Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
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

01

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 \(f(x) = 1 / x^2\).
02

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.
03

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, \(P_X(C_1 \cap C_2) = 0\).
04

Summarize findings

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X\) be a random variable such that \(R(t)=E\left(e^{t(X-b)}\right)\) exists for \(t\) such that \(-h

Let \(X\) be a random variable with mean \(\mu\) and variance \(\sigma^{2}\) such that the third moment \(E\left[(X-\mu)^{3}\right]\) about the vertical line through \(\mu\) exists. The value of the ratio \(E\left[(X-\mu)^{3}\right] / \sigma^{3}\) is often used as a measure of skeuness. Graph each of the following probability density functions and show that this measure is negative, zero, and positive for these respective distributions (which are said to be skewed to the left, not skewed, and skewed to the right, respectively). (a) \(f(x)=(x+1) / 2,-1

A French nobleman, Chevalier de Méré, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in 4 independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de Méré that the probabilities should be the same. Compute the probabilities of \((1)\) and \((2)\).

The random variable \(X\) is said to be stochastically larger than the random variable \(Y\) if $$P(X>z) \geq P(Y>z)$$ for all real \(z\), with strict inequality holding for at least one \(z\) value. Show that this requires that the cdfs enjoy the following property $$F_{X}(z) \leq F_{Y}(z)$$ for all real \(z\), with strict inequality holding for at least one \(z\) value.

For each of the following pdfs of \(X\), find \(P(|X|<1)\) and \(P\left(X^{2}<9\right)\). (a) \(f(x)=x^{2} / 18,-3

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free