Hello, Volga!

Here is part of the solution . . .

Consider three mutually exclusive and exhaustive events $\displaystyle A_0,\,A_1$ and $\displaystyle A_2$ where:

. . $\displaystyle P(A_0\cup{A_1})=p_0$

. . $\displaystyle P(A_1\cup{A_2})=p_1$

. . $\displaystyle P(A_2\cup{A_3})=p_2$.

What condition on $\displaystyle p_0, p_1, p_2$ must hold?

You are correct.

"Mutually exclusive" and "mutually exhaustive" means the sets have

. . no overlap, and together they "take up" the entire sample space.

The Venn diagram looks like this:

Code:

* - - - - - - - *
| |
* A0 *
| * * |
| * * |
| * * |
| A1 * A2 |
| | |
* - - - * - - - *

Let $\displaystyle a_0,a_1,a_2$ represent the *areas* of their respective regions in the unit square.

$\displaystyle \text{Then: }\;\begin{Bmatrix}P(A_0) \:=\:a_0 \\ P(A_1) \:=\:a_1 \\ P(A_2) \:=\:a_2 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix}P(A_0\cup A_1)\:=\:a_0+a_1 \:=\:p_0 & [1] \\ P(A_1\cup A_2) \:=\:a_1+a_2 \:=\:p_2 & [2] \\ P(A_2\cup A_0) \:=\:a_2+a_0 \:=\:p_2 & [3]\end{Bmatrix} $

$\displaystyle \text{Add [1], [2] and [3]: }\;2a_0 + 2a_1 + 2a_2 \:=\:p_0 + p_1+p_2$

. . . . . $\displaystyle \text{and we have: }\;2\underbrace{(a_0 + a_1 + a_2)}_{\text{This is 1}} \:=\:p_0 + p_1 + p_2$

Therefore: .$\displaystyle p_0+p_1+p_2 \;=\;2$