Question: Consider a finite population $~U=\{1,~2,~3\}~$ with $$p~\left(\{1,~2\}\right)=\dfrac 12~,~~~~p~\left(\{1,~3\}\right)=\dfrac 14~,~~~~p~\left(\{2,~3\}\right)=\dfrac 14~.$$ Then if $~\pi_k~$ denotes the first order inclusion probability of the $k^{\text{th}}$ unit, which of the following is not true ?
$(1)~~\pi_1=\dfrac 34$
$(2)~~\pi_2=\dfrac 34$
$(3)~~\pi_1+\pi_2+\pi_3=1$
$(4)~~\pi_1+\pi_2+\pi_3=2$
${}$
My opinion: We know that
First-order inclusion probability: The first-order inclusion probability $π_i$ refers to the probability that unit $i$ is included in the sample. $$π_i = Pr(i ∈ A) = \sum_{A; i∈A}P(A)$$
I came across example$~(2.1)$ and found accordingly to their procedure
$$\pi_1=\dfrac 12+\dfrac 14=\dfrac 34~,~~\pi_2=\dfrac 12+\dfrac 14=\dfrac 34~,~~\pi_3=\dfrac 14+\dfrac 14=\dfrac 12$$ and hence $$~\pi_1+\pi_2+\pi_3=\dfrac 34+\dfrac 34+\dfrac 12=2~$$
Hence option $\bf{(3)}$ is the correct answer here. But I am unable to understand how they did it (in which procedures they follow) ? Are they use some particular theorem or property ?
Please help to understand the fact.
Look at definition: $$\pi_i = Pr(i \in A) = \sum_{A : i\in A} P(A)$$ The right hand side is a sum of $P(A)$ over all sets $A$ which containes $i$ as element.
Let us find $\pi_1$. Here $i=1$. Which sets contain $i=1$ as element? $1\in\{1,2\}$ and $1\in\{1, 3\}$. Sum up probabilities of these sets and get $$\pi_1=P(\{1,2\})+P(\{1,3\})=\frac12+\frac14=\frac34.$$