Suppose we have a randomized algorithm Test for testing whether a string appears in a language $L$ that works as follows. Given an input $x$, the algorithm Test chooses a random integer $r$ uniformly from the set $S=\{0,1, \ldots, p-1\}$ for some prime $p$. If $x$ is in the language, then $\operatorname{Test}(x, r)=1$ for at least half of the possible values of $r$. A value of $r$ such that Test $(x, r)=1$ is called a witness for $x$. If $x$ is not in the language, then Test $(x, r)=0$ always.
If we run the algorithm Test twice on an input $x \in L$ by choosing two numbers $r_{1}$ and $r_{2}$ independently and uniformly from $S$ and evaluating Test $\left(x, r_{1}\right)$ and Test $\left(x, r_{2}\right)$, then we find a witness with probability at least $3 / 4$. Argue that we can obtain a witness with probability at least $1-1 / t$ using the same amount of randomness by letting $s_{i}=$ $r_{1} i+r_{2} \bmod p$ and evaluating Test $\left(x, s_{i}\right)$ for values $0 \leq i \leq t<p .$