Probability of a die; Expected Value as Infinite Sum?

The value of the random value $X$ is the number of throws which are needed to roll a 6 with a fair die.

a) What is the formula for the probability that you will roll your first 6 on the $nth$ throw? (So $P(X=n)$

b) What is the formula for $P(X<n)$, meaning that up to the nth throw at least one 6 is rolled.

c) Express the expectancy (expected value) $E(X)$ as an infinite sum.

• We have $\Pr(X\lt n)=1-\Pr(X\ge n)=1-(5/6)^{n-1}$. – André Nicolas Jun 21 '15 at 16:21
• The distribution of $X$ is the so-called geometric distribution – drhab Jun 21 '15 at 16:21

a) $P(X=n)=P($first 6 in $n$th throw$)=P($no 6 in $(n-1)$ throws, 6 in $n$th throw$)=\dfrac{5^{n-1}}{6^n}$, with $n\geq1$.
b) $P(X<n)=P(X\leq n-1)=\sum_{k=1}^{n-1}\dfrac{5^{k-1}}{6^k}$ which you can compute.
c) $E(X)=\sum_{n=1}^\infty n\dfrac{5^{n-1}}{6^n}$. To find the value of this sum, consider looking up how to sum Arithmetic-Geometric series.