/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q9SE Suppose that聽A,聽B, and聽C聽are... [FREE SOLUTION] | 91影视

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Chapter 2: Q9SE (page 90)

Suppose thatA,B, andCare three events such thatAandBare disjoint,AandCare independent, andBandCare independent. Suppose also that\({\bf{4Pr}}\left( {\bf{A}} \right){\bf{ = 2Pr}}\left( {\bf{B}} \right){\bf{ = Pr}}\left( {\bf{C}} \right)\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{A}} \cup {\bf{B}} \cup {\bf{C}}} \right){\bf{ = 5Pr}}\left( {\bf{A}} \right)\). Determine thevalue of\({\bf{Pr}}\left( {\bf{A}} \right)\)

Short Answer

Expert verified

The probability for event A is 1/6

Step by step solution

01

Given information

\(4\Pr \left( A \right) = 2\Pr \left( B \right) = \Pr \left( C \right)\,\,and\,\,\Pr \left( {A \cup B \cup C} \right) = 5\Pr \left( A \right)\)

02

Finding the probability of event A

\(\begin{aligned}{}\Pr \left( {A \cup B \cup C} \right) &= \Pr \left( A \right) + \Pr \left( B \right) + \Pr \left( C \right) - \Pr \left( {A \cap B} \right) - \Pr \left( {B \cap C} \right) - \Pr \left( {C \cap A} \right) + \Pr \left( {A \cap B \cap C} \right)\\5\Pr \left( A \right) &= \Pr \left( A \right) + 2\Pr \left( A \right) + 4\Pr \left( A \right) - 0 - \Pr \left( B \right)\Pr \left( C \right) - \Pr \left( A \right)\Pr \left( C \right) + 0\\5\Pr \left( A \right) &= 7\Pr \left( A \right) - \Pr \left( B \right)\Pr \left( C \right) - \Pr \left( A \right)\Pr \left( C \right)\\5\Pr \left( A \right) &= 7\Pr \left( A \right) - 2\Pr \left( A \right) \times 4\Pr \left( A \right) - \Pr \left( A \right) \times 4\Pr \left( A \right)\\5\Pr \left( A \right) &= 7\Pr \left( A \right) - 12{\left( {\Pr \left( A \right)} \right)^2}\end{aligned}\)

\(\begin{aligned}{}12{\left( {\Pr \left( A \right)} \right)^2} &= 2\Pr \left( A \right)\\6\Pr \left( A \right)& = 1\\\Pr \left( A \right) &= \frac{1}{6}\end{aligned}\)

The probability for event A is 1/6

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