Problem 48
Question
Sampling With and Without Replacement An urn contains \(K\) green and \(N-K\) blue balls. (a) You take \(n\) balls out of the urn. Find the probability that \(k\) of the \(n\) balls are green. (b) You take a ball out of the urn, note its color, and replace it. You repeat these steps \(n\) times. Find the probability that \(k\) of the \(n\) balls are green.
Step-by-Step Solution
Verified Answer
(a) Use the hypergeometric distribution formula; (b) Use the binomial distribution formula with \( p = \frac{K}{N} \).
1Step 1: Understanding the Problem
We need to find the probability of picking a certain number of green balls from an urn that contains both green and blue balls. In part (a), balls are drawn without replacement, while in part (b), balls are drawn with replacement.
2Step 1: Problem (a) - Identify the Hypergeometric Distribution
When drawing without replacement, the problem is described by the hypergeometric distribution. We are interested in finding the probability of drawing exactly \( k \) green balls from the \( n \) balls drawn, out of \( K \) green balls and \( N-K \) blue balls in the urn.
3Step 2: Calculate Probability for Hypergeometric Distribution
The probability of drawing \( k \) green balls in \( n \) trials without replacement is given by the formula: \[P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]where \( \binom{a}{b} \) is the binomial coefficient representing combinations.
4Step 3: Problem (b) - Identify the Binomial Distribution
When drawing with replacement, each draw is independent, and the problem can be described by a binomial distribution. We want to find the probability of getting exactly \( k \) green balls among \( n \) trials.
5Step 4: Calculate Probability for Binomial Distribution
The probability of obtaining \( k \) green balls in \( n \) trials with replacement is given by: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( p = \frac{K}{N} \) is the probability of drawing a green ball on any single trial.
Key Concepts
Hypergeometric DistributionBinomial DistributionSampling with ReplacementSampling without Replacement
Hypergeometric Distribution
The hypergeometric distribution is one of the fundamental concepts in probability theory that deals with scenarios where draws are made without replacement from a finite population. This distribution helps calculate the probability of drawing a certain number of successes in a subset of trials. Imagine an urn filled with a specific number of green and blue balls, and you want to determine the likelihood of extracting a particular number of green balls when each pick changes the remaining composition of the urn.
The formula for hypergeometric probability is given by: \[P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]Here, \(\binom{K}{k}\) denotes the number of ways to choose \(k\) green balls from the total green balls \(K\), and \(\binom{N-K}{n-k}\) represents the number of ways to choose the remaining \(n-k\) blue balls from the \(N-K\) blue balls.
To sum it up, the hypergeometric distribution gives you the probability of drawing a specific combination of items from a group, with each draw having an effect on the subsequent draws.
The formula for hypergeometric probability is given by: \[P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]Here, \(\binom{K}{k}\) denotes the number of ways to choose \(k\) green balls from the total green balls \(K\), and \(\binom{N-K}{n-k}\) represents the number of ways to choose the remaining \(n-k\) blue balls from the \(N-K\) blue balls.
To sum it up, the hypergeometric distribution gives you the probability of drawing a specific combination of items from a group, with each draw having an effect on the subsequent draws.
Binomial Distribution
The binomial distribution describes the probability of achieving a certain number of successful outcomes in a fixed number of independent trials, where each trial has two potential outcomes: success or failure.
When sampling with replacement, each observation remains unaffected by previous ones, maintaining constant probabilities throughout. This characteristic perfectly aligns with scenarios modeled by the binomial distribution. For instance, consider the urn example where after drawing a ball, it is returned to the urn before the next draw. This ensures each draw is independent and the probability of drawing a green ball (say \(p\)) remains constant.
The probability of drawing \(k\) green balls in \(n\) trials, with replacement, is given by: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]In this equation, \(\binom{n}{k}\) is the combination of \(n\) trials taken \(k\) at a time, \(p^k\) represents the probability of obtaining \(k\) successes, and \((1-p)^{n-k}\) accounts for the \(n-k\) failures.
When sampling with replacement, each observation remains unaffected by previous ones, maintaining constant probabilities throughout. This characteristic perfectly aligns with scenarios modeled by the binomial distribution. For instance, consider the urn example where after drawing a ball, it is returned to the urn before the next draw. This ensures each draw is independent and the probability of drawing a green ball (say \(p\)) remains constant.
The probability of drawing \(k\) green balls in \(n\) trials, with replacement, is given by: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]In this equation, \(\binom{n}{k}\) is the combination of \(n\) trials taken \(k\) at a time, \(p^k\) represents the probability of obtaining \(k\) successes, and \((1-p)^{n-k}\) accounts for the \(n-k\) failures.
Sampling with Replacement
Sampling with replacement is a technique where an item or sample is returned to the population after being drawn. This method ensures that the population remains the same size and composition, implying that each draw is independent.
Using the concept of an urn, consider drawing a ball, noting its color, and then putting it back in the urn before drawing again. This process allows each selection to have an identical probability distribution. Each draw does not affect the subsequent ones.
This method is commonly employed when dealing with problems related to the binomial distribution, as in situations where outcomes have two distinct possibilities and each trial remains isolated from the others. Sampling with replacement ensures that overall center statistics do not shift, which makes probabilistic predictions simpler and more convenient. Applications of this method are widespread, especially when each item in a population has an equal chance of being picked more than once.
Using the concept of an urn, consider drawing a ball, noting its color, and then putting it back in the urn before drawing again. This process allows each selection to have an identical probability distribution. Each draw does not affect the subsequent ones.
This method is commonly employed when dealing with problems related to the binomial distribution, as in situations where outcomes have two distinct possibilities and each trial remains isolated from the others. Sampling with replacement ensures that overall center statistics do not shift, which makes probabilistic predictions simpler and more convenient. Applications of this method are widespread, especially when each item in a population has an equal chance of being picked more than once.
Sampling without Replacement
Sampling without replacement involves drawing an item from a population and not returning it before the next draw. This is akin to drawing balls from an urn and leaving them out once they are drawn, thereby changing the population size and composition over time. This draws dependencies between trials because each draw affects the subsequent ones.
In probability terms, this situation is characterized by the hypergeometric distribution. The order and final makeup of draws alter the chances of each successive draw, reducing the population with each step.
Many real-life scenarios employ sampling without replacement for practicality, such as raffles, card games, or quality inspections in manufacturing processes where once an item is inspected, it remains non-returnable in the sampling pool. This method can help capture real-world complexities where drawn elements are permanently extracted, leaving a dynamic population to consider in probabilistic evaluations.
In probability terms, this situation is characterized by the hypergeometric distribution. The order and final makeup of draws alter the chances of each successive draw, reducing the population with each step.
Many real-life scenarios employ sampling without replacement for practicality, such as raffles, card games, or quality inspections in manufacturing processes where once an item is inspected, it remains non-returnable in the sampling pool. This method can help capture real-world complexities where drawn elements are permanently extracted, leaving a dynamic population to consider in probabilistic evaluations.
Other exercises in this chapter
Problem 47
Thirteen cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability that all are red?
View solution Problem 47
In the game of poker, determine the number of ways exactly two pairs can be picked.
View solution Problem 48
Let \(X\) be exponentially distributed with parameter \(\lambda\). Find \(\operatorname{var}(X)\).
View solution Problem 48
Four cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability that all are of different suits?
View solution