Let P(E)=0.85, P(F) = 0.23, and P(EUF) = 0.94. Find (a) P(EIF) and (b) P(FIE). (a) P(EIF) = (Type an integer or decimal rounded to two decimal places as needed.) (b) P(FIE) = (Type an integer or decimal rounded to two decimal places as needed.) ...
Let P(E)=0.85, P(F) = 0.23, and P(EUF) = 0.94. Find (a) P(EIF) and (b) P(FIE). (a) P(EIF) = (Type an integer or decimal rounded to two decimal places as needed.) (b) P(FIE) = (Type an integer or decimal rounded to two decimal places as needed.) ...
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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![Below is a transcription of the provided image, designed for educational purposes. The content includes probabilities and requires the application of conditional probability formulas.
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**Probability Problem**
Given:
- \( P(E) = 0.85 \)
- \( P(F) = 0.23 \)
- \( P(E \cup F) = 0.94 \)
Find:
- (a) \( P(E | F) \)
- (b) \( P(F | E) \)
---
(a) \( P(E | F) = \) [ ]
(Type an integer or decimal rounded to two decimal places as needed.)
---
(b) \( P(F | E) = \) [ ]
(Type an integer or decimal rounded to two decimal places as needed.)
---
**Explanation:**
- \( P(E | F) \) represents the probability of event E occurring given that event F has occurred.
- \( P(F | E) \) represents the probability of event F occurring given that event E has occurred.
To solve these problems, you will need to apply the definitions of conditional probability:
\[ P(E | F) = \frac{P(E \cap F)}{P(F)} \]
\[ P(F | E) = \frac{P(F \cap E)}{P(E)} \]
Since the union of E and F is provided, you may need to calculate the intersection \( P(E \cap F) \):
\[ P(E \cap F) = P(E) + P(F) - P(E \cup F) \]
---
There are no graphs or diagrams to explain in this image.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff138e77-4b21-4b3f-a581-600ea0faba85%2F5cfa61f1-a168-4df8-8ac9-2ced7f80c83e%2Fqnpxj3t_processed.png&w=3840&q=75)
Transcribed Image Text:Below is a transcription of the provided image, designed for educational purposes. The content includes probabilities and requires the application of conditional probability formulas.
---
**Probability Problem**
Given:
- \( P(E) = 0.85 \)
- \( P(F) = 0.23 \)
- \( P(E \cup F) = 0.94 \)
Find:
- (a) \( P(E | F) \)
- (b) \( P(F | E) \)
---
(a) \( P(E | F) = \) [ ]
(Type an integer or decimal rounded to two decimal places as needed.)
---
(b) \( P(F | E) = \) [ ]
(Type an integer or decimal rounded to two decimal places as needed.)
---
**Explanation:**
- \( P(E | F) \) represents the probability of event E occurring given that event F has occurred.
- \( P(F | E) \) represents the probability of event F occurring given that event E has occurred.
To solve these problems, you will need to apply the definitions of conditional probability:
\[ P(E | F) = \frac{P(E \cap F)}{P(F)} \]
\[ P(F | E) = \frac{P(F \cap E)}{P(E)} \]
Since the union of E and F is provided, you may need to calculate the intersection \( P(E \cap F) \):
\[ P(E \cap F) = P(E) + P(F) - P(E \cup F) \]
---
There are no graphs or diagrams to explain in this image.
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