Minimum Probability of Intersection of 3 events theory Say you have $P(A) = P(B) = P(C) = 0.9$, but they are not necessarily independent events.
What is the minimum probability of $P(A ∩ B ∩ C)$?
For example, we know that
$P(A ∪ B) = P(A) + P(B) - P(A ∩ B) \leq 1$, so
$P(A ∩ B) \geq P(A) + P(B) - 1 = 0.9 + 0.9 -1 = 0.8$
So the min probability of $P(A ∩ B)$ is $0.8$
I can only find the maximum probability being $0.8$ because
$P(A ∩ B ∩ C) \leq P(A ∩ B), P(B ∩ C), P(A ∩ C) = 0.8$
The working I've done so far:
$0 \leq P(A ∪ B ∪  C) \leq1$
$0 \leq P(A) + P(B) +P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A∩B∩C) \leq 1$
$-2.7 \leq P(A ∩ B ∩ C) - P(A ∩ B) -  P(A ∩ C) -  P(B ∩ C) \leq -1.7$
$1.7 \leq P(A ∩ B) +  P(A ∩ C) + P(B ∩ C) - P(A ∩ B ∩ C) \leq 2.7$
$P(A ∩ B ∩ C) + 2.7 \geq P(A ∩ B) +  P(A ∩ C) + P(B ∩ C)$
$P(A ∩ B ∩ C) \geq P(A ∩ B) +  P(A ∩ C) + P(B ∩ C) - 2.7$
Since, $0.8\leq P(A ∩ B)\leq 0.9$
$2.4 \leq P(A ∩ B) +  P(A ∩ C) + P(B ∩ C) \leq 2.7$
So $P(A ∩ B ∩ C) >= 2.4 - 2.7$? or $2.7 - 2.7$? Which is just $0$?
 A: You can solve the problem via linear programming, with a nonnegative decision variable $p_S$ for each of the possible subsets $S$ and linear constraints to enforce the given probabilities.  For your example, the problem is to minimize $p_{111}$ subject to
\begin{align}
p_{001} + p_{011} + p_{101} + p_{111} &= 0.9 \\
p_{010} + p_{011} + p_{110} + p_{111} &= 0.9 \\
p_{100} + p_{101} + p_{110} + p_{111} &= 0.9 \\
p_{000} + p_{001} + p_{010} + p_{011} + p_{100} + p_{101} + p_{110} + p_{111} &= 1 
\end{align}
The minimum objective value turns out to be $0.7$, attained by $p_{011}=p_{101}=p_{110}=0.1$, $p_{111}=0.7$, and all other $p_S = 0$.
The maximum objective value turns out to be $0.9$, attained by $p_{000}=0.1$, $p_{111}=0.9$, and all other $p_S = 0$.
A: This can be solved very simply using what I call a "line diagram"
$A: 0\,0\,|0\,0\,0\,0\,0\,0\,0|\bullet\quad$ Fill 9/10 of sample space from left end
$B: \bullet\,0\,|0\,0\,0\,0\,0\,0\,0\,|0\quad$ Fill 9/10 of sample space from other end
$C: 0\bullet|0\,0\,0\,0\,0\,0\,0|0\quad$ Fill 9/10 of sample space so as to minimize $A\cap B\cap C$
Ans $= 0.7$
A: The problem can be expressed as an LP and solved using symmetry.
Suppose $f$ is convex and $P$ is some permutation such that $f(Px) = f(x)$ and, if $x$ is feasible, then $Px$ is feasible. Then since $f$ is convex, we see that
$f({1 \over n} \sum_{k=0}^{n-1} P^k x) \le f(x)$. In particular, if $x$ is optimal, then so is the 'average'
${1 \over n} \sum_{k=0}^{n-1} P^k x$.

Choose $P$ to be the 'rotation' indicated in the diagram. We see that the problem can be reduced to three non negative variables $x$ for $x_{A \bar{B} \bar{C}}$, $y$ for $x_{A {B} \bar{C}}$ and $z$ for $x_{A BC}$ that satisfy
$x+2y+z = 0.9$, $3x+3y+z \le 1$.
Subtracting gives
$2x+y \le 0.1$, and since $z=0.9-2y-x$ we get
$z \ge 0.7+x$ and hence $0.7$ is a lower bound.
Setting $x=0$ gives $y=0.1$ which is feasible so
we see that $z=0.7$ is attained.
A: 
We can also find the answer using the Venn Diagram.

Let the probability of only $1$ regions be denoted by $I$ which would be $a+b+c$, 
Let $II$ denote only $2$ regions, i.e., $w+x+y$ and 
Let $III$ denote only $3$ region, i.e., $z$. 

Now we know that since $P(\text{Universe})=1$, so: $$\Rightarrow I+II+III+n=1$$ where $n$ represents the region outside the venn diagram, that is, belonging to none of $A$ or $B$ or $C$.
Since $P(A)=P(B)=P(C)=0.9$, so: $$I+2II+3III=2.7$$
Now, subtracting (1) from (2), we get:
$$II+2III=1.7+n$$

Now the last equation, eq (3) will help us get our answer as follows:
To minimize, let $n=0$
Then let $II=k$ and so from eq(1) we have $III=1-k$ assuming $I=0$. 
Now substituting in eq(3):
$$k+2(1-k)=1.7$$
$$\Rightarrow k=0.3$$
Thus, the $\boxed{\text{minimum value of }III=0.7}$.

To maximise $III$, the we would have let $n=k$ and assumed that both $I=II=0$ to get from eq(1) that $III=1-k$. 
Now substituting in eq(3):
$$2(1-k)=1.7+k$$
$$\Rightarrow k=0.1$$
Thus, the $\boxed{\text{maximum value of }III=0.9}$.
