can someone help answer this math question?

If this is a raffle with 800 tickets, of which 45 are picked at random to win prizes, then to guarantee (ie be 100% certain) that you hold at least one of the 45 winning tickets, you need to purchase 800-45+1 = 756 tickets.

Richard B is correct. If you are willing to be less than 100% certain of getting a prize, you can buy fewer tickets. That is a more difficult problem.

As Richard B states, you first need to decide how uncertain you are willing to be, eg 50%. However, Richard B's method solves the problem : If I buy n tickets, what is the probability of winning at least 1 of the 45 prizes? Your problem is the opposite : If I want to have probability x% of winning at least 1 of the 45 prizes, how many tickets do I need to buy?

Your problem can be solved by trial and error using Richard B's method on a spreadsheet. If you buy n tickets the probability of not winning any of the 45 prizes is

(800-n)/800 * (799-n)/799 * (798-n)/798... * (756-n)/756.

If you decide you want to have a probability of x% of winning at least 1 of the 45 prizes, vary n until this product is below 100-x.

To be 50% certain of winning at least 1 prize you need to buy 12 tickets.

To be 90% certain of winning at least 1 prize you need to buy 39 tickets.

To be 99% certain of winning at least 1 prize you need to buy 76 tickets.

756

cannot be answered. obviously if you have 766 entries you will be picked.

but that is not who to figure it. first decide is ½ or 50% 'good enough"

then 799/800 is a loser on the first draw, then 778/799 will lose on the second draw. calculate 45 draws by lose* lose * lose...45 times that is the chance you will lose all and never win once.