Derive which values $P(A\cap B \cap C)$ can take, using only the axioms of probability

For three events $A, B, C$ we know that $P(A\cap B) = 0.4, P(A \cap C)=0.5 \text{ and } P(B\cap C)=0.6$

(a) Derive which values $P(A\cap B \cap C)$ can take, with a clear explanation. Show in every step which axiom you use. You can only make use of the following axioms:

1. $P(A)\geq 0$
2. $P(S)=1$
3. $P(A_1\cup A_2 \cup \dots)=\sum\limits_{i=1}^\infty A_i, \text{ } \forall i\neq j$

(b) Let $D=(A\cap B^c \cap C^c) \cup (A^c\cap B \cap C^c) \cup (A^c\cap B^c \cap C)$. Which are the values that $P(D)$ can take? Give a clear explanation of your derivations.

What I attempted so far

(a)

I came this far: $A\cap B=(A\cap B \cap C) \cup (A\cap B \cap C^c)$ (these sets are disjoint). Hence $P(A\cap B \cap C) \leq P(A\cap B) = 0.4$, by axioms 3 and 2, since $P(A\cap B \cap C^c)\geq 0$.

By making a Venn-diagram I figured out that the lower boundary should probably be $0.25$, as this is $0.4+0.5+0.6-1=0.5$ (which is everything that should be in some intersection because the total probability cannot be greater than one) devided by $2$ (because the $0.5$ can be split over two intersections). However, I am not sure how to show this using the axioms provided. Could anyone please help?

(b) I think I will need the result from a. I am however not at all sure yet what I should do to solve this question. Any help would be appreciated.

1 Answer

Here’s a way to formalize your Venn diagram argument.

Let $X=(A\cap B)\cup(A\cap C)\cup(B\cap C)$, and let $Y=A\cap B\cap C$. Then $X$ is the disjoint union of the sets $Y$, $X_{AB}=(A\cap B)\setminus Y$, $X_{AC}=(A\cap C)\setminus Y$, and $X_{BC}=(B\cap C)\setminus Y$. Note that $A\cap B$ is the disjoint union of $X_{AB}$ and $Y$, so $0.4=P(A\cap B)=P(X_{AB})+P(Y)$, and $P(X_{AB})=0.4-P(Y)$. Similarly, $P(X_{AC})=0.5-P(Y)$ and $P(X_{BC})=0.6-P(Y)$. Thus,

\begin{align*} P(X)&=P(X_{AB})+P(X_{AC})+P(X_{BC})+P(X)\\ &=(0.4-P(Y))+(0.5-P(Y))+(0.6-P(Y))+P(Y)\\ &=1.5-2P(Y)\;, \end{align*}

or $P(Y)=\frac12(1.5-P(X))\ge\frac12(1.5-1.0)=0.25$, since $P(X)\le 1$.

To do a really thorough job on (a), for each $p\in[0.25,0.4]$ you should demonstrate that it’s possible to arrange $A,B$, and $C$ so that $P(Y)=p$. I would at least draw Venn diagrams illustrating the two extremes.

For (d), note that $D=(A\cup B\cup C)\setminus X$, where $X$ is as above: it’s the set of points belonging to exactly one of the sets $A,B$, and $C$. Thus,

$$P(D)=P(A\cup B\cup C)-P(X)=P(A\cup B\cup C)-1.5+2P(Y)\;.$$

Bearing in mind that $P(A\cup B\cup C)\le 1$, what can you say about the possible range of values of $P(D)$? Again, you should check to be sure that the apparent extremes are actually possible.

• Wow, what a great answer! Thank you very much for your great effort! To answer your last question, I think that using the fact that $P(A\cup B \cup C)\leq 1$ gives me the inequality $P(D)\leq 0.3$ (since using the other inequality that can be obtained using the range for $P(Y)$ gives contradicting inequality signs I think, and would thus not give you a definite inequality). Am I right? Jun 30, 2013 at 8:53
• @rbm: Looks good to me. You’re welcome! Jun 30, 2013 at 9:02