# Answer to Question #374 in Statistics and Probability for Leonard

Question #374
There are n tickets in the lottery, m of them are winning ones. Someone has bought k tickets. What is the probability that at least one of them is lottery loan?
1
2010-07-06T11:18:17-0400

The number of ways to select k different tickets from n is: A = (Cn)^k.
The number of ways to select k different tickets from (n-m) non-winning is: B = C(n-m)^k.
The probability of selecting all k non-winning tickets is: P = B/A = (C(n-m)^k)/(Cn)^k.

As it is the only case which doesn&rsquo;t suit us, the solution (probability to get at least one winning ticket) is: P&rsquo; = 1&ndash;P = 1&ndash;(B/A) = (1&ndash;C(n-m)^k)/(Cn)^k = 1&ndash;((n-m)!/((n-m-k)!*k!))/(n!/((n-k)!*k!)) =
= 1&ndash;((n-m)!*(n-k)!)/((n-m-k)!*n!).

Notice, if k&gt;(n-m) then P=1 (someone has bought more tickets than the amount of non-winning tickets).

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