# Probability of winning exactly one...

• Oct 11th 2010, 01:04 PM
Iceflash234
Probability of winning exactly one...
Given there were 50 tickets sold, you bought 5, and there are three prizes (1st, 2nd, and 3rd), what is the probability of winning exactly one prize.

I don't stand how to show the probability of an "exactly one" scenario. I know how to find the probability of "at least" or "none".

Help is greatly appreciated.
• Oct 11th 2010, 01:10 PM
Plato
I see that you have over twenty-five postings.
By now you should understand that this is not a homework service
So you need to either post some of your work on a problem or explain what you do not understand about the question.
• Oct 11th 2010, 01:13 PM
Iceflash234
I don't stand how to show the probability of an "exactly one" scenario. I know how to find the probability of "at least" or "none".
• Oct 11th 2010, 01:31 PM
Plato
There are $\dbinom{3}{1}\dbinom{47}{4}$ ways to win exactly one prize with five tickets.
• Oct 11th 2010, 02:27 PM
Iceflash234
Quote:

Originally Posted by Plato
There are $\dbinom{3}{1}\dbinom{47}{4}$ ways to win exactly one prize with five tickets.

So the probability of getting exactly one would be $\frac{1}{\dbinom{3}{1}\dbinom{47}{4}}$ ?

That makes sense. Thanks! I appreciate it.
• Oct 11th 2010, 02:32 PM
Plato
Good grief. NO! This totally wrong.
• Oct 11th 2010, 02:43 PM
Iceflash234
Quote:

Originally Posted by Plato
Good grief. NO! This totally wrong.

Wait what? How? $\dbinom{3}{1}\dbinom{47}{4}$ is the number of ways to win exactly one with 5 tickets.

Oh wait a minute, I need to figure out the number of total overall possible winning combinations and then divide the number of ways to win by the total number of winning combinations. But regardless of whether I do 50C3 or 50P3, both of which are smaller that the above combinations so they wouldn't work. Now I'm confused.
• Oct 11th 2010, 03:17 PM
Plato
The answer is $\dfrac{\binom{3}{1}\binom{47}{4}}{\binom{50}{5}}$