Regarding this question: If k-universal hash family then (k-1)-universal hash family
It's been answered and I dont understand the answer:
\begin{align}
Pr[h(x_1) = i_1 &\land \cdots \land h(x_{k-1}) = i_{k-1}]
\\&= Pr[h(x_1)=i_1 \land \cdots \land h(x_{k-1})=i_{k-1}\land h(x_k)=1]
\\&+ Pr[h(x_1)=i_1 \land \cdots \land h(x_{k-1})=i_{k-1}\land h(x_k)=2]+\ldots
\\&+Pr[ h(x_1)=i_1 \land \cdots \land h(x_{k-1})=i_{k-1}\land h(x_k)=m]
\\&= \frac{1}{m^k}+\frac{1}{m^k}+\ldots +\frac{1}{m^k}=\frac{m}{m^k}=\frac{1}{m^{k-1}}
\end{align}
Why is the first equality correct?
Best Answer
By definition, $h$ takes one of the values $1$ through $m$. Thus the event that it takes certain values at $x_1$ through $x_{k-1}$ is the disjoint union over $1\le j\le m$ of the events that it takes those values and also the value $j$ at $x_k$. The probability of the disjoint union of events is the sum of the probabilities of those events.