$10 \%$ of people in a population suffer from high blood pressure. Calculate the probability that the twelfth person tested will be the second one with high blood pressure. So this is a negative binomial distribution. Now $P(X=k) = \binom{k+r-1}{k} (1-p)^{r}p^{k}$. So this is the probability of $k$ successes after $r$ failures. So $P(X=2) = \binom{11}{2}(0.9)^{10}(0.1)^2$? 
At least to me, it is helpful to think in terms of bases.
(I'll only be talking about tensor products of finite-dimensional vector
spaces here.)
This makes the universal mapping property that Zach Conn talks about
a bit less abstract (in fact, almost trivial).

First recall that if $L: V \to U$ is a linear map, then $L$ is completely determined
by what it does to a basis $\{ e_i \}$ for $V$:
$$L(x)=L\left( \sum_i x_i e_i \right) = \sum_i x_i L(e_i).$$
(The coefficients of $L(e_i)$ in a basis for $U$ give the $i$th column
in the matrix for $L$ with respect to the given bases.)

Tensors come into the picture when one studies multilinear maps.
If $B: V \times W \to U$ is a bilinear map, then $B$
is completely determined by the values $B(e_i,f_j)$ where
$\{ e_i \}$ is a basis for $V$
and $\{ f_j \}$ is a basis for $W$:
$$B(x,y) = B\left( \sum_i x_i e_i,\sum_j y_j f_j \right) = \sum_i \sum_j x_i y_j B(e_i,f_j).$$
For simplicity, consider the particular case when $U=\mathbf{R}$;
then the values $B(e_i,f_j)$
make up a set of $N=mn$ real numbers (where $m$ and $n$ are the
dimensions of $V$ and $W$), and these numbers are all that we need to keep
track of in order to know everything about the bilinear map $B:V \times W \to \mathbf{R}$.

Notice that in order to compute $B(x,y)$ we don't really need to know the
individual vectors $x$ and $y$, but rather the $N=mn$ numbers $\{ x_i y_j \}$.
Another pair of vectors $v$ and $w$ with $v_i w_j = x_i y_j$ for all $i$ and $j$
will satisfy $B(v,w)=B(x,y)$.

This leads to the idea of splitting the computation of $B(x,y)$ into two stages.
Take an $N$-dimensional vector space $T$ (they're all isomorphic so it doesn't matter
which one we take) with a basis $(g_1,\dots,g_N)$.
Given $x=\sum x_i e_i$ and $y=\sum y_j f_j$,
first form the vector in $T$
whose coordinates with respect to the basis $\{ g_k \}$ are given by the column vector
$$(x_1 y_1,\dots,x_1 y_m,x_2 y_1,\dots,x_2 y_m,\dots,x_n y_1,\dots,x_n y_m)^T.$$
Then run this vector through the linear map $\tilde{B}:T\to\mathbf{R}$ whose matrix
is the row vector
$$(B_{11},\dots,B_{1m},B_{21},\dots,B_{2m},\dots,B_{n1},\dots,B_{nm}),$$
where $B_{ij}=B(e_i,f_j)$.
This gives, by construction, $\sum\sum B_{ij} x_i y_j=B(x,y)$.

We'll call the space $T$ the tensor product of the vector spaces $V$ and $W$
and denote it by $T=V \otimes W$;
it is “uniquely defined up to isomorphism”,
and its elements are called tensors.
The vector in $T$ that we formed from $x\in V$ and $y\in W$ in the first stage above
will be denoted $x \otimes y$;
it's a “bilinear mixture” of $x$ and $y$ which doesn't allow us to
reconstruct $x$ and $y$ individually,
but still contains exactly all the information needed
in order to compute $B(x,y)$ for any bilinear map $B$;
we have $B(x,y)=\tilde{B}(x \otimes y)$.
This is the “universal property”; any bilinear map $B$ from $V \times W$
can be computed by taking a “detour” through $T$, and this detour
is unique, since the map $\tilde{B}$ is constructed uniquely from
the values $B(e_i,f_j)$.

To tidy this up, one would like to make sure that the definition is
basis-independent. One way is to check that everything transforms
properly under changes of bases. Another way is to do the construction
by forming a much bigger space and taking a quotient with respect to
suitable relations (without ever mentioning bases).
Then, by untangling definitions, one can for
example show that a bilinear map $B:V \times W \to \mathbf{R}$ can be
canonically identified with an element of the space $V^* \otimes W^*$,
and dually an element of $V \otimes W$ can be identified with a
bilinear map $V^* \times W^* \to \mathbf{R}$.
Yet other authors find this a convenient starting point, so that they
instead define $V \otimes W$ to be the space of bilinear maps $V^*
\times W^* \to \mathbf{R}$.
So it's no wonder that one can become a little confused when trying
to compare different definitions...
