Given $Y$ as a geometrically distributed random variable with $p\in (0,1)$, what is $E[Y|Y\geq 10]?$
I got:$$E(Y|Y\geq 10)=\frac{E(\mathbf{1}_{Y\geq 10}Y)}{P(Y\geq 10)}=\frac{E(\mathbf{1}_{Y\geq 10}Y)}{1-P(Y\leq 9)}=\frac{E(\mathbf{1}_{Y\geq 10}Y)}{(1-p)^9}$$ But what is $E(\mathbf{1}_{Y\geq 10}Y)$ equal to?
$$E[1_{Y \ge 10} Y] = \sum_{y \ge 10} yp(1-p)^{y-1}$$ $$(1-p) E[1_{Y \ge 10} Y] = \sum_{y \ge 11} (y-1)p(1-p)^{y-1}$$ Subtracting the two equations yields $$pE[1_{Y \ge 10} Y] = 10p(1-p)^9 + p\sum_{y \ge 11}(1-p)^{y-1} = 10p(1-p)^9 + (1-p)^{10} = (1-p)^9 (9p+1)$$ and thus $$E[1_{Y \ge 10} Y] = \frac{(1-p)^9 (9p+1)}{p}.$$ Dividing by $(1-p)^9$ yields $E[Y \mid Y \ge 10] = \frac{9p+1}{p} = 9 + \frac{1}{p}$.
If you know the "memorylessness" property of the Geometric distribution, you can arrive at the answer much quicker by noting that the conditional distribution of $Y \mid \{Y \ge 10\}$ is the same as the distribution of $Y + 9$, so $E[Y \mid Y \ge 10] = E[Y] + 9 = \frac{1}{p} + 9$.