Question

Let the probability set function \(P_{X}(D)\) of the random variable \(X\) be \(P_{X}(D)=\) \(\int_{D} f(x) d x\), where \(f(x)=2 x / 9, x \in \mathcal{D}=\\{x: 0

Short answer

The probabilities are obtained as: \(P_{X}(D_{1})= \int_{0}^{1} \frac{2x}{9} dx = \frac{1}{9}\), \(P_{X}(D_{2})= \int_{2}^{3} \frac{2x}{9} dx = \frac{5}{9}\), \(P_{X}(D_{1} \cup D_{2})= \frac{1}{9}+ \frac{5}{9}= \frac{2}{3}\)

Step-by-step solution

Computing Probability \(P_{X}(D_{1})\)

To compute this probability, one needs to evaluate the integral of \(f(x)\) over the interval defined by \(D_{1}\). This involves integrating the function from 0 to 1. \(P_{X}(D_{1})= \int_{0}^{1} f(x) dx = \int_{0}^{1} \frac{2x}{9} dx\)

Computing Probability \(P_{X}(D_{2})\)

By following the same process, the probability for \(D_{2}\) is computed as the integral of \(f(x)\) over the interval defined by \(D_{2}\) (from 2 to 3). \(P_{X}(D_{2})= \int_{2}^{3} f(x) dx = \int_{2}^{3} \frac{2x}{9} dx\)

Computing Probability of the Union \(P_{X}(D_{1} \cup D_{2})\)

Considering that \(D_{1}\) and \(D_{2}\) are disjoint sets (their intersection is empty), the probability of their union is equal to the sum of their individual probabilities. \(P_{X}(D_{1} \cup D_{2})=P_{X}(D_{1})+P_{X}(D_{2})\)

Key concepts

Continuous Random Variable

In the realm of probability and statistics, a continuous random variable is a variable that can take an infinite number of possible values. Unlike a discrete random variable which can only take specific values, a continuous random variable can assume any value within a given range. One common example is measuring time or temperature, where the values can be fine-tuned to an arbitrary level of precision.

In our exercise scenario, the continuous random variable is denoted as X and it takes values in the range from 0 to 3. This means that X can be any number between 0 and 3, such as 0.5, 2.718, or even an irrational number like \(\pi\). It's essential to understand that we cannot list all possible values that X can take; they are uncountably infinite. To handle such variables, we rely on probability density functions and integrals to find the probabilities of certain events.

Integral of Probability Density Function

The probability density function (PDF), represented as \(f(x)\) in our exercise, is the function that describes the likelihood of a continuous random variable to take on a specific value. To find out how likely a random variable is to be within a specific range, we integrate the PDF over that range. This integral gives us the probability that the random variable will fall within the interval we're interested in.

For instance, the exercise presents a PDF \(f(x) = \frac{2x}{9}\), where \(x\) is within the range \(0 < x < 3\). To compute the probability that \(X\) falls within a specific interval, such as \(D_1\), we calculate the integral of \(f(x)\) from the lower bound to the upper bound of that interval. Therefore, the integration \(\int_{0}^{1} \frac{2x}{9} dx\) performed in the solution tells us the probability of \(X\) being between 0 and 1, which is a fundamental application of integrals in probability theory for continuous variables.

Disjoint Sets in Probability

In probability theory, disjoint sets, also known as mutually exclusive sets, are sets that have no elements in common. In simpler terms, they cannot happen at the same time. For any two disjoint sets A and B, the intersection A ∩ B would be the empty set.

In our exercise, D1 and D2 represent two disjoint intervals with no overlapping values. D1 is the interval from 0 to 1, and D2 is from 2 to 3. Since there is no value that lies in both intervals, they are disjoint. When dealing with disjoint sets, as shown in the solution, the probability of their union (D1 ∪ D2) is simply the sum of their probabilities. This characteristic simplifies many problems in probability, as it allows the separate calculation of probabilities when dealing with non-overlapping events.