# Help understanding conditional probability

Hi I'm having a hard time wrapping my head around this particular problem.

Suppose the lifetime of a shirt bought from Sears, in days, is a non-negative random variable $L$ with probability mass function $p(n) = 2^{-n}$ for $n = 1,2,...$ and $p = 0$ otherwise. What is the conditional probability mass function of $L$ given that $L > n$

Is the answer not 0?

$p_{L}(n | L > n) = Pr(L = n | L > n) = \frac{Pr(L = n \cap L > n)}{Pr(L > n)} = 0$, L cannot be equal to n and be greater than n.

Thanks for your help.

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You should be computing $P(L=k\mid L>n)$ for every $k$. You would find $0$ for every $k\leqslant n$ and something nonzero for each $k\geqslant n+1$. –  Did Oct 17 '11 at 5:21
Thank you for the clarification. –  Jonathan Oct 17 '11 at 5:40

You are asked to find $p_L(x|L>n)$ for $x=1,2,\dots$, not $p_L(n|L>n)$. Of course $p_L(n|L>n)=0$. In fact $p_L(x|L>n)=0$ for $x=1,2,\dots, n$.