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Let X~geometric (1/3), and let Y=|X-5|. Find the range n and PMF of Y.

Here is my trial

If $x=0$, $P(Y=|0-5|)=P(Y=5)=\left(\frac{2}{3}\right)^5 \frac{1}{3}$

If $x=1$, $P(Y=|1-5|)=P(Y=4)=\left(\frac{2}{3}\right)^4 \frac{1}{3}$

If $x=2$, $P(Y=|2-5|)=P(Y=3)=\left(\frac{2}{3}\right)^3 \frac{1}{3}$

If $x=3$, $P(Y=|3-5|)=P(Y=2)=\left(\frac{2}{3}\right)^2 \frac{1}{3}$

If $x=4$, $P(Y=|4-5|)=P(Y=1)=\left(\frac{2}{3}\right)^1 \frac{1}{3}$

If $x=5$, $P(Y=|5-5|)=P(Y=0)= \frac{1}{3}$

Range of $Y={0,1,2,3,4,5}$ and PMF is I $P(Y=y)=\left(\frac{2}{3}\right)^y \frac{1}{3}$ for $y=0,1,2,3,4,5$

Am I correct. If not please correct me.

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    $X$ has geometric distribution (not $Y$). This leads to answers like: $\{Y=3\}=\{X=2\vee X=8\}$ so that $P(Y=3)=P(X=2)+P(X=8)$ et cetera. To find the range ask yourself: what values can $X$ take (having geometric distribution)? Then what values can $Y:=|X-5|$ take? – 2017-02-28

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