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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: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \text { outcome } & a & b & c & d & e & f \\ \hline x & 0 & 0 & 1.5 & 1.5 & 2 & 3 \\ \hline \end{array} $$ Determine the probability mass function of \(a\). 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
P(X=1.5)=1/3, P(0.53)=0, P(0≤X<2)=2/3, P(X=0 or X=2)=1/2.

Step by step solution

01

Determine Probability of Each Outcome

Since the sample space contains six outcomes {a, b, c, d, e, f} and each is equally likely, the probability of each outcome is \( \frac{1}{6} \).
02

Define Probability Mass Function (PMF)

The PMF is defined as:\[P(X = x) = \begin{cases} \frac{1}{3}, & \text{if } x = 0 \ \frac{1}{3}, & \text{if } x = 1.5 \ \frac{1}{6}, & \text{if } x = 2 \ \frac{1}{6}, & \text{if } x = 3 \end{cases}\]The values 0, 1.5, 2, and 3 have probabilities computed by counting occurrences and dividing by the sample space size.
03

Compute P(X=1.5)

Outcomes 'c' and 'd' corresponds to \(X = 1.5\), thus \(P(X = 1.5) = 2 \times \frac{1}{6} = \frac{1}{3}\).
04

Compute P(0.5

The values \(X = 1.5\) and \(X = 2\) fall in this interval. Hence, \(P(0.5
05

Compute P(X>3)

There are no outcomes for \(X\) greater than 3, so \(P(X>3) = 0\).
06

Compute P(0 ≤ X

The values \(X = 0\) and \(X = 1.5\) satisfy this condition. Therefore, \(P(0 \leq X<2) = P(X=0) + P(X=1.5) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}\).
07

Compute P(X=0 or X=2)

These values correspond to outcomes where \(X = 0\) and \(X = 2\), thus \[P(X=0 \text{ or } X=2) = P(X=0) + P(X=2) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\].

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Key Concepts

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

Random Variables
A random variable is a crucial concept in probability and statistics and can feel a bit abstract at first. It is essentially a function that associates a numerical value with each possible outcome of a random event. For instance, in our exercise, each letter from the sample space \( \{a, b, c, d, e, f\} \) is mapped to a numerical value through a random variable \( X \). This association can be due to any characteristic observed in the outcomes, such as measurement or counting.

In our particular problem, the random variable is defined with specific values: outcomes 'a' and 'b' both map to \( 0 \), 'c' and 'd' to \( 1.5 \), 'e' to \( 2 \), and 'f' to \( 3 \). These mappings help us assess and predict the likelihood of different events by assigning a probability to the numeric outputs of our random variable.
Probability Distributions
Probability distributions describe how probabilities are distributed over the values of the random variable. When we talk about a probability mass function (PMF) in the context of discrete random variables like in our exercise, we are describing the probability distribution.

The PMF gives us a function that assigns each outcome a probability. Our PMF is as follows:
  • \( P(X = 0) = \frac{1}{3} \)
  • \( P(X = 1.5) = \frac{1}{3} \)
  • \( P(X = 2) = \frac{1}{6} \)
  • \( P(X = 3) = \frac{1}{6} \)
These probabilities are calculated based on how often each value of \( X \) appears relative to the total number of outcomes.
Understanding this distribution is instrumental in answering questions about the random variable's behavior and making informed predictions based on that.
Sample Space
The sample space of a random experiment encompasses all possible outcomes. It's the foundation upon which random variables and probability distributions are built. In our exercise, the sample space is \( \{a, b, c, d, e, f\} \).

This sample space is assumed to be equally likely. Hence, each outcome has a probability of \( \frac{1}{6} \) before considering the effect of the random variable \( X \). The sample space acts as the universal set from which we draw conclusions, analyze events, and apply random variables to give a numerical form to our probability questions.
Being aware of the sample space is crucial, as it dictates the possible outcomes and forms the basis for determining probabilities in any probabilistic model.
Discrete Probability
Discrete probability deals with situations where the set of possible outcomes is finite or countable. In our problem, we are working within a discrete probability framework with the sample space \( \{a, b, c, d, e, f\} \).

Each outcome has a specific probability, and the random variable \( X \) takes on discrete values (0, 1.5, 2, 3) with associated probabilities mapped by the PMF. The beauty of discrete probability lies in its ability to precisely determine probability values for specific outcomes, making it applicable to real-world scenarios where distinct and countable outcomes are observed. This contrasts with continuous probability, where outcomes could take any value within a range, requiring different techniques of analysis.
These concepts of discreteness provide clarity and structure when delving into probability-related questions, especially in exercises like ours where each part asks for probability calculations connected to these specific discrete values.

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