Sample space S is partitioned into E₁, E₂, E3, and E4 such that P(E₁)=0.12, P(E₂)=0.09, and P(E3)=0.49. a. Find P(E4). b. Find the odds in favor of and the odds against E4 occurring. a. P(E4)= (Simplify your answer.) b. The odds in favor of E4 occurring, in lowest terms, are: (Type whole numbers.) The odds against E4, in lowest terms, are:0· (Type whole numbers.)
Sample space S is partitioned into E₁, E₂, E3, and E4 such that P(E₁)=0.12, P(E₂)=0.09, and P(E3)=0.49. a. Find P(E4). b. Find the odds in favor of and the odds against E4 occurring. a. P(E4)= (Simplify your answer.) b. The odds in favor of E4 occurring, in lowest terms, are: (Type whole numbers.) The odds against E4, in lowest terms, are:0· (Type whole numbers.)
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 in a Partitioned Sample Space
In this exercise, we consider a sample space \( S \) that is partitioned into four events: \( E_1, E_2, E_3 \), and \( E_4 \). The probabilities for the first three events are given as follows:
- \( P(E_1) = 0.12 \)
- \( P(E_2) = 0.09 \)
- \( P(E_3) = 0.49 \)
#### Problem:
a. Determine \( P(E_4) \).
b. Find the odds in favor of and the odds against \( E_4 \) occurring.
---
**Solution:**
a. To find \( P(E_4) \), use the fact that the total probability of the sample space \( S \) must equal 1.
\[
P(E_1) + P(E_2) + P(E_3) + P(E_4) = 1
\]
Given:
\[
0.12 + 0.09 + 0.49 + P(E_4) = 1
\]
Solving for \( P(E_4) \):
\[
0.70 + P(E_4) = 1 \\
P(E_4) = 1 - 0.70 \\
P(E_4) = 0.30
\]
Thus, \( P(E_4) = 0.30 \).
b. **Odds in Favor of \( E_4 \):**
The odds in favor of an event is the ratio of the probability of the event occurring to the probability of the event not occurring.
\[
\text{Odds in favor of } E_4 = \frac{P(E_4)}{1 - P(E_4)}
\]
Here,
\[
\frac{0.30}{1 - 0.30} = \frac{0.30}{0.70} = \frac{3}{7}
\]
The odds in favor of \( E_4 \) occurring, in lowest terms, are \( \boxed{3} : \boxed{7} \).
**Odds Against \( E_4 \):**
The odds against an event is the ratio of the probability of the event not occurring to the probability of the event occurring.
\[
\text{Odds against } E_4 = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b1d7e88-5924-4190-9ef7-53bc83ebfef1%2Fec49c3e6-775c-4304-b8cf-6d99333e3e24%2Fsv15xmn_processed.png&w=3840&q=75)
Transcribed Image Text:### Probability and Odds in a Partitioned Sample Space
In this exercise, we consider a sample space \( S \) that is partitioned into four events: \( E_1, E_2, E_3 \), and \( E_4 \). The probabilities for the first three events are given as follows:
- \( P(E_1) = 0.12 \)
- \( P(E_2) = 0.09 \)
- \( P(E_3) = 0.49 \)
#### Problem:
a. Determine \( P(E_4) \).
b. Find the odds in favor of and the odds against \( E_4 \) occurring.
---
**Solution:**
a. To find \( P(E_4) \), use the fact that the total probability of the sample space \( S \) must equal 1.
\[
P(E_1) + P(E_2) + P(E_3) + P(E_4) = 1
\]
Given:
\[
0.12 + 0.09 + 0.49 + P(E_4) = 1
\]
Solving for \( P(E_4) \):
\[
0.70 + P(E_4) = 1 \\
P(E_4) = 1 - 0.70 \\
P(E_4) = 0.30
\]
Thus, \( P(E_4) = 0.30 \).
b. **Odds in Favor of \( E_4 \):**
The odds in favor of an event is the ratio of the probability of the event occurring to the probability of the event not occurring.
\[
\text{Odds in favor of } E_4 = \frac{P(E_4)}{1 - P(E_4)}
\]
Here,
\[
\frac{0.30}{1 - 0.30} = \frac{0.30}{0.70} = \frac{3}{7}
\]
The odds in favor of \( E_4 \) occurring, in lowest terms, are \( \boxed{3} : \boxed{7} \).
**Odds Against \( E_4 \):**
The odds against an event is the ratio of the probability of the event not occurring to the probability of the event occurring.
\[
\text{Odds against } E_4 = \
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