Question 1024798: The events A and B are contained in a sample space S; the Pr (ANB) = 1/2. Pr (A|B)= 1/3. and Pr (B|A)=1/3. Explain why this is impossible?
Answer by mathmate(429)   (Show Source):
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Question:
The events A and B are contained in a sample space S; the Pr (ANB) = 1/2. Pr (A|B)= 1/3. and Pr (B|A)=1/3. Explain why this is impossible?
Solution:
Much of these problems are based on the definitions.
A and B are contained in a sample space S, so
P(A∩B)=P(A)+P(B)-P(A∪B)
P(A|B)=P(A∩B)/P(B), and
P(B|A)=P(A∩B)/P(A)
Since we're given
P(A∩B)=P(A)+P(B)-P(A∪B)=1/2........(1)
P(A|B)=P(A∩B)/P(B)=1/3.................(2)
P(B|A)=P(A∩B)/P(A)=1/3.................(3)
We can solve for each of P(A), P(B), and P(A∪B) by elimination:
From (2), we get by cross multiplication
P(B)=3P(A∩B).............(2A)
Similarly,from (3), we get by cross multiplication
P(A)=3P(A∩B).............(3A)
Substitute (2A) and (3A) into (1) to get:
1/3+1/3-P(A∪B)=1/2, or
P(A∪B)=1/2-1/3-1/3=-1/6.
By Kolmogorov's Axioms, 0≤P(E)≤1 for any event E, therefore the given numerical values are impossoble.
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