12. Suppose that P(E)=0.82, P(F)=0.63, and P(EF) = 0.12. Find the following:

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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A) 0.800

B) 0.940

C) 0.794

D) 0.410

E) 0.060

**Problem 12:**

Suppose that \( P(E) = 0.82 \), \( P(F) = 0.63 \), and \( P(E^c \cap F) = 0.12 \). Find the following:

(No graphs or diagrams are present in this image. It's a simple mathematical problem related to probability involving events \( E \) and \( F \).)
Transcribed Image Text:**Problem 12:** Suppose that \( P(E) = 0.82 \), \( P(F) = 0.63 \), and \( P(E^c \cap F) = 0.12 \). Find the following: (No graphs or diagrams are present in this image. It's a simple mathematical problem related to probability involving events \( E \) and \( F \).)
F. \( P(E^c \cap F^c) \)

This expression represents the probability of the intersection of the complements of events E and F. In probability, the complement of an event E (denoted as \( E^c \)) includes all outcomes in the sample space that are not in E. Similarly, \( F^c \) is the complement of event F.

The intersection of \( E^c \) and \( F^c \) (denoted as \( E^c \cap F^c \)) includes all outcomes that are neither in event E nor in event F. The probability, \( P(E^c \cap F^c) \), is thus the likelihood that neither event E nor event F occurs.
Transcribed Image Text:F. \( P(E^c \cap F^c) \) This expression represents the probability of the intersection of the complements of events E and F. In probability, the complement of an event E (denoted as \( E^c \)) includes all outcomes in the sample space that are not in E. Similarly, \( F^c \) is the complement of event F. The intersection of \( E^c \) and \( F^c \) (denoted as \( E^c \cap F^c \)) includes all outcomes that are neither in event E nor in event F. The probability, \( P(E^c \cap F^c) \), is thus the likelihood that neither event E nor event F occurs.
Expert Solution
Step 1: Given that

P(E)=0.82 ,P(F)=0.63

and P(Ec ∩ F)=0.12

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