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
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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) \( P(F | E) \)

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(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 \( P(E \cap F) \):
\[ P(E \cap F) = P(E) + P(F) - P(E \cup F) \]

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There are no graphs or diagrams to explain in this image.
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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