<s>[INST] If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20? 
If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?

Is there someway to figure this out without manually figuring out each possible scenario in which the 9 digits don't sum to 20?
 [/INST] A generating function approach yields the result without too much computation.
If we choose 9 integers from 1 to 10, there are $10^9$ possible outcomes, all of which we assume are equally likely.  We want to count the outcomes in which the total is less than or equal to 19, i.e. the number of solutions in integers to 
$$x_1 + x_2 + x_3 + \dots +x_9 \le 19$$
subject to $1 \le x_i \le 10$ for $i = 1,2,3, \dots ,9$.
An equivalent problem is to count the number of solutions to
$$x_1 + x_2 + x_3 + \dots +x_9 + y = 19$$
subject to $1 \le x_i \le 10$ for $i = 1,2,3, \dots ,9$ and $0 \le y$.
The generating function for the number of solutions to 
$$x_1 + x_2 + x_3 + \dots +x_9 + y = r$$ subject to the above constraints is
$$\begin{align}
f(z) &= (z + z^2 + z^3 + \dots + z^{10})^9 \; (1 + z + z^2 + \dots) \\
&= z^9 \; \left( \frac{1-z^{10}}{1-z} \right)^9 \frac{1}{1-z}\\
&= z^9 \; (1-z^{10})^9 \; (1-z)^{-10} \\
&= z^9 \cdot \sum_{i=0}^9 (-1)^i \binom{9}{i}z^{10 i} \cdot \sum_{j=0}^{\infty} \binom{10+j-1}{j} z^j
\end{align}$$
From the last equation we can easily find the coefficient of $z^{19}$, which is
$$[z^{19}] f(z) = 1 \cdot \binom{19}{10} - \binom{9}{1} \cdot 1 = 92,369$$
So the probability of getting a sum less than or equal to 19 is 
$$\frac{92,369}{10^9} = 9.2369 \times 10^{-5}$$
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