TITLE: What is the probability that $7$ cards are chosen and no suit is missing?
QUESTION [6 upvotes]: Cards are drawn one by one from a regular deck ($13$ cards for each of the
  $4$ suits). If $7$ cards are drawn, what is the probability that no suit will be
  missing?

Ok, so I tried the approach where I choose the $1$ suit out of $4$ and then I don't know what do next. I dont know how am I supposed to arrange the cards in such a random manner, and I found the total which is obvious, $52$ choose $7$.

REPLY [4 votes]: There are $4$ conditions, one for each suit that needs to be included. The number of (unordered) draws that violate $k$ particular conditions is $\binom{13(4-k)}7$, so by inclusion-exclusion the desired probability is
\begin{align}
\binom{52}7^{-1}\sum_{k=0}^3(-1)^k\binom4k\binom{13(4-k)}7
&=
\binom{52}7^{-1}\left(\binom{52}7-4\binom{39}7+6\binom{26}7-4\binom{13}7\right)
\\
&=
\frac{63713}{111860}
\\
&\approx57\%\;.
\end{align}