Answer on Statistics and Probability Question for Leonard
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’t suit us, the solution (probability to get at least one winning ticket) is: P’ = 1–P = 1–(B/A) = (1–C(n-m)^k)/(Cn)^k = 1–((n-m)!/((n-m-k)!*k!))/(n!/((n-k)!*k!)) =
Notice, if k>(n-m) then P=1 (someone has bought more tickets than the amount of non-winning tickets).
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