Given the following discrete probability distribution: -2 -1 1 2 f(x) 1/8 2/8 2/8 2/8 1/8 Then P(X < -1 or X = 2) is:
Given the following discrete probability distribution: -2 -1 1 2 f(x) 1/8 2/8 2/8 2/8 1/8 Then P(X < -1 or X = 2) is:
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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1/2 ,7/8 , 3/4 , 1
![**Discrete Probability Distribution**
Given the following discrete probability distribution:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & \frac{1}{8} & \frac{2}{8} & \frac{2}{8} & \frac{2}{8} & \frac{1}{8} \\
\hline
\end{array}
\]
**Explanation of the Table:**
- The table represents a discrete probability distribution with random variable \( X \) taking on values -2, -1, 0, 1, and 2.
- The probabilities associated with these values, \( f(x) \), are \(\frac{1}{8}\) for \( x = -2 \), \(\frac{2}{8}\) for \( x = -1 \), \(\frac{2}{8}\) for \( x = 0 \), \(\frac{2}{8}\) for \( x = 1 \), and \(\frac{1}{8}\) for \( x = 2 \).
**Problem Statement:**
Then \( P(X \leq -1 \text{ or } X = 2) \) is:
To find this probability, sum the probabilities for the values satisfying \( X \leq -1 \) (which includes \( x = -2 \) and \( x = -1 \)) and for \( X = 2 \).
\[ P(X \leq -1 \text{ or } X = 2) = f(-2) + f(-1) + f(2) = \frac{1}{8} + \frac{2}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9a59ff3-32cb-4a72-806c-172db4d1cb49%2Fddf4bfe0-207c-423c-8198-761d9069e840%2F3jsaogl_processed.png&w=3840&q=75)
Transcribed Image Text:**Discrete Probability Distribution**
Given the following discrete probability distribution:
\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & \frac{1}{8} & \frac{2}{8} & \frac{2}{8} & \frac{2}{8} & \frac{1}{8} \\
\hline
\end{array}
\]
**Explanation of the Table:**
- The table represents a discrete probability distribution with random variable \( X \) taking on values -2, -1, 0, 1, and 2.
- The probabilities associated with these values, \( f(x) \), are \(\frac{1}{8}\) for \( x = -2 \), \(\frac{2}{8}\) for \( x = -1 \), \(\frac{2}{8}\) for \( x = 0 \), \(\frac{2}{8}\) for \( x = 1 \), and \(\frac{1}{8}\) for \( x = 2 \).
**Problem Statement:**
Then \( P(X \leq -1 \text{ or } X = 2) \) is:
To find this probability, sum the probabilities for the values satisfying \( X \leq -1 \) (which includes \( x = -2 \) and \( x = -1 \)) and for \( X = 2 \).
\[ P(X \leq -1 \text{ or } X = 2) = f(-2) + f(-1) + f(2) = \frac{1}{8} + \frac{2}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]
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