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Chapter 8: Problem 21

Calculate the expected value of the given random variable \(X .\) [Exercises \(23,24,27\), and 28 assume familiarity with counting arguments and probability (Section 7.4).] \(\nabla X\) is the higher number when two dice are rolled.

Short Answer

Expert verified
The expected value of the higher number when two dice are rolled is approximately \(E(X) = \frac{141}{36} \approx 3.92\).

Step by step solution

01

Compute the probability of each value of \(X\)

First, let's determine the probability of each possible outcome. There are 36 total outcomes when rolling two dice (since each die has 6 sides), and we want to find the number of favorable outcomes for each possible value of \(X\). - \(P(X = 1)\): Both dice show 1; only one favorable outcome, so the probability is \(\frac{1}{36}\). - \(P(X = 2)\): The possible pairs are (1, 2), (2, 1), and (2, 2), resulting in 3 favorable outcomes. Therefore, the probability is \(\frac{3}{36} = \frac{1}{12}\). - \(P(X = 3)\): The possible pairs are (1, 3), (2, 3), (3, 1), (3, 2), and (3, 3), resulting in 5 favorable outcomes. Therefore, the probability is \(\frac{5}{36}\). - \(P(X = 4)\): The possible pairs are (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), and (4, 4), resulting in 7 favorable outcomes. Therefore, the probability is \(\frac{7}{36}\). - \(P(X = 5)\): The possible pairs are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), and (5, 5), resulting in 9 favorable outcomes. Therefore, the probability is \(\frac{9}{36} = \frac{1}{4}\). - \(P(X = 6)\): There are 11 favorable outcomes: (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), and (6, 6). Therefore, the probability is \(\frac{11}{36}\).
02

Compute the expected value of \(X\)

Now that we have the probabilities for each possible value of \(X\), we can compute the expected value using the formula: $$E(X) = \sum_{i = 1}^n x_i P(x_i) = 1 \cdot \frac{1}{36} + 2 \cdot \frac{1}{12} + 3 \cdot \frac{5}{36} + 4 \cdot \frac{7}{36} + 5 \cdot \frac{1}{4} + 6 \cdot \frac{11}{36}.$$ Calculating this expression gives: $$E(X) = \frac{1}{36} + \frac{1}{6} + \frac{15}{36} + \frac{28}{36} + \frac{30}{36} + \frac{66}{36} = \frac{141}{36} \approx 3.92.$$ So, the expected value of the higher number when two dice are rolled is approximately 3.92.

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Key Concepts

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

Probability
When we discuss probability, we're looking at the likelihood of a particular event occurring out of all possible events. In the context of rolling dice, each die has 6 sides, which means there are 36 possible outcomes when two dice are rolled together. Probabilities are expressed as fractions, where the numerator represents the number of favorable outcomes and the denominator represents the total number of possible outcomes. It's essential to ensure that each outcome has an equal chance of occurring, which is certainly the case when using fair dice.

Calculating the probability involves identifying the scenarios that align with our event of interest. For example, when calculating the probability of rolling a certain number on a die, you would count the number of ways that number can appear and divide by the total number of possible outcomes. Understanding how to evaluate probabilities correctly is vital because it lays the groundwork for more advanced concepts like the expected value of a random variable.
Random Variable
A random variable can be thought of as a numerical outcome of a random process. In our dice-rolling scenario, the random variable X represents the higher number that appears when two dice are rolled. It assigns a numerical value to each possible outcome of the random process, which in this case, is between 1 and 6. Thinking of a random variable as a function may help make sense of it; this function takes the dice outcomes and translates them into a number that is of interest to us.

Random variables can be either discrete or continuous. The random variable we're dealing with here is discrete because it takes on a finite or countably infinite number of possible values. This distinction is vital, as the methods to calculate probabilities and expected values can differ for discrete and continuous random variables.
Counting Arguments
Counting arguments refer to the techniques used to count the number of ways certain events can occur, which are essential in determining probabilities in scenarios like rolling dice. A common method is to create a systematic list or a diagram, such as a tree diagram or a grid, to visually represent and count outcomes. For instance, we can visualize all the pairs of outcomes from two dice rolls on a 6x6 grid.

The counting argument often employed with dice is the multiplication rule of counting. If you have a series of events, and the first event can occur in 'm' ways and the second in 'n' ways, then the total number of ways both events can occur is the product 'm * n'. Note that the counting arguments become more complex when the events are not independent, but in the case of independent dice rolls, the multiplication rule simplifies the process significantly.
Dice Outcomes
The outcomes of rolling dice are a practical example of random variables and probabilities at work. Each die, by itself, has six faces, numbered from 1 to 6. When two dice are rolled, there are 36 different outcomes. Since dice are symmetric and fair, each of these 36 outcomes is equally likely.

For our exercise, we're interested in the higher number of the two dice, which means we need to count each scenario where a given number is the highest. A comprehensive approach would enumerate all the pairs that result in each possible higher number, which we did in the step-by-step solution. It's important to factor in all possible pairs, including the cases where both dice show the same number, to guarantee an accurate probability calculation for our random variable X.

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