# Total risk in multiple risky actions

I have a minefield with $c \times r$ cells. There are $n$ mines randomly placed in cells (no cell has more than one mine, just none, or one), and $n \lt c \times r$. How do I calculate the risk of exploding at least one mine by probing $m$ distinct cells? Probing does not stop if a mine explodes.

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The number of ways to select $m$ cells from the minefield is the binomial coefficient ${c r} \choose m$. The number of ways to select $m$ of the $c r - n$ non-mined cells is ${c r - n} \choose m$. So if the selection is random, the probability of not finding a mine is ${{c r - n} \choose m}/{{c r} \choose m}$.
The probability of finding at least one mine is $1 -$ that.
Thanks! From the upvotes I can say your answer is correct, but I have no clue what ${{c r - n} \choose m}/{{c r} \choose m}$ stands for ... is there a way to use simple math functions (+,-,*,/,^) to state the formula? –  Majid Fouladpour May 3 '11 at 21:47