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)
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Chapter1: Combinatorial Analysis
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Question
A) 0.800
B) 0.940
C) 0.794
D) 0.410
E) 0.060

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 \( F \).)

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
Step by step
Solved in 3 steps with 1 images

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