Problem 23
Question
Urn 1 contains 5 white and 6 black balls, while urn 2 contains 8 white and 10 black balls. Two balls are randomly selected from urn 1 and are put into urn \(2 .\) If 3 balls are then randomly selected from urn \(2,\) compute the expected number of white balls in the trio. Hint: Let \(X_{i}=1\) if the \(i\) th white ball initially in urn 1 is one of the three selected, and let \(X_{i}=0\) otherwise. Similarly, let \(Y_{i}=1\) if the \(i\) th white ball from urn 2 is one of the three sclected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{J} X_{i}+\sum_{1}^{\delta} Y_{i}\)
Step-by-Step Solution
Verified Answer
The expected number of white balls in the final selection is \(\frac{577}{99}\).
1Step 1: Compute the probability of selecting a white ball from urn 1 and urn 2
First, we need to find the probability of selecting a white ball from urn 1 and urn 2. We know that there are 5 white balls and 6 black balls in urn 1, and 8 white balls and 10 black balls in urn 2 before the transfer.
Probability of selecting a white ball from urn 1:
\(p(X_i = 1) = \frac{5}{11}\)
Probability of selecting a black ball from urn 1:
\(p(X_i = 0) = \frac{6}{11}\)
Probability of selecting a white ball from urn 2:
\(p(Y_i = 1) = \frac{8}{18}\)
Probability of selecting a black ball from urn 2:
\(p(Y_i = 0) = \frac{10}{18}\)
2Step 2: Find the expected values of the individual random variables
To find the expected values of the random variables \(X_i\) and \(Y_i\), we will use their probabilities.
Expected value of \(X_i\):
\(E(X_i) = p(X_i = 1) = \frac{5}{11}\)
Expected value of \(Y_i\):
\(E(Y_i) = p(Y_i = 1) = \frac{8}{18}\)
3Step 3: Use the linearity of expectation
The linearity of expectation states that the expected value of a sum of random variables is the sum of the expected values of the individual random variables. Hence, we can find the expected number of white balls in the final selection by adding the expected values of \(X_i\) and \(Y_i\).
Expected number of white balls in the final selection:
\(E(\sum_1^{5} X_i + \sum_1^{8} Y_i) = \sum_1^{5} E(X_i) + \sum_1^{8} E(Y_i) = 5 \times \frac{5}{11} + 8 \times \frac{8}{18}\)
4Step 4: Compute the final result
Now let's compute the expected number of white balls in the final selection using the expression we found in the previous step.
Expected number of white balls in the final selection:
\(E(\sum_1^{5} X_i + \sum_1^{8} Y_i) = 5 \times \frac{5}{11} + 8 \times \frac{8}{18} = \frac{25}{11} + \frac{32}{9}\)
Compute the common denominator and find the sum:
Expected number of white balls in the final selection:
\(E(\sum_1^{5} X_i + \sum_1^{8} Y_i) = \frac{225 + 352}{99} = \frac{577}{99}\)
The expected number of white balls in the final selection is \(\frac{577}{99}\).
Key Concepts
Expected ValueRandom VariablesLinearity of ExpectationCombinatorics
Expected Value
Expected value is a fundamental concept in probability and statistics. It represents the average outcome of a random variable after many repetitions of an experiment. Think of it as a weighted average, where outcomes have different probabilities. Expected value gives us an idea of the central tendency of the variable.
For instance, in the exercise, we calculate the expected number of white balls selected by finding the expected values of two sets of variables, \(X_i\) and \(Y_i\). Each of these variables either indicates the selection of a white ball or not. The expected value of each \(X_i\) (from urn 1) is determined by the probability of selecting a white ball from urn 1, which is \(\frac{5}{11}\). Similarly, the expected value of \(Y_i\) (from urn 2) relies on the probability of selecting a white ball from urn 2, which is \(\frac{8}{18}\).
By understanding expected value, you can predict the average number of white balls expected over many trials of reaching into the urns.
For instance, in the exercise, we calculate the expected number of white balls selected by finding the expected values of two sets of variables, \(X_i\) and \(Y_i\). Each of these variables either indicates the selection of a white ball or not. The expected value of each \(X_i\) (from urn 1) is determined by the probability of selecting a white ball from urn 1, which is \(\frac{5}{11}\). Similarly, the expected value of \(Y_i\) (from urn 2) relies on the probability of selecting a white ball from urn 2, which is \(\frac{8}{18}\).
By understanding expected value, you can predict the average number of white balls expected over many trials of reaching into the urns.
Random Variables
Random variables are integral to probability. They help model stochastic, or random, processes and can take on various values based on an underlying random phenomenon. In simple terms, a random variable is a numerical summary of outcomes from a random process.
In our example, \(X_i\) and \(Y_i\) are random variables. Here, \(X_i = 1\) if the \(i\)-th white ball from urn 1 is chosen, and \(X_i = 0\) otherwise. Similarly, \(Y_i = 1\) for the \(i\)-th white ball from urn 2 being chosen, and \(Y_i = 0\) otherwise. These variables help quantify the random selection process.
In this example, the random variables followed by their probabilities allow us to compute expected values, forming the foundation to solve this probability puzzle.
In our example, \(X_i\) and \(Y_i\) are random variables. Here, \(X_i = 1\) if the \(i\)-th white ball from urn 1 is chosen, and \(X_i = 0\) otherwise. Similarly, \(Y_i = 1\) for the \(i\)-th white ball from urn 2 being chosen, and \(Y_i = 0\) otherwise. These variables help quantify the random selection process.
In this example, the random variables followed by their probabilities allow us to compute expected values, forming the foundation to solve this probability puzzle.
Linearity of Expectation
The linearity of expectation is a powerful tool that simplifies the computation of expected values. It states that the expected value of a sum of random variables equals the sum of their expected values. This property holds true regardless of any dependence between the variables.
In our exercise, we calculate the expected number of white balls by leveraging this concept. We consider the total number of selected white balls as the sum of all \(X_i\) and \(Y_i\) indicators. According to the linearity of expectation, we can compute this by simply adding the expected values of \(X_i\) and \(Y_i\). This step converts a potentially complex computation into a straightforward arithmetic operation. Here, the expected value becomes \(5 \times \frac{5}{11} + 8 \times \frac{8}{18}\). The formula simplifies due to the linearity of expectation, making the task much more manageable.
This principle is a cornerstone of probability theory and is especially handy when dealing with large numbers of random variables.
In our exercise, we calculate the expected number of white balls by leveraging this concept. We consider the total number of selected white balls as the sum of all \(X_i\) and \(Y_i\) indicators. According to the linearity of expectation, we can compute this by simply adding the expected values of \(X_i\) and \(Y_i\). This step converts a potentially complex computation into a straightforward arithmetic operation. Here, the expected value becomes \(5 \times \frac{5}{11} + 8 \times \frac{8}{18}\). The formula simplifies due to the linearity of expectation, making the task much more manageable.
This principle is a cornerstone of probability theory and is especially handy when dealing with large numbers of random variables.
Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arranging, and finding patterns. It is crucial when analyzing scenarios with multiple possible outcomes, like selecting balls from urns.
In this exercise, combinatorics is used implicitly through the selection and transfer of balls between urns. Understanding the number of ways to select and position the balls in urn 2 after moving them from urn 1 is essential. This forms the basis for calculating probabilities of each possible configuration of white and black balls.
For example, initially, the number of ways to choose two balls from urn 1 or three from urn 2 involves computing combinations. These combinations set the stage for determining probabilities, which then connect to expected values. While the exercise doesn't explicitly walk through these step-by-step in the solution, the underlying use of combinatorial reasoning is key to understanding the calculations.
By learning about combinatorics, students gain powerful techniques to approach similar problems in probability and statistics.
In this exercise, combinatorics is used implicitly through the selection and transfer of balls between urns. Understanding the number of ways to select and position the balls in urn 2 after moving them from urn 1 is essential. This forms the basis for calculating probabilities of each possible configuration of white and black balls.
For example, initially, the number of ways to choose two balls from urn 1 or three from urn 2 involves computing combinations. These combinations set the stage for determining probabilities, which then connect to expected values. While the exercise doesn't explicitly walk through these step-by-step in the solution, the underlying use of combinatorial reasoning is key to understanding the calculations.
By learning about combinatorics, students gain powerful techniques to approach similar problems in probability and statistics.
Other exercises in this chapter
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