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Let \(\mathrm{X}\) be a random variable whose value is determined by the flip of a fair coin. If the coin lands heads up \(\mathrm{X}=1\), if tails then \(\mathrm{X}=0\). Find the expected value of \(\mathrm{X}\).

Short Answer

Expert verified
The expected value of the random variable X, determined by the flip of a fair coin, is \(E(X) = 0.5\).

Step by step solution

01

Identify the values and their probabilities

The random variable X can take two values: 0 (if the coin lands tails) and 1 (if the coin lands heads). Since the coin is fair, the probability of getting a head (X = 1) is 0.5 and the probability of getting a tail (X = 0) is 0.5.
02

Calculate the expected value E(X)

We can use the formula E(X) = Σ [x * P(x)] to find the expected value. In this case, there are two values for X, 0 and 1, with equal probabilities of 0.5 each. Therefore, E(X) = (0 * 0.5) + (1 * 0.5) = 0 + 0.5 = 0.5. The expected value of the random variable X is 0.5.

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Key Concepts

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

Probability and Statistics
Probability and statistics are branches of mathematics concerned with collecting, analyzing, interpreting, and presenting empirical data. In particular, probability theory deals with the likelihood of an event occurring. It is a measure that quantifies the uncertainty involved in a game of chance, a scientific experiment, or everyday life.

In the context of the given exercise, understanding probability is essential to determine the expected outcome of flipping a fair coin. The exercise revolves around a basic but fundamental concept in statistics known as the 'expected value', which is the long-term average value of repetitions of the experiment it represents. This concept is widely used in various fields such as finance, insurance, economics, and game theory to predict future events and their implications.
Random Variables
A random variable is a key concept in probability and statistics that describes a numerical outcome of a random phenomenon. To put it simply, it's a variable that can take on different values, with each value associated with a probability.

In discrete mathematics, random variables are often categorized into two types: discrete and continuous. Discrete random variables, like the one in the provided exercise involving a coin flip, take on a set of distinct values. Here, the variable \(X\) represents the outcome of the coin toss and can be either 0 or 1. The probabilities associated with these outcomes are calculated using probability mass functions, which in this case, designate a probability of 0.5 to each outcome due to the nature of a fair coin.
Discrete Mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. This means that it deals with countable, distinct elements that have a finite number of possibilities. Concepts from discrete mathematics are widely applicable in computer science and information theory, where logical analysis and the counting of objects are essential.

In the context of the given exercise, the concept of expected value, which is a part of discrete mathematics, is calculated for a random variable representing the outcome of a simple activity – flipping a coin. To calculate the expected value for such discrete scenarios, we sum over the possible outcomes, multiplying each outcome by its probability, to determine the average or 'expected' result over an infinite number of trials. This allows us to understand and analyze scenarios where outcomes are not continuous but have distinct separate values.

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Most popular questions from this chapter

Consider the joint distribution of \(\mathrm{X}\) and \(\mathrm{Y}\) given in the form of a table below. The cell (i,j) corresponds to the joint probability that \(\mathrm{X}=\mathrm{i}, \mathrm{Y}=\mathrm{j}\), for \(\mathrm{i}=1,2,3, \mathrm{j}=1,2,3\) $$ \begin{array}{|c|c|c|c|} \hline \mathrm{Y}^{\mathrm{X}} & 1 & 2 & 3 \\ \hline 1 & 0 & 1 / 6 & 1 / 6 \\ \hline 2 & 1 / 6 & 0 & 1 / 6 \\ \hline 3 & 1 / 6 & 1 / 6 & 0 \\ \hline \end{array} $$ Check that this is a proper probability distribution. What is the marginal distribution of \(\mathrm{X} ?\) What is the marginal distribution of \(\mathrm{Y}\) ?

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Find the expected values of the random variables \(\mathrm{X}\) and \(\mathrm{Y}\) if \(\quad \operatorname{Pr}(\mathrm{X}=0)=1 / 2 \quad\) and \(\operatorname{Pr}(\mathrm{X}=1)=1 / 2\) and \(\operatorname{Pr}(\mathrm{Y}=1)=1 / 4 \quad\) and \(\operatorname{Pr}(\mathrm{Y}=2)=3 / 4\). Compare the sum of \(\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})\) with \(\mathrm{E}(\mathrm{X}+\mathrm{Y})\) if \(\operatorname{Pr}(\mathrm{X}=\mathrm{x}, \mathrm{Y}=\mathrm{y})=\operatorname{Pr}(\mathrm{X}=\mathrm{x}) \operatorname{Pr}(\mathrm{Y}=\mathrm{y})\)

Consider a simple random variable \(\mathrm{X}\) having just two possible values \(\operatorname{Pr}(\mathrm{X}=1)=\mathrm{p}\) and \(\operatorname{Pr}(\mathrm{X}=0)=1-\mathrm{p}\). Find the moment generating function of \(\mathrm{X}\) and \(\mathrm{E}\left(\mathrm{X}^{\mathrm{k}}\right)\) for all \(\mathrm{k}=1,2,3, \ldots\)

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