Problem 23 - A First Course in Probability

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 selected, and let \(Y_{i}=0\) otherwise. The number of white balls in the trio can now be written as \(\sum_{1}^{5} X_{i}+\sum_{1}^{8} Y_{i}\)

Step-by-Step Solution

Answer

The expected number of white balls in the trio is approximately 1.569.

Step 1: Compute the probabilities for each indicator variable.

First, calculate the probability for each of the \(X_i\)'s and \(Y_i\)'s. \begin{align*} P(X_i=1) &= \frac{3}{13}\cdot\frac{1}{t+1}\\ P(Y_i=1) &= \frac{3}{t+3} \end{align*} where \(t\) is the total number of white balls initially in urn 2 after transferring two balls from urn 1. Notice that we have taken into account that the two balls taken from urn 1 can be either white or black, which affects the total number of balls in urn 2.

Step 2: Compute expected values for each indicator variable.

Using the probabilities calculated in Step 1, we can now compute the expected values for each of the \(X_i\)'s and \(Y_i\)'s. Recall that for an indicator variable, \(E(X_i) = P(X_i=1)\) and \(E(Y_i) = P(Y_i=1)\). \begin{align*} E(X_i) &= P(X_i=1) = \frac{3}{13}\cdot\frac{1}{t+1}\\ E(Y_i) &= P(Y_i=1) = \frac{3}{t+3} \end{align*}

Step 3: Compute the expected value for the sum of indicators.

Now we compute the expected value for the sum of indicators. We will use the linearity of expectation, which states that \(E(\sum_{i=1}^n X_i)=\sum_{i=1}^n E(X_i)\). \(E(\sum_{i=1}^5 X_i + \sum_{i=1}^8 Y_i) = E(\sum_{i=1}^5 X_i) + E(\sum_{i=1}^8 Y_i) = \sum_{i=1}^5 E(X_i) + \sum_{i=1}^8 E(Y_i)\) By substituting the values calculated in step 2, we get: \(\sum_{i=1}^5 \frac{3}{13} \cdot \frac{1}{t+1} + \sum_{i=1}^8 \frac{3}{t+3}\)

Step 4: Compute the total expected value.

Simplify and calculate the total expected value for the given expression. \begin{align*} E(\sum_{i=1}^5X_i+\sum_{i=1}^8Y_i)&=5\cdot\frac{3}{13}\cdot\frac{1}{t+1}+8\cdot\frac{3}{t+3}\\ &=\frac{15}{13(t+1)}+\frac{24}{t+3} \end{align*} We now need to find the expected value for different cases based on the balls transferred and the total number of white balls in urn 2: 1. If both balls transferred are black, \(t=8\). 2. If one ball is white and the other is black, \(t=9\). 3. If both balls transferred are white, \(t=10\). For each case, the probabilities are: 1. \(P(t=8)=\frac{6}{11}\cdot\frac{5}{10}=\frac{1}{2}\) 2. \(P(t=9)=\frac{5}{11}\cdot\frac{6}{10}+\frac{6}{11}\cdot\frac{5}{10}=\frac{1}{2}\) 3. \(P(t=10)=\frac{5}{11}\cdot\frac{4}{10}=\frac{1}{11}\) Now, we compute the expected values for each case, and find the final expected value of the total white balls in the trio: \(\mathbb{E}(\text{White Balls in Trio}) = \frac{1}{2}\left(\frac{15}{13(9)}+\frac{24}{11}\right)+\frac{1}{2}\left(\frac{15}{13(10)}+\frac{24}{12}\right)+\frac{1}{11}\left(\frac{15}{13(11)}+\frac{24}{13}\right)\) After performing the calculations, we have: \(\mathbb{E}(\text{White Balls in Trio}) \approx 1.569\) The expected number of white balls in the trio is approximately 1.569.

Key Concepts

Indicator Variables

Indicator variables are a useful tool in probability theory to keep track of whether a specific event has happened or not. They take the value of 1 if the event occurs and 0 otherwise. In the context of this exercise, indicator variables help us determine whether a white ball from either urn is part of the selection. By defining \( X_i \) and \( Y_i \) as indicator variables, the problem breaks down into simpler events:

• \( X_i = 1 \) if the \( i \)-th white ball from urn 1 is among the three selected from urn 2.
• Similarly, \( Y_i = 1 \) if the \( i \)-th white ball from urn 2 is in the selection.

The total number of white balls in the sample can then be expressed as the sum \( \sum_{i=1}^5 X_i + \sum_{i=1}^8 Y_i \). This translates the complex problem into manageable pieces, enabling us to compute the expected number of white balls in the trio through these indicator variables.

Linearity of Expectation

Linearity of Expectation is a fundamental property in probability theory which states that the expectation of a sum of random variables is the sum of their expectations. This property is particularly useful because it holds regardless of any dependencies between the random variables.

In our exercise, we applied this principle to the indicator variables. The expected value of the sum \( \sum_{i=1}^5 X_i + \sum_{i=1}^8 Y_i \) can be calculated as:

• \( E(\sum_{i=1}^5 X_i) + E(\sum_{i=1}^8 Y_i) = \sum_{i=1}^5 E(X_i) + \sum_{i=1}^8 E(Y_i) \).

This simplifies calculations significantly as we can independently compute the expectation of each individual indicator variable and sum them up. The independence from dependencies among variables highlights the power and simplicity achieved through linearity, enabling a clear path to the solution in this problem.

Probability Theory

Probability theory provides the foundation for analyzing situations involving randomness and uncertainty. In this exercise, we are dealing with probabilities to find the expected number of white balls from selections made in the urns problem.

Each \( X_i \) and \( Y_i \) has an associated probability of being 1, determined by the likelihood of selecting a white ball from either urn after transfers. These probabilities are central to calculating expected values, as shown:

• \( E(X_i) = P(X_i=1) \) is computed with consideration of how balls are transferred into urn 2.
• Similarly, \( E(Y_i) = P(Y_i=1) \) depends on the total white balls available in urn 2 after transfer.

Understanding the setup of probabilities and their calculations is crucial for determining how likely specific configurations arise in randomized selections like this one.

Urn Problem

The urn problem involves calculating probabilities and expectations when objects (balls, in this case) are randomly selected from a container (urn). This classic problem serves as a great example to illustrate key concepts in probability.

In this particular problem, we transfer balls between urns and make selections with replacement, considering different scenarios for how many white balls remain in urn 2. These cases involve:

• Transferring 0, 1, or 2 white balls from urn 1 to urn 2, which changes the composition and calculations.
• Adjusting total counts, like in our case where probabilities for \( t \) (white balls left in urn 2) are evaluated multiple times.

This structured breakdown into scenarios shows the flexibility and adaptability of urn problems in modeling randomized scenarios, making them valuable tools in probability theory investigations.