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Let a and b be events.]

Max0815  Sep 3, 2018
edited by Max0815  Sep 5, 2018
edited by Max0815  Sep 5, 2018
edited by Max0815  Sep 15, 2018
 #1
avatar+194 
0

Someone please help!!!

Max0815  Sep 3, 2018
 #2
avatar+194 
0

Somebody please help!

Max0815  Sep 4, 2018
 #3
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+3

it won't be true as long as \(P(A) \neq P(B)\)

 

\(P(A \wedge B) = P(A|B)P(B) = P(B|A)P(A) \\ \\ \text{clearly } P(A) \neq P(B) \Rightarrow P(A|B) \neq P(B|A)\)

Guest Sep 4, 2018
 #4
avatar+194 
0

Thanks!!!

Max0815  Sep 5, 2018

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