Construct a copy of figure 3.10 in your text, where the first outcome is one of (A,B,C) and the second outcome in each case is one of (1,2). Use the following probabilities instead of those given in your text:Pr[A]=13Pr[A]=13Pr[A1]=112Pr[A1]=112Pr[A2]=14Pr[A2]=14Pr[1|B]=34Pr[1|B]=34Pr[2|B]=14Pr[2|B]=14Pr[B1]=14Pr[B1]=14Pr[B2]=112Pr[B2]=112Pr[C1]=16Pr[C1]=16Pr[C2]=16Pr[C2]=16Find the following missing probabilities:(1) Pr[1|A]=Pr[1|A]= (2) Pr[B]=Pr[B]= (3) Pr[C]=Pr[C]= (4) Pr[2|C]=Pr[2|C]=

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Asked Oct 1, 2019
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Construct a copy of figure 3.10 in your text, where the first outcome is one of (A,B,C) and the second outcome in each case is one of (1,2). Use the following probabilities instead of those given in your text:

Pr[A]=13Pr[A]=13
Pr[A1]=112Pr[A1]=112
Pr[A2]=14Pr[A2]=14
Pr[1|B]=34Pr[1|B]=34
Pr[2|B]=14Pr[2|B]=14
Pr[B1]=14Pr[B1]=14
Pr[B2]=112Pr[B2]=112
Pr[C1]=16Pr[C1]=16
Pr[C2]=16Pr[C2]=16

Find the following missing probabilities:

(1) Pr[1|A]=Pr[1|A]= 

(2) Pr[B]=Pr[B]= 

(3) Pr[C]=Pr[C]= 

(4) Pr[2|C]=Pr[2|C]= 

 

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Expert Answer

Step 1

Given probabilities are,

P(A)=13,P(Aand1)=112, P(A and 2)=14, P(1/B)=34, P(2/B)=14, P(B and 1)=14,

P(B and 2)=112, P(C and 1)=

Given probabilities are,

P(A)=13,P(Aand1)=112, P(A and 2)=14, P(1/B)=34, P(2/B)=14, P(B and 1)=14,

P(B and 2)=112, P(C and 1)=16, P(C and 2)=16

Note that probability cannot be greater than or equal to 1. Also the probability cannot be less than or equal to 0. So given values of the probability is invalid.

Step 2

But using given values the probability of B that is P(B) is,

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Р(B) - Р(Вn1) Р(1/ B) 14 34 7 17

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Step 3

Probability P(1...

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P(1/A)P(A1) P(4) 112 13

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