Can we estimate the minimum and maximum value of the probability of union of three dependent events (each with probability $\frac12$)?

Question

I have three dependent events $A,B,C$, each one happens with probability $\frac 12$. I want to estimate the probability of the union of these events, ie. $$ P_{tot}=P (A\cup B\cup C)=P (A) + P (B) + P (C) -P (A \cap B) - P (A \cap C) - P (B \cap C) +P (A \cap B \cap C ) \\ =\frac 32 -P (A \cap B) - P (A \cap C) - P (B \cap C) +P (A \cap B \cap C ) $$

If I do not know anything about the probability of the intersection of these events (I mean the last four terms in the above equation), can we estimate the minimum and maximum value that $P_{tot}$ might have?

Answer 1

The minimum would be if they were all the same event, in which case $P(A \cup B \cup C) = \frac 12$. The maximum would be if they covered the sample space, in which case $P(A \cup B \cup C) = 1$. The only bounds you can really get are $$P(A \cup B \cup C) \ge \max(P(A),P(B),P(C))$$ and $$P(A \cup B \cup C) \le P(A) + P(B) + P(C),$$ besides the obvious $P(A \cup B \cup C) \le 1$.

Answer 2

I don't know whether you want an estimate in terms of fixed numbers or in terms of other probabilities. If you want fixed numbers, then as said in user6247850's answer, the minimum is $1/2$ and the maximum is $1$.

But if you want an estimate in terms of other probabilities, then an obvious maximum would be \begin{align*} P(A∪B∪C) &= \frac{3}{2}−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C) \\ &\leq \frac{3}{2}−P(A∩B∩C)−P(A∩B∩C)−P(A∩B∩C)+P(A∩B∩C) \\ &= \frac{3}{2}−2P(A∩B∩C) \end{align*}

For an estimate of minimum, let $E = (A∩B) ∪ (A∩C) ∪ (B∩C)$. Then we have $P(A∩B∩C) \leq P(E)$ and $$ P(A∩B)+P(A∩C)+P(B∩C)=P(E)+2P(A∩B∩C) $$ (this is obvious if you look at the Venn diagram)

Thus, \begin{align*} P(A∪B∪C) &= \frac{3}{2}−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C) \\ &= \frac{3}{2}−(P(E)+2P(A∩B∩C))+P(A∩B∩C) \\ &= \frac{3}{2}−P(E)-P(A∩B∩C) \\ &\geq \frac{3}{2}−2P(E) \end{align*}

In conclusion, we have

$$ \frac{3}{2} - 2 P((A∩B) ∪ (A∩C) ∪ (B∩C)) \leq P(A∪B∪C) \leq \frac{3}{2}−2P(A∩B∩C) $$