# Find Probability of one event out of three when all of them can't happen together

STATEMENT

Three events E, F , and G cannot occur simultaneously. Further it is known that P(E ∩ F ) = P(F ∩ G) = P(E ∩ G) = 1/3. Can you determine P(E)?

I made this diagram:

$$P(E \cup F \cup G) = P(E) + P(F) + P(G) - P(E \cap F) - P(E \cap G) - P(F \cap G)$$

$$\implies$$ $$P(E) = P(E \cup F \cup G) - P(F) - P(G) + P(E \cap F) + P(E \cap G) + P(F \cap G)$$

$$\implies$$

$$P(E) = P(E \cup F \cup G) - P(F) - P(G) + \frac 13 + \frac 13 + \frac 13$$

$$\implies$$

$$P(E) = P(E \cup F \cup G) - P(F) - P(G) + 1$$

Now what to do next?

Looks like this diagram matches better with the problem description:

• FWIW, your first diagram is a correct Euler Diagram. The second diagram is not valid for the situation at hand. May 15, 2020 at 12:43

## 5 Answers

This Venn diagram displays a situation where the chance of mutual intersection is zero:

From $$\Pr(E\cap F) = 1/3$$ we deduce all this probability lies in the overlap of the $$E$$ and $$F$$ disks, but not in the mutual overlap of all three disks. That permits us to update the diagram:

Applying the same reasoning to $$\Pr(F\cap G) = \Pr(E\cap G) = 1/3,$$ we obtain a Venn diagram displaying all the information in the question:

The Axiom of Total Probability asserts the sum of all the probabilities (including the probability for the complement of $$E\cup F\cup G,$$ shown at the bottom left) is $$1.$$

An even more basic probability axiom asserts all probabilities must be non-negative. But since $$1/3+1/3+1/3+0=1,$$ all the possible probability already appears. The remaining probabilities must be zero, meaning the picture can be completed only like this:

Finally, a third axiom (the same one used in the second step of filling in the Venn diagram) asserts the probability of $$E$$ equals the sum of the probabilities of its four parts, because they are disjoint. Thus, beginning with the central probability and moving counterclockwise around the disk that portrays $$E,$$

$$\Pr(E) = 0 + 1/3 + 0 + 1/3 = 2/3.$$

One moral worth remembering:

Draw Venn diagrams in full generality so they show all possible intersections of the sets, even when you know some of the probabilities are zero.

This helps you keep track of all the information systematically. (It's also conceptually more accurate, because sets of probability zero do not have to be nonempty!)

• Definitely the simplest way to think about it. You have 3 mutually exclusive events that sum to 1, therefore the probability of everything else is 0. May 14, 2020 at 17:44
• (+1) I don't see any reason to downvote this answer May 16, 2020 at 22:04
• @gunes It's not the only downvoted answer here and I can't see any reason for that, either, so I have presumed somebody is disgruntled about something. These things happen.
– whuber
May 16, 2020 at 22:09
• Apparently, yes. But, it does more good than harm. My upvote goes there as well. May 16, 2020 at 22:27

If you try to fill in the Venn diagram, you can't put non-zero entries inside regions other than represented by pairwise intersections. They'll form up the sample space by themselves, which means $$\mathbb P(E)=\mathbb P(E\cap F)+\mathbb P(E\cap G)=2/3$$

• How did you come up with this formula? May 14, 2020 at 12:10
• Since $E\cap F$,$E\cap G$ and $F\cap G$ are mutually exclusive, and their probabilities add up to $1$. So, they form up the sample space, and all other events' probabilities can be written as sum of the probs of these. An easier way to think is trying to put numbers inside the sets in the Venn diagram. For example, put $x$ to all pairwise intersections and $y,z,t,u$ to remaining regions. And, try to calculate the probability of the intersections, then equate to $1/3$. May 14, 2020 at 12:14
• I added a pic for the sample space. Does it match? May 14, 2020 at 12:38
• Yes, in order to calculate $P(E)$, you need to marginalize wrt all events:$$P(E)=\sum_{\mathcal A} P(E\cap \mathcal A)$$ You can do this because $E\cap \mathcal A$ are mutually exclusive. May 14, 2020 at 12:40
• @probabilityislogic yes. However, here since the sample space can be divided into three mutually exclusive events and $E$ can not be strictly inside these events, we can find the probability by summing the suitable ones, i.e. $E\cap F$ and $E\cap G$. I still think the Venn diagram approach is far more easier to understand. May 14, 2020 at 13:20

The answer to the question "Can you determine $$P(E)$$?" is Yes.

Given events $$E, F, G$$ defined on a sample space $$\Omega$$, we know that \begin{align} &E\cap F\cap G\\ &E\cap F\cap G^c\\ &E\cap F^c\cap G\\ &E\cap F^c\cap G^c\\ &E^c\cap F\cap G\\ &E^c\cap F\cap G^c\\ &E^c\cap F^c\cap G\\ &E^c\cap F^c\cap G^c\\ \end{align} are $$8$$ mutually exclusive events whose union is $$\Omega$$. Thus, the sum of the probabilities of these $$8$$ events is $$1$$. Now, we are told that $$E, F, G$$ cannot occur simultaneously, that is, $$E\cap F\cap G = \emptyset$$ and so $$P(E\cap F\cap G) = 0$$. We are also told that \begin{align} P(E\cap F) &= P(E\cap F\cap G) + P(E\cap F\cap G^c) = \frac 13\\ P(E\cap G) &= P(E\cap F\cap G) + P(E\cap F^c \cap G) = \frac 13\\ P(F\cap G) &= P(E\cap F\cap G) + P(E^c\cap F \cap G) = \frac 13 \end{align} where we can feel comfortable about the sum in the middle in each equation by musing on the fact that the probability of the union of two mutually exclusive events is the sum of the probabilities of the two events. Since $$P(E\cap F\cap G)=0$$, we conclude that \begin{align}P(E\cap F) &= P(E\cap F\cap G^c) = \frac 13\\ P(E\cap G) &= P(E\cap F^c \cap G) = \frac 13\\ P(F\cap G) &= P(E^c\cap F \cap G) = \frac 13 \end{align}

But, of the $$8$$ mutually exclusive events listed above whose union is $$\Omega$$, we have identified three events whose probabilities add up to $$1$$ and so the other $$5$$ events (one of which is $$E\cap F\cap G$$) must have probability $$0$$. Consequently, \begin{align} P(E) &= P(E\cap F\cap G) + P(E\cap F\cap G^c) + P(E\cap F^c\cap G) + P(E\cap F^c\cap G^c)\\ &= 0 + \frac 13 + \frac 13 + 0\\ &= \frac 23 \end{align} By symmetry (or by a brute force repetition of the above arguments mutatis mutandis), we can conclude that $$E, F, G$$ all have probabiity $$\frac 23$$.

• And what someone finds objectionable enough in my answer to downvote by clicking on "This answer is not useful", I don't know because the down voter was not willing to leave a comment. May 16, 2020 at 19:51
• (+1) I don't see a reason for a downvote here as well. May 16, 2020 at 22:17

Can we think of it that way?

P(E ∩ F ) = P(F ∩ G) = P(E ∩ G) = 1/3

P(E ∩ F ) + P(F ∩ G) + P(E ∩ G) = 1

Meaning that The probability of event E happening by itself is zero, which means it can only happen with either F or G and it can't happen with both.

P(E) = P(E ∩ F ) + P(E ∩ G) = 1/3 + 1/3 = 2/3

Since the events $$(E,F)$$, $$(E,G)$$ $$(F,G)$$ are mutually exclusive and sum to one we can use the law of total prob: $$P(E) = P(E, F) + P(E, G) = \tfrac{2}{3}$$ Since $$P(E \mid E,F)P(E, F) = P(E, F)$$, ditto for $$E,G$$ and $$P(E \mid F, G) = 0$$.

• What are the $A_i$? and do they need to be disjoint (mutually exclusive) events in order for the first equality to hold? May 16, 2020 at 19:53
• You seem to be assuming that $\{F,G\}$ is an exhaustive partition of the sample space, but it's not. You're missing the term $P(E\mid (F\cup G)^c) P((F\cup G)^c)$ needed to make this a generally correct equation (and you are implicitly assuming $F\cap G$ is empty, but that at least is a fact in evidence).
– whuber
May 16, 2020 at 22:19