Question: Problem states:

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> 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.
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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 $0$...

Answer: 
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 $.1\le P(A\cap B) \le .4$
$$
\begin{align\*}
0\leq P(A \cup B) \leq 1 & \implies 0 \leq P(A)+P(B)-P(A \cap B) \leq 1\\
& \implies 0\leq 0.4 + 0.7 - P(A\cap B) \leq 1 \\
& \implies -1.1\lt -P(A\cap B)\leq -0.1 \\
& \implies 0.1 \leq P(A\cap B) \lt 1.1.
\end{align\*}
$$

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 $P(A \cup B)^c = 0.$ 

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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 $A \subset B$.