The probabilities of events E, F, and EnF are given below. Find (a) P(EIF), (b) P(FIE), (c) P (EIF'), and (d) P (FIE'). 1 P(E)=7, P(F) = P(EnF)= 1/1₁ ₁ =1/11

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Chapter2: Second-order Linear Odes
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The probabilities of events E, F, and E∩F are given below. Find (a) P(E|F), (b) P(F|E), (c) P(E|Fʹ), and (d) P(F|Eʹ).

\[ P(E) = \frac{1}{7}, P(F) = \frac{1}{4}, P(E∩F) = \frac{1}{8} \]

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a. \( P(E|F) = \)
(Type an integer or a simplified fraction.)

b. \( P(F|E) = \)
(Type an integer or a simplified fraction.)

c. \( P(E|Fʹ) = \)
(Type an integer or a simplified fraction.)

d. \( P(F|Eʹ) = \)
(Type an integer or a simplified fraction.)

---

This problem involves finding conditional probabilities given basic probability values of events E, F, and their intersection E∩F. The provided information includes \( P(E) \), \( P(F) \), and \( P(E∩F) \). The tasks are to calculate various conditional probabilities derived from these values.

To solve these problems, one would typically use the following formulas:
- \( P(E|F) = \frac{P(E∩F)}{P(F)} \)
- \( P(F|E) = \frac{P(E∩F)}{P(E)} \)
- \( P(E|Fʹ) = \frac{P(E) - P(E∩F)}{P(Fʹ)} \text{ where } P(Fʹ)=1-P(F) \)
- \( P(F|Eʹ) = \frac{P(F) - P(E∩F)}{P(Eʹ)} \text{ where } P(Eʹ)=1-P(E) \)

Please enter your answers, simplified as much as possible, into the corresponding text boxes.
Transcribed Image Text:The probabilities of events E, F, and E∩F are given below. Find (a) P(E|F), (b) P(F|E), (c) P(E|Fʹ), and (d) P(F|Eʹ). \[ P(E) = \frac{1}{7}, P(F) = \frac{1}{4}, P(E∩F) = \frac{1}{8} \] --- a. \( P(E|F) = \) (Type an integer or a simplified fraction.) b. \( P(F|E) = \) (Type an integer or a simplified fraction.) c. \( P(E|Fʹ) = \) (Type an integer or a simplified fraction.) d. \( P(F|Eʹ) = \) (Type an integer or a simplified fraction.) --- This problem involves finding conditional probabilities given basic probability values of events E, F, and their intersection E∩F. The provided information includes \( P(E) \), \( P(F) \), and \( P(E∩F) \). The tasks are to calculate various conditional probabilities derived from these values. To solve these problems, one would typically use the following formulas: - \( P(E|F) = \frac{P(E∩F)}{P(F)} \) - \( P(F|E) = \frac{P(E∩F)}{P(E)} \) - \( P(E|Fʹ) = \frac{P(E) - P(E∩F)}{P(Fʹ)} \text{ where } P(Fʹ)=1-P(F) \) - \( P(F|Eʹ) = \frac{P(F) - P(E∩F)}{P(Eʹ)} \text{ where } P(Eʹ)=1-P(E) \) Please enter your answers, simplified as much as possible, into the corresponding text boxes.
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