If there are 10 outcomes, what is the probability that ANY outcome is selected 3 times after 15 tries? Each of the 10 outcomes has the same chance of being selected, and you choose 15 times. What is the probability that any of the outcomes is selected 3 times?
I get how to get the probability for any distinct outcome to be selected 3 times using binomial distribution, but I have no clue how to do it for all outcomes...
 A: With ten possible outcomes and instances of a outcomes appearing three
times being marked we get the species
$$\mathfrak{S}_{=10}(\mathfrak{P}_{=0}(\mathcal{Z})
+ \mathfrak{P}_{=1}(\mathcal{Z})
+ \mathfrak{P}_{=2}(\mathcal{Z})
+ \mathcal{U}\mathfrak{P}_{=3}(\mathcal{Z})
+ \mathfrak{P}_{=4}(\mathcal{Z})
+ \cdots).$$
This has EGF
$$f(z, u) = \left(\exp(z)-\frac{z^3}{6}+u\frac{z^3}{6}\right)^{10}.$$
Now the  complementary count of  no outcome appearing three  times has
EGF
$$f(z, 0) = \left(\exp(z) - \frac{z^3}{6}\right)^{10}.$$
Extracting coefficients we find
$$15! [z^{15}] \left(\exp(z) - \frac{z^3}{6}\right)^{10}
\\ = 15! [z^{15}] \sum_{q=0}^{10} {10\choose q} 
(-1)^q \frac{z^{3q}}{6^q} \exp((10-q)z)
\\ = 15!\sum_{q=0}^{5} {10\choose q} 
(-1)^q [z^{15-3q}] \frac{1}{6^q} \exp((10-q)z)
\\ = 15!\sum_{q=0}^{5} {10\choose q} 
(-1)^q \frac{1}{6^q} \frac{(10-q)^{15-3q}}{(15-3q)!}.$$
We thus obtain
$$1-10^{-15} \times 208406304739450
\approx 0.7915936953.$$
With  the exact  meaning of  the  question being  somewhat unclear  we
present a Maple program to clarify the interpretation being used which
will compute the first six values  for $k$ rolls with ten outcomes and
some outcome occuring three times.

ENUM :=
proc(k)
option remember;
    local ind, d, res, mset, adm;

    res := 0;

    for ind from 10^k to 2*10^k-1 do
        d := convert(ind, base, 10);

        mset := convert(d[1..k], `multiset`);

        adm := select(p->p[2]=3, mset);

        if nops(adm) > 0 then
            res := res + 1;
        fi;
    od;

    res/10^k;
end;

X := k -> 1 - k!*coeftayl((exp(z)-z^3/6)^10, z=0, k)/10^k;

A: This problem is unfortunately tedious and still ambiguously worded.
I shall interpret the problem to be that of the fifteen draws, you want at least one object to have been drawn exactly three times.
We approach via inclusion-exclusion:  Letting $A_1,A_2,\dots,A_{10}$ be the events that the first, second,... tenth object occurs exactly three times, we wish to count how many different length fifteen strings exist satisfying the desired condition.  I.e. we wish to count:
$$|\bigcup\limits_{i=1}^{10}A_i|=\sum\limits_{i=1}^{10}|A_i|-\sum\limits_{1\leq i<j\leq 10}|A_i\cap A_j|+\sum\limits_{1\leq i<j<k\leq 10} |A_i\cap A_j\cap A_k|-\dots$$
As each of the sequences are equally likely and there are $10^{15}$ possible sequences, this will let us calculate the probability.
You say you already know how to find the probability for a specific outcome to be selected exactly three times.  That's good.  To recap, we pick which three spaces in the sequence are occupied by that outcome and for the remaining spaces we pick which outcome takes its place.
That tells us that $|A_1|=\binom{15}{3}9^{12}$.  Equivalently, $Pr(A_1)=\binom{15}{3}(\frac{1}{10})^3(\frac{9}{10})^{12}$ which is what you might have thought to look for.  I'm just going to divide by my $10^{15}$ at the very end of my calculations instead of right now.
By symmetry, $|A_1|=|A_2|=\dots=|A_{10}|$ so the first summation will equal $10\cdot\binom{15}{3}\cdot 9^{12}$
We continue by calculating $|A_1\cap A_2|$.  We do so similarly.  Pick the three spaces used by object 1, pick the three spaces used by object 2, and then for each remaining space pick an object other than object1 and object2 to be placed there.
We get $|A_1\cap A_2|=\binom{15}{3}\binom{12}{3}8^{9}$.  Similarly by symmetry we get $|A_1\cap A_2|=|A_1\cap A_3|=\dots=|A_9\cap A_{10}|$, so each of the $\binom{10}{2}$ terms in the second summation will be this most recently calculated amount, showing that the second summation alltogether equals $\binom{10}{2}\binom{15}{3}\binom{12}{3}8^9$
We continue using the same method to calculate the remaining terms and recognize that we can stop at the fifth summation because it is impossible that in fifteen draws to get six or more objects drawn exactly three times each (as that would require at least 18 draws to do so).
We have a final total then of:
$$|\bigcup\limits_{i=1}^{10}A_i|=10\cdot\binom{15}{3}\cdot 9^{12}-\binom{10}{2}\binom{15}{3}\binom{12}{3}8^9+\binom{10}{3}\binom{15}{3}\binom{12}{3}\binom{9}{3}7^6\\-\binom{10}{4}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}6^3+\binom{10}{5}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}5^0$$
Which wolfram calculates as being approximately $7.9159\cdot 10^{14}$ different valid sequences.
Dividing then by $10^{15}$ we get a final answer that the probability is approximately $0.79159$
A: To me, this seems to be asking that if we put $15$ marbles into $10$ urns, what is the probability that some urn has at least $3$ marbles in it. This is different than the way that JMoravitz and Marko Riedel interpreted the question, and why my probability is higher than what they got.

To count the number of ways that urns can contain only $0$, $1$, or $2$ marbles, we let $k_1$ be the number of urns with one marble and $k_2$ be the number of urns with $2$ marbles. Then we have
$$
k_1+2k_2=15
$$
For a given $k_2$, there are 
$$
\frac{10!}{\underbrace{\left(15-2k_2\right)}_{k_1}!\,k_2!\underbrace{\,\left(k_2-5\right)}_{10-k_1-k_2}!}
$$
ways to choose the urns. For each such choice of urns, there are
$$
\frac{15!}{1!^{15-2k_2}\,2!^{k_2}}
$$
ways to fill the urns.
This gives
$$
\sum_{k=5}^7\frac{10!}{(15-2k)!\,k!\,(k-5)!}\frac{15!}{2^k}=31\,138\,995\,888\,000
$$
out of $10^{15}$ possible arrangements or $3.1139\%$ chance of not getting $3$ marbles in any urn.  This gives a probability of $96.8861\%$ chance of getting at least $3$ marbles in some urn.
