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Suppose 2 events $A$ and $B$ are there. Can we say that $P(A\mid B) \ge P(A)$ is true always ?

Intuitively I think that if event $B$ has happened already, it may have reduced sample space, so conditional probability should be greater than unconditional probability. Is it correct reasoning ?

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3 Answers 3

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If $A\cap B = \varnothing$ then $\Pr(A\mid B) = 0.$

If $A=\{1,2,3,4\}$ and $B=\{4,5,6,7,8\}$ then $\Pr(A)= \dfrac 4 8 = \dfrac 1 2$ and $\Pr(A\mid B) = \dfrac 1 4 < \dfrac 1 2.$

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Suppose that $B$ is some event and the event $A$ happens if and only if the event $B$ happens. Then $P(A\mid B)=1$, but $P(A)$ can be any number from $[0,1]$. So your inequality is incorrect.

If you want to show that the reversed inequality cannot be true, take two mutually exclusive events.

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  • $\begingroup$ Sorry, I have mistaken in equality, please refer to edited question. $\endgroup$
    – MeetR
    Apr 27, 2017 at 10:39
  • $\begingroup$ Got it. Means there doesn't exist this kind of inequality. $\endgroup$
    – MeetR
    Apr 27, 2017 at 10:43
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For a concrete example, suppose you flip two coins and let $A$ be the event that you get the same side ($HH$ and $TT$), and let $B$ be the event that you get two heads..

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  • $\begingroup$ Sorry, I have mistaken in equality, please refer to edited question. $\endgroup$
    – MeetR
    Apr 27, 2017 at 10:39
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    $\begingroup$ @MeetRayvadera suppose A = event that you get two heads, and B = event that you get opposite sides $\endgroup$
    – Alain
    Apr 27, 2017 at 10:42

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