Probability of all my raffle tickets being winners

[belvet]

Hello,

I'm taking an introductory course on calculating probabilities, and I have a homework problem, the text of which is below. So far we've learned about classical probability, combinations, and permutations. I must be missing something obvious, because I'm completely confused as to where to even begin with this even though it should be pretty basic stuff. I would greatly appreciate any help in putting me on the right path.

A raffle has 100 tickets for sale. 10 of these tickets have prizes, and I have bought 3 tickets. What is the probability that all 3 of my tickets have prizes?

 

[Steven G]

What is the probability that any one ticket wins? What is the probability that your ticket is the winner (assume you only have one ticket)? After answering these questions try to think about your problem.

Please post back with the answers to my questions above so that we can move on from there.

 

[Dr.Peterson]

I'm taking an introductory course on calculating probabilities, and I have a homework problem, the text of which is below. So far we've learned about classical probability, combinations, and permutations. I must be missing something obvious, because I'm completely confused as to where to even begin with this even though it should be pretty basic stuff. I would greatly appreciate any help in putting me on the right path.

There are various ways to approach such a problem, but based on what you've learned, this one may be appropriate:

Imagine you lay out all 100 tickets in a row, with the 10 winners first. Now you randomly choose 3 of the 100 locations to be your tickets.

How many ways are there to choose 3 of the 100 tickets? How many ways are there to choose 3 of the 10 winners? How can you use these to calculate the probability that your 3 are all winners?

A very different approach would be to take it one ticket at a time: Pick one ticket; what is the probability that it is a winner? Now put that aside and pick another ticket; what is the probability that it, too, is a winner? How can you combine these probabilities? Then repeat with a third ticket, selecting again without replacement.

 

[belvet]

How many ways are there to choose 3 of the 10 winners?

I think this was the obvious thing I was missing, and I'm fairly certain that the answer is \( \frac{{10 \choose 3}}{{100 \choose 3}} = \frac{2}{2695} \). I think in my head I was somehow trying to force this to be more complicated than it needed to be.

A very different approach would be to take it one ticket at a time: Pick one ticket; what is the probability that it is a winner? Now put that aside and pick another ticket; what is the probability that it, too, is a winner? How can you combine these probabilities? Then repeat with a third ticket, selecting again without replacement.

The probability of the first ticket being a winner is \(\frac{10}{100}\), the next \(\frac{9}{99}\) and the third \(\frac{8}{98}\), assuming all three are winners. Putting these together according to the multiplication principle, we get \(\frac{10}{100} \cdot \frac{9}{99} \cdot \frac{8}{98} = \frac{2}{2695}\), which suggests the above answer was correct.

Thanks so much for your help.

 

[Dr.Peterson]

I think this was the obvious thing I was missing, and I'm fairly certain that the answer is \( \frac{{10 \choose 3}}{{100 \choose 3}} = \frac{2}{2695} \). I think in my head I was somehow trying to force this to be more complicated than it needed to be.

The probability of the first ticket being a winner is \(\frac{10}{100}\), the next \(\frac{9}{99}\) and the third \(\frac{8}{98}\), assuming all three are winners. Putting these together according to the multiplication principle, we get \(\frac{10}{100} \cdot \frac{9}{99} \cdot \frac{8}{98} = \frac{2}{2695}\), which suggests the above answer was correct.

Thanks so much for your help.

This is typical of probability problems: both making things too complicated, and being able to find two very different ways to the same answer, in order to confirm that what you thought was right really is. I never quite trust my work until I can do that!

Good work.