Bingo probability question

Hello all,

I have visited the wizzard of odds to get some answers on bingo probabilities but I have a new version that isn't included on that site. I was wandering if anyone could help me with my problem.

First of all, how bingo is played (90 ball). Each player has a card with 15 numbers on it. The bingo caller starts calling numbers and when one player has all his 15 numbers on his card called he wins. So the earliest someone can win is on the 15th ball call.

This new variant I am trying to figure out has 91 balls, numbers 1-90 and one extra ball (let's call it the X ball). To win a special prize, you have to match 15 numbers on your card and have the X ball being called the second before last winning call. Therefore, the earliest someone can win is on the 16th ball call (15 numbers + X ball).

What I am trying to figure out is: The probability in a game of getting Bingo AND the X ball being called the second before last.

I thought that even if you have 1 player with 1 card, he will eventually get bingo, even if it gets to the 91st ball call. Therefore the probability of getting Bingo in a bingo game is 100%. The probability of getting the X ball in a specific calling position (before the last ball call) = 1/91. Therefore the overall probability is 1/91 (1 * 1/91). What do you think?

For the probability of getting this winning combination by a specific number of ball calls, I used the folllowing formula:

Combin(16,16)*(combin(75,ballcall-16)/combin(91,ballcall))

So for example, for the 30th ball call the formula would be:

Combin(16,16)*(combin(75,30-16)/combin(91,30)

Am I on the right track? Should be a piece of cake for anyone with basic knowledge on statisics.

Thanks in advance guys!