Solutions - 5110-03

(1) Define random variable:

Let $X\sim \text{Geom}\left(\frac{1}{6}\right)$.

We wish to find $P[X\ge 10]$.

For all $X=k$, $k\ge 10$, the first $k-1$ trials result in failure, and the $k^{\text{th}}$ trial is a success.

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(2) Compute probability:

$$\begin{array}{c}P[X\ge 10]\gg \gg \sum _{k=10}^{\infty }\frac{1}{6}{(1-\frac{1}{6})}^{k-1}\\ \\ \gg \gg \frac{1}{6}\sum _{k=10}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\gg \gg \frac{1}{6}\left(\frac{{\left(\frac{5}{6}\right)}^9}{1-\frac{5}{6}}\right)\\ \\ \gg \gg {\left(\frac{5}{6}\right)}^9\gg \gg \approx 0.1938\end{array}$$

Note that the summation is simplified using the formula for a geometric series.