Probability: geometric random variables

Let $X$ be a geometric random variable with parameter $p$. Find the probability that $X\ge10$. Express your answer in terms of $p$

so I said:$$P(X\ge10)=\sum_{k=10}^\infty p^{10}(1-p)^{k-10}=p^{10}\sum_{k=0}^\infty (1-p)^k=p^9$$but this is wrong. Where is my mistake?

Question

Probability: geometric random variables

Let $X$ be a geometric random variable with parameter $p$. Find the probability that $X\ge10$. Express your answer in terms of $p$

so I said:$$P(X\ge10)=\sum_{k=10}^\infty p^{10}(1-p)^{k-10}=p^{10}\sum_{k=0}^\infty (1-p)^k=p^9$$but this is wrong. Where is my mistake?

ybarrap · Answer

$$
P(X\ge10)=\sum_{k=10}^\infty q^{k-1}p=p\sum_{k=9}^\infty q^{k}\
=1-p\sum_{k=0}^8 q^{k}\
=1-\cfrac{1-q^{9}}{1-q}\
=1-\cfrac{1-(1-p)^9}{p}\
=\cfrac{p-1-(1-p)^9}{p}\
=\cfrac{-(1-p)^9}{p}-\cfrac{1}{p}+1
$$
Does this make sense?

turingtest · Answer

why yes, indeed it does. Thank you!

ybarrap · Answer

Oops, I found an error:
$$
P(X\ge10)=\sum_{k=10}^\infty q^{k-1}p=p\sum_{k=9}^\infty q^{k}\
=1-p\sum_{k=0}^8 q^{k}\
=1-p\cfrac{1-q^{9}}{1-q}\
=1-p\cfrac{1-q^{9}}{p}\
=1-(1-q^9)\
=q^9\
=(1-p)^9
$$