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Let $X$ and $Y$ be independent random variables, each uniformly distributed on $\{1,2,3,\ldots,11\}.$ I want to find $\mathrm{P}(X+Y=16).$

Well, the joint probability mass function of $X$ and $Y$ is give by $$ P(X=x, Y=y)=\frac{1}{11^2}.$$ To compute the above probability, I find all combinations of $X$ and $Y$ whose sum is 16. Doing that I get 7 pairs, and hence $$P(X+Y=16)=\frac{7}{11^2}.$$

My question is this: Is there a better way of approaching these types of problems other than the procedure I've outlined above?

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This method is fine, though you could say something like $P(X + Y = 16) = \sum_x P(X + Y = 16|X = x)P(X = x)$. The conditional probability is $1/11$ when $x\ge 5$ and zero otherwise, and $P(X = x)$ is $1/11$, which gives the same answer. – Louis Jan 1 '12 at 20:41

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