# Talk:Indecomposable distribution

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Maybe Typo ???

${\displaystyle {\begin{matrix}V=\left\{{\begin{matrix}1&\mathrm {with} \ \mathrm {probability} &1,\\0&\mathrm {with} \ \mathrm {probability} &1-q,\end{matrix}}\right.\end{matrix}}}$

I feel this should be

${\displaystyle {\begin{matrix}V=\left\{{\begin{matrix}1&\mathrm {with} \ \mathrm {probability} &q,\\0&\mathrm {with} \ \mathrm {probability} &1-q,\end{matrix}}\right.\end{matrix}}}$

-hgkamath

Yes. The "q" and "1" keys are close together. Thanks. I've fixed it. Michael Hardy 01:30, 28 August 2006 (UTC)

## Don't understand example

Suppose a random variable Y has a geometric distribution
${\displaystyle \Pr(Y=n)=(1-p)^{n}p\,}$
... now let Dn be the nth binary digit of Y ... then the Ds are independent and
${\displaystyle Y=\sum _{n=1}^{\infty }{D_{n} \over 2^{n}},}$

Umm. What?? I don't know how to fix this. --God made the integers (talk) 20:29, 13 January 2017 (UTC)