Can pmf of discrete random variables vary? Can probability mass function be in different form even though when we plug in $k =1,2,3,\cdots$ for $P(X=k)$, the probability for each $k$ between pmf A and pmf B would still be the same. For example :
$1.6.2$. Let a bowl contain $10$ chips of the same size and shape. One and only one of these chips is red. Continue to draw chips from the bowl, one at a time and at random and without replacement, until the red chip is drawn.
(a) Find the pmf of $X$, the number of trials needed to draw the red chip. (b) Compute $P(X \leq 4)$. (from Introduction to Mathematical Statistics by C. Hogg)
The key answer would be $p(x) = \dfrac{\binom{9}{x-1}}{\binom{10}{x-1}}\times\dfrac1{11-x}$ for $x = 1,2,...,10$. What if i write $p(x) = \dfrac{\binom{x}{1}}{\binom{10}{1}} \times\frac1x$ for $x = 1,2,...,10$?

Note that \begin{align*}
\frac{\binom9{x-1}}{\binom{10}{x-1}}\times \frac{1}{11-x}&=\frac{\frac{9!}{(x-1)!(10-x)!}}{\frac{10!}{(x-1)!(11-x)!}}\times \frac{1}{11-x}=\frac{9!}{10!}\times \frac{(11-x)!}{(10-x)!}\times \frac{1}{11-x}\\
&= \frac{1}{10}\times \frac{(10-x)!}{(10-x)!}=\frac{1}{10}
\end{align*}
and your answer is equal to $$\frac{\binom x1}{\binom{10}{1}}\times \frac1x=\frac x{10}\times \frac 1x=\frac{1}{10} $$
The two expressions are equivalent.
