Question

Let \(f(x)=1 / x^{2}, 1

Short answer

The probability \(P_X(C_1 \cup C_2)\) is the sum of the two integral results and the probability of \(P_X(C_1 \cap C_2)\) is 0 as the two intervals are disjoint.

Step-by-step solution

Identify the Intervals and Functions

Identify the intervals \(C_1 = \{x: 1 < x < 2\}\) and \(C_2 = \{x: 4 < x < 5\}\). The probability density function is given as \(f(x) = 1 / x^2\).

Calculate \(P_X(C_1 \cup C_2)\)

To find the probability of \(X\) lying in either \(C_1\) or \(C_2\), sum the probabilities of the two intervals. This is done by integrating the pdf from 1 to 2 and from 4 to 5, then adding the results, as \(P_X(C_1 \cup C_2) = P_X(C_1) + P_X(C_2)\). So, calculate the integral:\[\int_{1}^{2} f(x) dx + \int_{4}^{5} f(x) dx\]and simplify.

Calculate \(P_X(C_1 \cap C_2)\)

Since the sets \(C_1\) and \(C_2\) are disjoint (they have no elements in common), their intersection is empty. Therefore, \(P_X(C_1 \cap C_2) = 0\).

Summarize findings

Summarize the calculated probabilities for \(P_X(C_1 \cup C_2)\) and \(P_X(C_1 \cap C_2)\).

Key concepts

Disjoint Sets

Disjoint sets are collections of elements that have no shared members, meaning they do not overlap in any way. In probabilistic terms, if you have two events represented by sets which are disjoint, the probability of their intersection is zero. Essentially, they cannot both occur at the same time. This is a simple yet powerful concept.

In the context of our exercise, the sets \(C_1\) and \(C_2\) are disjoint because they represent two different, non-overlapping intervals: \(C_1 = \{x: 1 < x < 2\}\) and \(C_2 = \{x: 4 < x < 5\}\). Since these intervals do not share common values, the probability of simultaneously being in both is zero. This is expressed mathematically as:

• \( P_X(C_1 \cap C_2) = 0 \)

Understanding disjoint sets simplifies calculations, especially when determining probabilities, as there is no need to consider overlapping elements.

Integration

Integration is a mathematical tool used to calculate areas, volumes, central points, and various useful concepts. In probability, integration is essential when dealing with continuous random variables and their probability density functions (pdf).

When given a pdf, integrating over an interval gives the probability that the random variable falls within that interval. For our exercise, \(f(x) = 1/x^2\) is the pdf, and the task is to find the probability over intervals \(C_1\) and \(C_2\). This is performed as follows:

• Calculate \(\int_{1}^{2} \frac{1}{x^2} \, dx\) for interval \(C_1\).
• Calculate \(\int_{4}^{5} \frac{1}{x^2} \, dx\) for interval \(C_2\).

After finding these integrals, you can determine the likelihood of \(X\) falling within each specific interval.

Probability Calculation

Probability calculation involves determining the likelihood of a random variable falling into a particular set of values. When dealing with a probability density function, this involves integration as previously mentioned.

For disjoint sets like \(C_1\) and \(C_2\), to find the probability of the union \(C_1 \cup C_2\), you simply add their individual probabilities since they do not overlap. Mathematically, this is expressed as:

• \(P_X(C_1 \cup C_2) = P_X(C_1) + P_X(C_2)\)

The exercise involves adding the integrals of the pdf over each defined interval:

• \(\int_{1}^{2} \frac{1}{x^2} \, dx + \int_{4}^{5} \frac{1}{x^2} \, dx\)

By calculating these, you get the overall probability of \(X\) lying in either one of these intervals.