Please prove P(A|B) is not equal to P(A and B) algebraically

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The learning sections states that:

P(A|B) is not equal to P(A and B).

I would like to understand this mathematically, so can someone please prove this algebraically?

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  P(A|B)=P(A and B)/P(B)

=> P(A|B) * P(B) = P(A and B)
=> P(A and B) = P(A|B) * P(B)
So, P(A|B) != P(A and B) :smiley:

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If you don’t understand anything @ LuukvanVliet Luukvan Vliet feel free to ask me.

Hi @raisa.jerin.sristy79,

Thanks for explaining.

You state: P(A and B) = P(A|B) / P(B) and the learning sections states:
P(A and B) = P(A|B) * P(B)

Which one is true?

Sorry for the silly mistake, my bad!
P(A|B)=P(AB)/P(B)
=> P(A|B) * P(B) = P(AB)
=> P(AB) = P(A|B) * P(B)
And thanks @ LuukvanVliet

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