/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 For each of the following exerci... [FREE SOLUTION] | 91Ó°ÊÓ

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For each of the following exercises, determine the range (possible values) of the random variable. The sample space of a random experiment is \(\\{a, b, c, d,\) \(e, f\\},\) and each outcome is equally likely. A random variable is defined as follows: $$ \begin{array}{ccccccc} \text { outcome } & a & b & c & d & e & f \\ x & 0 & 0 & 1.5 & 1.5 & 2 & 3 \end{array} $$ Determine the probability mass function of \(X\). Use the probability mass function to determine the following probabilities: a. \(P(X=1.5)\) b. \(P(0.53)\) d. \(P(0 \leq X<2)\) e. \(P(X=0\) or \(X=2\) )

Short Answer

Expert verified
a. \( \frac{1}{3} \) b. \( \frac{1}{2} \) c. 0 d. \( \frac{2}{3} \) e. \( \frac{1}{2} \)

Step by step solution

01

Define Probability Mass Function (PMF)

The PMF, denoted as \( P(X = x) \), lists the probabilities of the random variable \( X \) taking each possible value. First, identify the distinct values that \( X \) can take: 0, 1.5, 2, and 3. Since each sample outcome \( \{a, b, c, d, e, f\} \) is equally likely and there are 6 total outcomes, each has probability \( \frac{1}{6} \). Count the occurrences of each value to find their probabilities:- \( P(X=0) \) for outcomes \{a, b\}: \( \frac{2}{6} = \frac{1}{3} \)- \( P(X=1.5) \) for outcomes \{c, d\}: \( \frac{2}{6} = \frac{1}{3} \)- \( P(X=2) \) for outcome \{e\}: \( \frac{1}{6} \)- \( P(X=3) \) for outcome \{f\}: \( \frac{1}{6} \).
02

Determine \( P(X=1.5) \)

From the PMF, we found that \( X = 1.5 \) corresponds to two outcomes, \{c, d\}. Each has a probability of \( \frac{1}{6} \). Thus, the probability that \( X = 1.5 \) is \( \frac{1}{3} \).
03

Determine \( P(0.5

The range \( 0.5 < X < 2.7 \) implies all possibilities except 0 and 3. From the PMF, this includes \( X = 1.5 \) and \( X = 2 \). Thus, the probability is the sum:- \( P(X=1.5) = \frac{1}{3} \)- \( P(X=2) = \frac{1}{6} \)So, \( P(0.5<X<2.7) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \).
04

Determine \( P(X>3) \)

Check the PMF for values of \( X \) that are greater than 3. There are none, so \( P(X>3) = 0 \).
05

Determine \( P(0 \leq X

From the values of \( X \), \( 0 \leq X < 2 \) includes \( X = 0 \) and \( X = 1.5 \):- \( P(X=0) = \frac{1}{3} \)- \( P(X=1.5) = \frac{1}{3} \)Therefore, \( P(0 \leq X<2) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \).
06

Determine \( P(X=0 \text{ or } X=2) \)

Using the PMF, find \( P(X=0) \) and \( P(X=2) \):- \( P(X=0) = \frac{1}{3} \)- \( P(X=2) = \frac{1}{6} \)Thus, \( P(X=0 \text{ or } X=2) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a fundamental concept that acts as a numeric representation of a random outcome from a certain experiment. It essentially maps the outcomes of a random experiment to numerical values. For example, in the given exercise, we have outcomes labeled from "a" to "f."
Each of these corresponds to specific values of the random variable, say 0, 1.5, 2, and 3. Random variables can be categorized as discrete or continuous, with our current focus being on discrete types.
  • Discrete Random Variable: Takes on a countable number of distinct values. These values can be explicitly listed as shown in our problem.
  • Continuous Random Variable: Can take any value within a given range, not applicable to our current discussion.
Sample Space
The sample space is the collection of all possible outcomes of a random experiment. It is the foundation for defining related probabilities and random variables. In our exercise, the sample space is defined as \( \{a, b, c, d, e, f \} \) representing different possible scenarios that can arise.
This sample space allows us to systematically account for every possibility during the experiment. By knowing all the possible outcomes, we can assign probabilities to each, with the help of a random variable that maps these outcomes to real numbers. This process makes it easier to work out probabilities for various events.
Understanding the sample space is crucial as it lays the groundwork for determining how often specific events occur, and what numerical values are to be assigned through random variables.
Equally Likely Outcomes
Equally likely outcomes imply that each outcome in a sample space has the same probability of occurring. This scenario often simplifies the calculation of probabilities because each event is treated the same. In the exercise, it is mentioned that our outcomes \( \{a, b, c, d, e, f \} \) are all equally likely.
With six outcomes present, each must therefore have a probability of \( \frac{1}{6} \). This assumption allows us to directly calculate the probability of any related event by multiplying the probability of a single outcome by the number of outcomes contributing to this event.
This assumption of equally likely outcomes often accompanies practical examples such as rolling a fair die or flipping a fair coin, where each face or side has an equal chance of occurrence.
Discrete Probability
Discrete probability focuses on discrete random variables, where possible values form a finite or countable set. In discrete probability, we utilize probability mass functions (PMFs) to assign probabilities to these values. The term "mass" refers to how likelihood "accumulates" at specific points in the set.
A probability mass function (PMF), denoted as \( P(X = x) \), allows us to tabulate probabilities for each potential value of a discrete random variable. In the current exercise, each value from the discrete set \( \{0, 1.5, 2, 3\} \) has its probability determined based on the occurrence in the sample space.
  • \( P(X=0) = \frac{1}{3} \), because outcomes \( {a, b} \) map to 0.
  • \( P(X=1.5) = \frac{1}{3} \), because outcomes \( {c, d} \) map to 1.5.
  • \( P(X=2) = \frac{1}{6} \), because only {e} maps to 2.
  • \( P(X=3) = \frac{1}{6} \), because only {f} maps to 3.
Discrete probability helps us compute and understand events for specific partitions or categories within our data set, making it a vital concept in probability studies.

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