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Description: Suppose P(F)=1/3, P(E and F)=3/4 and P(E or F)=4/5Find P(E).Can this really be a probability? If not, explain why not.

Math

Probability

Suppose P(F)=1/3, P(E and F)=3/4 and P(E or F)=4/5 Find P(E). Can this really be a probability? If not, explain why not.

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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Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Suppose P(F)=1/3, P(E and F)=3/4 and P(E or F)=4/5

Find P(E).
Can this really be a probability? If not, explain why not.

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Expert Solution

Step 1

$P (E or F) = P (E) + P (F) - P (E and F) \frac{4}{5} = P (E) + \frac{1}{3} - \frac{3}{4} [Plug the given values] P (E) = \frac{4}{5} - \frac{1}{3} + \frac{3}{4} P (E) = \frac{73}{60} P (E) = 1.21666666667$

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