Sample space S is partitioned into E₁, E2, and E3 such that P(E1. a) Find P(E3). b) Find the odds in favor of and the odds against E3 occurring. a) P(E3) = (Simplify your answer.) b) The odds in favor of E3 occurring, in lowest terms, are (Simplify your answer. Type whole numbers.) The odds against E3, in lowest terms, are (Simplify your answer. Type whole numbers.) co and P(E
Sample space S is partitioned into E₁, E2, and E3 such that P(E1. a) Find P(E3). b) Find the odds in favor of and the odds against E3 occurring. a) P(E3) = (Simplify your answer.) b) The odds in favor of E3 occurring, in lowest terms, are (Simplify your answer. Type whole numbers.) The odds against E3, in lowest terms, are (Simplify your answer. Type whole numbers.) co and P(E
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Probability and Odds Calculation**
**Problem Statement:**
Given that a sample space \( S \) is partitioned into \( E_1 \), \( E_2 \), and \( E_3 \) such that \( P(E_1) = \frac{1}{8} \) and \( P(E_2) = \frac{1}{6} \):
a) Find \( P(E_3) \).
b) Find the odds in favor of and the odds against \( E_3 \) occurring.
**Solution Steps:**
---
**a) \( P(E_3) = \) [Blank space provided for the answer]**
(Simplify your answer.)
**b) The odds in favor of \( E_3 \) occurring, in lowest terms, are: \(\, \) [Blank space provided for the ratio]
(Simplify your answer. Type whole numbers.)
The odds against \( E_3 \), in lowest terms, are: \(\, \) [Blank space provided for the ratio]
(Simplify your answer. Type whole numbers.)
To find \( P(E_3) \):
1. Recall that the total probability of the sample space \( S \) equals 1:
\[ P(E_1) + P(E_2) + P(E_3) = 1 \]
2. Substitute \( P(E_1) \) and \( P(E_2) \):
\[ \frac{1}{8} + \frac{1}{6} + P(E_3) = 1 \]
3. Calculate the common denominator for \(\frac{1}{8}\) and \(\frac{1}{6}\):
The common denominator is 24. Converting the probabilities:
\[ \frac{1}{8} = \frac{3}{24} \]
\[ \frac{1}{6} = \frac{4}{24} \]
4. Add these fractions:
\[ \frac{3}{24} + \frac{4}{24} = \frac{7}{24} \]
5. Subtract this sum from 1 to find \( P(E_3) \):
\[ 1 - \frac{7}{24} = \frac{24}{24} - \frac{7}{24} = \frac{17](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b1d7e88-5924-4190-9ef7-53bc83ebfef1%2F474c38c9-60a8-4f5c-9243-49d7f157f027%2Fudwrbx8_processed.png&w=3840&q=75)
Transcribed Image Text:**Probability and Odds Calculation**
**Problem Statement:**
Given that a sample space \( S \) is partitioned into \( E_1 \), \( E_2 \), and \( E_3 \) such that \( P(E_1) = \frac{1}{8} \) and \( P(E_2) = \frac{1}{6} \):
a) Find \( P(E_3) \).
b) Find the odds in favor of and the odds against \( E_3 \) occurring.
**Solution Steps:**
---
**a) \( P(E_3) = \) [Blank space provided for the answer]**
(Simplify your answer.)
**b) The odds in favor of \( E_3 \) occurring, in lowest terms, are: \(\, \) [Blank space provided for the ratio]
(Simplify your answer. Type whole numbers.)
The odds against \( E_3 \), in lowest terms, are: \(\, \) [Blank space provided for the ratio]
(Simplify your answer. Type whole numbers.)
To find \( P(E_3) \):
1. Recall that the total probability of the sample space \( S \) equals 1:
\[ P(E_1) + P(E_2) + P(E_3) = 1 \]
2. Substitute \( P(E_1) \) and \( P(E_2) \):
\[ \frac{1}{8} + \frac{1}{6} + P(E_3) = 1 \]
3. Calculate the common denominator for \(\frac{1}{8}\) and \(\frac{1}{6}\):
The common denominator is 24. Converting the probabilities:
\[ \frac{1}{8} = \frac{3}{24} \]
\[ \frac{1}{6} = \frac{4}{24} \]
4. Add these fractions:
\[ \frac{3}{24} + \frac{4}{24} = \frac{7}{24} \]
5. Subtract this sum from 1 to find \( P(E_3) \):
\[ 1 - \frac{7}{24} = \frac{24}{24} - \frac{7}{24} = \frac{17
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