Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 5

A random variable \(x\) takes the values \(0,1,2,3\), with probabilities \(\frac{5}{12}, \frac{1}{3}, \frac{1}{12}, \frac{1}{6}\)

Short Answer

Expert verified
The probabilities sum to 1, and are correctly assigned.

Step by step solution

01

- Verify Probabilities

Sum all given probabilities to ensure they equal 1: \[\frac{5}{12} + \frac{1}{3} + \frac{1}{12} + \frac{1}{6}\] Convert all fractions to a common denominator to sum them.
02

- Convert Fractions to a Common Denominator

Find the least common denominator (LCD) for the fractions. The denominators are 12, 3, 12, and 6. The LCD is 12. Convert each fraction: \[\frac{1}{3} = \frac{4}{12} \] and \[\frac{1}{6} = \frac{2}{12} \].
03

- Add the Converted Fractions

Sum each converted fraction: \[\frac{5}{12} + \frac{4}{12} + \frac{1}{12} + \frac{2}{12} = \frac{12}{12} = 1 \]. Since the sum is 1, the probabilities are correctly assigned.

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.

Start your free trial

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.

Probability Distribution
In this exercise, we see a random variable, often denoted by \( x \), which can take on specific values \(0, 1, 2,\) and \(3\). Each value is associated with a probability. A probability distribution shows how the probabilities are distributed over the values this random variable can take. For a valid probability distribution:
  • All probabilities must be non-negative.
  • The sum of all probabilities must equal 1.
This ensures that all potential outcomes are accounted for appropriately.
Common Denominator
When working with fractions, especially in problems about probabilities, it's essential to have a common denominator. A common denominator allows us to add or subtract fractions more easily by ensuring they share the same base. Imagine you have different sized pieces of pie and you want to combine them, you need to make sure they all relate to a similar sized piece for accurate calculation. In this exercise, you need to convert each probability so they share a common denominator before summing them up. This step ensures precision in our work.
Fraction Addition
Once all fractions have the same denominator, you can proceed to the addition. Fraction addition involves adding the numerators while keeping the denominator constant. Think of it like combining slices of the same pie. In the exercise, after converting each probability to a fraction with a denominator of 12, you end up with: \(\frac{5}{12} + \frac{4}{12} + \frac{1}{12} + \frac{2}{12}\). Adding these gives: \( \frac{12}{12} \), which simplifies to 1. This result is crucial as it confirms the probabilities are correctly assigned.
Sum of Probabilities
A vital check for any probability distribution is that the sum of all probabilities must equal 1. This ensures the random variable accounts for all potential outcomes, forming a complete scenario. In this specific exercise, after summing the fractions: \(\frac{5}{12} + \frac{4}{12} + \frac{1}{12} + \frac{2}{12}\), we verified that the total is indeed 1. This confirms that all probabilities were correctly distributed among the possible values \( \{0, 1, 2, 3\} \). This is a key property in probability that helps guarantee the model is correctly representing real-world randomness.

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

Plot a graph of the binomial frequency function \(f(x)\) for the case \(n=6, p=\frac{1}{6}, q=\frac{5}{6}\), representing the probability of, say, \(x\) aces in 6 throws of a die. Also draw graphs of \(n f(x)\) as a function of \(x / n\), and of \(F(x)\). What is the probability of at least 2 aces out of 6 tosses of a die?

Suppose that Martian dice are t-sided (tetrahedra) with points labeled 1 to \(4 .\) When a pair of these dice is tossed, let \(x\) be the product of the two numbers at the tops of the dice if the product is odd; otherwise \(x=0\).

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) \cdot p(B) .\) Similarly, two random variables \(x\) and \(y\) with probability functions \(f(x)\) and \(g(y)\) are called independent if the probability of \(x=x_{i}\) and \(y=y_{j}\) is \(f\left(x_{i}\right) \cdot g\left(y_{j}\right)\) for every pair of values of \(x\) and \(y\), that is, if the joint probability function for \(x, y\) is \(f(x) g(y)\). Show that if \(x\) anct \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) \cdot E(y)=\mu_{x} \mu_{y}\).

What is the probability that you and a friend have different birthdays? (For simplicity, let a year have 365 days.) What is the probability that three people have three different birthdays? Show that the probability that \(n\) people have \(n\) different birthdays is $$ p=\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right) \cdots\left(1-\frac{n-1}{365}\right) $$ Estimate this for \(n \notin 365\) by calculating in \(p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x<1]\). Find the smallest (integral) \(n\) for which \(p<\frac{1}{2}\). Hence show that for a group of 23 people or more, the probability is greater than \(\frac{1}{2}\) that two of them have the same birthdit? (Try it with a group of friends or a list of people such as the presidents of the United States.)

Two dice are thrown. Given the information that the number on the first die is even, and the number on the second is \(<4\), set up an appropriate sample space and answer the following questions. (a) What are the possible sums and their probabilities? (b) What is the most probable sum? (c) What is the probability that the sum is even?
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
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