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If given probability of $A$ is $a$ and probability of $B$ is $b$, how do I find min/max probability of intersection? Max value of intersection would be $\min(a,b)$, how do I find the min?

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2 Answers 2

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if $a+b \le 1$, then presumably one can find disjoint sets $A$ and $B$ with ${\rm P}A = a$ and ${\rm P}B = b$. so in this case, the min is 0.

if $a+b > 1$, we get a smallest intersection by choosing $B$ to contain all of $A^C$, which has probability $1-a$ and then adding to that a piece of $A$ to bring ${\rm P}B$ up to $b$. so the piece of $A$ added in has to have probability $b - (1-a) = a+b - 1$. this last quantity is then the min probability for the intersection when $a+b > 1$.

so in any case. the min is $(a+b-1)^+$.

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  • $\begingroup$ your last explanation on finding the smallest intersection can be applied to all cases. Min A∩B is max(a+b-1,0) $\endgroup$
    – user862
    Aug 31, 2010 at 3:57
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The min is the smaller of two values: $\min(a,b) = a$ if $a < b$ and $b$ otherwise. Though I do not think this is what you are asking for...

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  • $\begingroup$ what? the question says 'blah blah blah would be $min(a,b)$, how do I find the min?' I answered the question. $\endgroup$
    – shabbychef
    Aug 30, 2010 at 16:38
  • $\begingroup$ @shabbychef I agree; bad questions should be punished with no answers. $\endgroup$
    – user88
    Aug 31, 2010 at 14:21
  • $\begingroup$ No, that's not what he's asking for. He asked for the min of the intersection, as can be deducted by careful reading of the first of the two sentences in the opening post. $\endgroup$
    – Joris Meys
    Aug 31, 2010 at 16:51
  • $\begingroup$ @mbq - whats wrong with the question? $\endgroup$
    – user862
    Sep 1, 2010 at 6:00
  • $\begingroup$ @saminny You could have formulated it better so that it would be less ambiguous and easier to understand. shabbychef just pointed it out with his answer. $\endgroup$
    – user88
    Sep 1, 2010 at 7:46

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