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Like if I was given a P(A) of .5 and a P(B) value of .4 how would I get the minimum of the P(A∩B)?

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    $\begingroup$ A and B could be disjoint, so the minimum possible value of the probability of their intersection is zero. $\endgroup$
    – mark999
    Commented Sep 12, 2011 at 0:26
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    $\begingroup$ Is this homework? If so, should be marked as such. $\endgroup$
    – user88
    Commented Sep 12, 2011 at 9:07
  • $\begingroup$ @mark999 Can you post this as an answer? $\endgroup$
    – user88
    Commented Sep 12, 2011 at 9:07

2 Answers 2

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Quickly generalising from cardinal's comment, we have:

$$\mathbb P(A \cap B) = \mathbb P(A) + \mathbb P(B) - \mathbb P( A \cup B )$$

Now the unknown term on the right must be between $0$ and $1$. So a loose bound is given by using these limits.

$$\mathbb P(A) + \mathbb P(B) \geq \mathbb P(A \cap B) \geq \mathbb P(A) + \mathbb P(B) - 1$$

However, we can do a little better than this. Firstly, the intersection cannot be larger than the smallest marginal probability, as we have

$$\mathbb P(A \cap B)=\mathbb P(A)\mathbb P(B|A)\leq \mathbb P(A)$$ $$\mathbb P(A \cap B)=\mathbb P(B)\mathbb P(A|B)\leq \mathbb P(B)$$

Also, the probability cannot be negative. Hence we have:

$$min\left[\mathbb P(A),\mathbb P(B)\right] \geq \mathbb P(A \cap B) \geq max\left[\mathbb P(A) + \mathbb P(B) - 1,0\right]$$

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$A$ and $B$ could be disjoint, so the minimum possible value of $P(A \cap B)$ is zero.

Example: Suppose a number is chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with each number equally likely to be chosen. Let $A$ be the event that the chosen number is less than or equal to 5; let $B$ be the event that the chosen number is greater than or equal to 7. Then $P(A)=0.5$, $P(B)=0.4$, and $A \cap B$ is empty, so $P(A \cap B)=0$.

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    $\begingroup$ (+1) Perhaps you can provide a more general statement, since the OP's seems to be an example. I'm thinking along the lines of $\mathbb P(A \cap B) = \mathbb P(A) + \mathbb P(B) - \mathbb P( A \cup B )$. So, for example, if $\mathbb P(B) = 0.6$, instead, then the first statement in the answer no longer holds. :) $\endgroup$
    – cardinal
    Commented Sep 12, 2011 at 12:12

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