If P(B)=0.2,P(A | B)=0.7,P(B)=0.8, and P(A|B)=0.5, find P(B | A).
Answer:
Given P(B)=0.2,P(A|B)=0.7,P(B')=0.8, and
P(A|B')=0.5
P(A)= P(A∩S)
= P[A∩(BUB')]
= P[(A∩B) U (A∩B')]
= P(A∩B) + P(A∩B')
= P(A|B)P(B) + P(A|B')P(B')
= (0.7)(0.2) + (0.5)(0.8)
= 0.54
P(B|A) = P(A∩B)/P(A)
= 0.7*0.2/0.54
=0.2592
If P(B)=0.2,P(A | B)=0.7,P(B)=0.8, and P(A|B)=0.5, find P(B | A).
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Check the time reversibility π,B- π, P,
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0.7 0.2 0.1 0.2 0.6 0.2 0.1 0.4 0.5
0.7 0.2 0.1 0.2 0.6 0.2 0.1 0.4 0.5