Expected valued of Random sums about dice and jar problem A six-sided die is rolled , and the number N on the uppermost face is recorded.
From a Jar containing 10 tag numbered 1,2,,,,10 , we then select N tags at random without replacement. Let X be the smallest number on the drawn tags.
Determine p(X=2) and expected value.
Actually, I found out P(X=2) with long calculation using conditional probability. 
But, do I have to find out every probability of X for X=1,2,3,,,10 to find out expected 
value?
Is there any way that I can use property of random sum? 
 A: What we want to do is obtain a general expression for the (conditional) probability distribution of the random variable $X$ representing the minimum number drawn among $N = n$ numbers chosen from $\{1, 2, \ldots, 10\}$ without replacement.  To this end, we make the following critical observation:  If the minimum number is $X = x$, then the remaining $n-1$ numbers must be drawn from the set $\{x+1, x+2, \ldots, 10\}$.  Thus, the desired probability is given by $$\Pr[X = x \mid N = n] = \frac{\binom{10-x}{n-1}}{\binom{10}{n}}, \quad x = 1, 2, \ldots, 11-n,$$ where $N \sim \operatorname{Uniform}[1,6]$ is the discrete uniform on the set $\{1, 2, \ldots, 6\}$.  This easily gives us $$\Pr[X = 2] = \sum_{n=1}^6 \Pr[X = 2 \mid N = n] \Pr[N = n] = \frac{1}{6} \sum_{n=1}^6 \frac{\binom{8}{n-1}}{\binom{10}{n}} = \frac{1}{6 \cdot 90} \sum_{n=1}^6 n(10-n) = \frac{119}{540}.$$  To compute the expectation, we simply use the law of total expectation (iterated expectations):  $$\begin{align*} \operatorname{E}[X] &= \operatorname{E}[\operatorname{E}[X \mid N]] \\ &= \operatorname{E}\left[ \sum_{x=1}^{11-N} x \cdot \Pr[X = x \mid N] \right] \\ &= \operatorname{E}\left[ \frac{11}{1+N} \right] \\ &= \frac{2453}{840}. \end{align*}$$
