Finding the Probability function of the remainder. Let X have a geometric distribution with $f(x) = p(1−p)x$, $x = 0,1,2,\ldots$ Find the probability function of $R$, the remainder when $X$ is divided by $4$.


I am stuck on the above practice question: If I say $R= 0, 1/4, 2/4, \ldots$ and then use the continuous case equivalent of geometric distribution to find the probability function of $R$, which is exponential distribution, will I be correct?
I'm not sure what you mean by "using the continuous case equivalent of geometric distribution to find the probability function of R".  However, here's how I would approach the problem:

When you divide $X$ by $R$, there are three possible remainders: $0,1,2,$ and $3$.  You will get a remainder of $0$ for multiples of $4$, that is, when $X = 0,4,8,\dots$.  You will get a remainder of $1$ for numbers one more than the multiples of $4$, such as $1,5,9,\dots$.  Hopefully, you get the idea by now.

With that in mind, the probabilities are as follows: clearly, $P(R=r) = 0$ for all numbers $r$ besides $0,1,2,3$.  For $0,1,2,3,$ we have:
$$
\begin{align}
P(R = 0) &amp;= 
p(1-p)^0 + p(1-p)^4+p(1-p)^8+\cdots\\
&amp;= \sum_{k=0}^\infty p(1-p)^{4k}\\
P(R = 1) &amp;= 
p(1-p)^1 + p(1-p)^5+p(1-p)^9+\cdots\\
&amp;= \sum_{k=0}^\infty p(1-p)^{4k+1}\\
P(R = 2) &amp;= 
p(1-p)^2 + p(1-p)^6+p(1-p)^10+\cdots\\
&amp;= \sum_{k=0}^\infty p(1-p)^{4k+2}\\
P(R = 3) &amp;= 
p(1-p)^3 + p(1-p)^7+p(1-p)^11+\cdots\\
&amp;= \sum_{k=0}^\infty p(1-p)^{4k+3}\\
\end{align}
$$
From there, you could use the formula for the sum of a geometric series.  Alternatively, you could simply make the following observations:
$$
P(R = 3) = (1-p)P(R = 2)\\
P(R = 2) = (1 -p)P(R = 1)\\
P(R = 1) = (1 - p)P(R = 0)\\
P(R = 0) + P(R = 1) + P(R = 2) + P(R = 3) = 1
$$
And solve the above system of equations to find the required probabilities.
If  0≤ r ≤3, then  P(R = r) = P(X = r) + P(X = 4 + r) + P(X = 8 + r) + .... = p(1 - p)^r + p(1 - p)^(4+r) + p(1 - p)^(8+r) + ......, which is a geometric series.
Then we get the answer,p(1-p)^r /(1-(1-p)^4)