavily influences probability calculations, as it becomes the key "success" parameter in our binomial distribution formula. When considering batting averages, one also thinks about factors that might influence these numbers, such as the player's form, opposing team's strategies, playing environment, and even luck. Recognizing these influences helps to address randomness and variability in predictive models of sports performances.
Statistical Distribution
Statistical distributions are essential in understanding how probabilities spread across possible outcomes. The binomial distribution is specifically tailored for scenarios like this baseball problem, where results are binary—hit or no hit. In a binomial distribution: - Each trial is independent, meaning the result of one doesn't affect another. - Each trial has only two possible outcomes (hit or miss here). - A fixed number of trials are considered, which are Tony's 5 at-bats in each game. Distributions like these enable us to predict not just single event probabilities, like those of exactly 3 or at least 3 hits, but a full range of outcomes. By summing the probabilities of different numbers of hits (3, 4, or 5), we find the probability of getting at least 3 hits. Understanding distribution shapes also offers insights into variability and expectation. In this exercise, calculating the cumulative probability of Tony getting at least 3 hits involved summing lower probabilities calculated from the binomial formula applied to 4 and 5 hits. These calculations utilize patterns within the chosen distribution to reveal more nuanced insights into likely events in a given set of trials.

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