# probability problem that's been messing with my head. Please at least look at it

#### sergio1er

##### New Member
I'm taking a liberal arts math at my college and the teacher gave us a take home test.
There is one probability problem that's just not clicking.
Here is the problem and what I have done so far.

A person purchased 10 of the 1000 tickets sold for a raffle.
To determine the five prize winners, 5 tickets are to be drawn at random and without replacement.
Compute the probability that this person will win at least one prize.

Ok, first our teacher told us to approach the problem by finding the opposite probability of winning at least one prize. So the opposite of winning at least one prize is winning zero prices? Or winning all 5 prizes?

So I calculated with winning no prizes so it's 0C0, 0 chose 0, or in the ti 84, 0>MATH>PRB>3, which is nCr>0
All that is divided by 1000 choose 5, 1000 tickets and choosing 5 at random.

The teacher told us to do 1 minus (0c0/1000c5)
That gave me 1 which is basically 100 percent the probability of winning no prizes at all.

And that is where I'm stuck, I found the opposite probability, but does that mean that the probability of winning at least one prize is 0%?
That doesn't seem right to me.

Maybe there is a different approach to this problem.
Any help will be greatly appreciated Thank you.

#### asterisk

##### New Member
The probability that you do not win the first prize is $$\frac{990}{1000}$$

The probability that you do not win first or second prize is $$\frac{990}{1000}*\frac{989}{999}$$

#### X-man

##### New Member
There are 5C1000 ways to pick five winning tickets from the 1000 sold.

There are 5C990 ways to pick five winning tickets from the 990 that were sold to persons other than the person of interest.

Thus the probability that the person of interest does NOT win any of the five prizes is

5C990/5C1000 = 990*989*988*987*986 / 1000*999*998*997*996

and the probability that he wins at least one prize is this subtracted from one (unity).