Probablity of winning full house at Bingo The probability of getting a full house (all 15 numbers) depends on the number of Numbers (between 1 and 90) that are called out. If $N$ numbers are called, what is the probability of getting a full house as a function of $N$. Obviously $P(N) = 0$ for $N<15$ and $P(90)=1$.
$P(N)$ will rise gradually from 0 to 1 as $N$ goes up from 14 to 90. Find $P(N)$  
 A: As you said the probability is zero for $N<15$. For $N\ge 15$ we need from these $N$ draws 15 specific numbers and the rest doesn't matter. We calculate the probabilty by fraction of combinatorical possibilities:
$$
\frac{\binom{90-15}{N-15}}{\binom{90}{N}}.
$$
Explanation: There are $\binom{90}{N}$ ways to draw $N$ from 90 balls. From those $N$ balls we want 15 specific numbers, so these are fixed. The number of ways to draw the rest is $\binom{90-15}{N-15}$.
A: There are in total $\frac{90!}{N!(90-N)!}$ ways to draw N from 90 balls.
The winning criteria requires 15 of those N balls to be specific numbers; and there are $\frac{75!}{(N-15)!(90-N)!}$ ways to draw the rest.
The probability of drawing a full house on or before the $N^{th}$ call is:$$\frac{75!N!}{90!(N-15)!}$$
A: The existing answers give the probability of getting a full house in a given column. I interpret the question to be asking for the probability of getting a full house in any of the $m$ columns. This can be calculated using inclusion-exclusion:
$$
\sum_{k=1}^m(-1)^{k+1}\binom mk\binom{t - 15k}{n - 15k}\;,
$$
where $t$ denotes the total number of balls drawn, which is given as $90$ in the question but which I believe is $75$ in the style of bingo under consideration; in standard $90$-ball bingo there are no columns with $15$ numbers. The number $m$ of columns would be $5$ for $75$-ball bingo and $9$ for $90$-ball bingo under standard rules.
