Given $P(E_1' \cap E_2)=a$, $P(E_2' \cap E_1)=b$, find $P(E_1)$ in terms of $a$ and $b$

CUCET 2021

Came in a test I gave today, here is what I did:

$$|P(E_1' \cap E_2) - P(E_2' \cap E_1)| =| P(E_1) - P(E_2)|$$

The above result comes from looking at the ven diagrams, I also know that:

$$ P(E_1' \cap E_2) + P(E_2'\cap E_1)=P(E_1 \cup E_2) - P(E_1 \cap E_2)$$

Again a venn diagram result, but I can't see how to move forward from here...

  • $\begingroup$ The question is wrong. You cannot find $P(E_1)$ from the given information. $\endgroup$ Sep 23, 2021 at 12:09
  • $\begingroup$ Ok, could you show a reason as to why? I went mad trying to solve this in the exam hall @KaviRamaMurthy $\endgroup$ Sep 23, 2021 at 12:13

4 Answers 4


We have

$$P(E_{1}'\cap E_2)=P(E_2)-P(E_1\cap E_{2})$$ $$P(E_{2}'\cap E_1)=P(E_1)-P(E_1\cap E_{2})$$

Indeed consider the Venn diagram of the symmetric difference of $E_1$ and $E_2$. We have the shaded regions

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ enter image description here

However, you are not given $P(E_1\cap E_2)$ or $P(E_2)$, so there is not enough information to determine $P(E_1)$. If $E_1$ and $E_2$ are disjoint, then you would have $P(E_1)=b.$


Draw a Venn diagram, with events $E_1$ and $E_2$ represented by circles that have an unknown intersection. That is, the two events might not intersect, and if they do intersect, the size of this intersection is unknown.

Based on the constraints, you know the size of the portion of the circle that represents $E_1$ that is outside of $E_2$. You also know the size of the portion of the circle that represents $E_2$ that is outside of $E_1$.

However, in order to determine the $p(E_1)$, since you know the size of the portion of $E_1$ that is outside of $E_2$, you must (somehow) then determine the size of the portion of $E_1$ that is also inside $E_2$.

There is no way to determine that information from the given information. Therefore, the $p(E_1)$ can not be determined.


Example to show that $P(E_1)$ is not uniquely determined by $P(E_1'\cap E_2)$ and $P(E_1 \cap E_2')$: Let $\Omega =\{1,2,3\}$ and $P$ be the uniform measure.

Case 1) $E_1=\{1\}, E_2=\{2\}$

Case 2) $E_1=\{1,2\}, E_2=\{2,3\}$

In both cases $P(E_1'\cap E_2)=\frac 1 3$ and $P(E_1 \cap E_2')=\frac 1 3$. However $P(E_1)$ is not the same in these two cases. So the values of $a$ and $b$ do not determine $P(E_1)$ uniquely.


A quicker way to do this is by noticing that $P(A\cup B)=P(A)+P(A'\cap B)$

$$P(E_1\cup E_2)=P(E_1)+a$$

$$P(E_2\cup E_1)=P(E_2)+b$$

$$\implies P(E_1)=P(E_2)+b-a$$

It follows that we need the value $P(E_2)$ to evaluate $P(E_1)$, as noted by the other answers

  • $\begingroup$ This "quantity" seems to be out of nowhere. Can you please elaborate a bit more? $\endgroup$ Sep 23, 2021 at 17:25
  • $\begingroup$ @Buraian are you talking about the first line? If yes, then it is quite straightforward to notice: $P(A\cup B)=P(A\cup (A'\cap B))=P(A)+P(A'\cap B)$, since $A$ and $A'$ are disjoint $\endgroup$
    – DatBoi
    Sep 24, 2021 at 3:01
  • $\begingroup$ It is clear now! $\endgroup$ Sep 24, 2021 at 3:19

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