It goes like this: Let $X$ be the number of random number selected from $\{0,1,2,\ldots,9\}$ independently until $0$ is chosen. Find the probability mass functions of $X$ and $Y=2X+1$.
I know that the probability mass function of $X$ is $(\frac{9}{10})^{i-1}(\frac{1}{10}):i=1,2,3,4,\ldots$ because I know that $i$ is the number of picks and we keep picking a non-zero number $i-1$ times and the $i^{th}$ pick is 0 with probability $(\frac{1}{10})$.
But for the second part, why is $(\frac{9}{10})^\frac{(j-1)}{2}(\frac{1}{10})$ the answer? More importantly how is $\frac{(j-1)}{2}$ the exponent on the $(\frac{9}{10})$?
Sorry for any confusion, I know what the exponent is representing I'm just wondering how it was found.