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Problem states:

Consider two events $A$ and $B$, with $P(A) = 0.4$ and $Pr(B) = 0.7$. Determine the maximum and the minimum possible values for $P(A \& B)$ and the conditions under which each of these values is attained.

To solve, I considered the event with the lowest probability $A$ to be a subset of the other, so maximum value is attained under that circumstance giving a probability of $0.4$. But the book states that the minimum is $0.1$, if $P(A \cup B) = 1$.

I don't understand why! Because I thought that the minimum value is get when the two events are disjoint... So the minimum value must be $0$...

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4  
Two events of probability $0.4$ and $0.7$ cannot be disjoint. Draw a Venn diagram. –  Qiaochu Yuan Jun 14 '12 at 23:31
    
Your intuition is roughly correct, we want to separate $A$ and $B$ as much as possible. When we do, there is still an overlap of $0.1$. –  André Nicolas Jun 15 '12 at 0:37
    
another simple question: how do I include those fancy math symbols when asking questions? Thank you! :D –  Umar Jamil Jun 15 '12 at 14:43

2 Answers 2

up vote 4 down vote accepted

Use the fact that $$P(A\cup B) = P(A) + P(B) - P(A\cap B).$$ This gives you $$P(A\cup B) = 1.1 - P(A\cap B).$$ Since $P(A\cup B) \le 1$ we have $$1\ge 1.1 - P(A\cap B),$$ yielding $P(A\cap B)\ge .1$. If $A\subseteq B$, then $P(A\cap B) = P(A) = .4.$ This represents the largest possible value for $P(A\cap B)$. so $.1\le P(A\cap B) \le .4$

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$$ \begin{align*} 0\leq P(A \cup B) \leq 1 & \implies 0 \leq P(A)+P(B)-P(A \cap B) \leq 1\\ & \implies 0\leq 0.4 + 0.7 - P(A\cap B) \leq 1 \\ & \implies -1.1\lt -P(A\cap B)\leq -0.1 \\ & \implies 0.1 \leq P(A\cap B) \lt 1.1. \end{align*} $$

So, the minimum possible value of $P(A\cap B)$ is $0.1$ and this is possible if $P(AB)=0.1, P(A^c \cap B) = 0.1, P(A \cap B^c)= 0.8$ and $P(A \cup B)^c = 0.$


For the maximum value, $P(AB)\leq P(A)= 0.4$ and $P(AB)\leq P(B)= 0.7$. Taking the most restrictive of these two values gives the maximum of $P(AB)$ as $0.4$.

This upper bound is possible if $A \subset B$.

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