Question

Question about summation with Geometric Distribution.

See below.

Answer 1

So my geometric distribution is as follows
\[P[X=k]=(\frac{ 5 }{ 6 })^{k-1}(\frac{ 1 }{ 6 })\]

Answer 2

What I want to find is
\[P[X \le x]=\sum_{k=1}^{x}P[X=k]=???????\]

Answer 3

\[\begin{align*}
P(X\le x)&=\sum_{k=1}^xP(X=k)\\[1ex]
&=\sum_{k=1}^x\left(\frac{5}{6}\right)^{k-1}\frac{1}{6}\\[1ex]
&=\frac{1}{5}\sum_{k=1}^x\left(\frac{5}{6}\right)^{k-1}
\end{align*}\]Recall that
\[\sum_{k=1}^n ar^{k-1}=\frac{a\left(1-r^n\right)}{1-r}\]So to compute your desired probability, you can use the formula with \(a=\dfrac{1}{5}\) and \(r=\dfrac{5}{6}\).

Answer 4

thank you... as I'm sure you meant to put a = 1/6, but I got it.