Raffle with 20 tickets and 2 prize

[AlonzoN]

In a raffle, 20 tickets are sold and there are 2 prizes. If you buy 5 tickets, what is the probability that you win at least 1 of the prizes?

There's an another question with 30 tickets but the same principles, but asking for the pribability of winning A) neither prize B) Both prizes C) at least one prize.

 

[Harry_the_cat]

What do you think? Where do you get stuck?

 

[pka]

In a raffle, 20 tickets are sold and there are 2 prizes. If you buy 5 tickets, what is the probability that you win at least 1 of the prizes?

You have done eighty post, so you should know the rules. We help those who show some effort,
Questions that ask for at least one often are found easiest as one minus none.
So please post some work.

 

[AlonzoN]

Well the work I've done is this:

The chances of you having one of the winning tickets is 1/10, and the odds of you having the losing tickets is 9/10, so I constructed a tree diagram showing the odds:

 

[AlonzoN]

But then I thpugh, since you have 5 tickets, you would need to times the first set of "W" by 5, so 5 * 1/10, but then if you win, you'd have 4 tickets left, so for the "WW" probability I did this maths; 5 * 1/10 * 4 * 1/19, then I figured it would be the same for the others:

5 * 1/10 * 4 * 1/19 = 2/19

5 * 1/10 * 4 * 18/19 = 36/19

5 * 9/10 * 4 * 2 /19 = 36/19

I think for the last one though, if somebody else won the prize, it should be 1/18, but I'm not too sure.

 

[Deleted member 4993]

In a raffle, 20 tickets are sold and there are 2 prizes. If you buy 5 tickets, what is the probability that you win at least 1 of the prizes?

Calculating "NONE" of the five tickets are winner:

18/20 * 17/19 * 16/18 * 15/17 * 14/16 = (18!/13!) / (20!/15!)

 

[pka]

In a raffle, 20 tickets are sold and there are 2 prizes. If you buy 5 tickets, what is the probability that you win at least 1 of the prizes?

Picking upon on reply #6, I would do it this way: [imath]\dfrac{\dbinom{18}{5}}{\dbinom{20}{5}}[/imath] SEE HERE
Use that link to see that the probability of no prizes in five tickets is about 55.2%.
You can try other combinations.

 

[AlonzoN]

Calculating "NONE" of the five tickets are winner:

18/20 * 17/19 * 16/18 * 15/17 * 14/16 = (18!/13!) / (20!/15!)

So would I need to do the same for if one of the tickets is a winner? What about if two of the five are winning tickets? How would I do that? I looked at the back for an answer and it said that to get at least one of the prizes, it's 17/38. I don't know if that's including "WW" or not with "WL" and "LW".

 

[AlonzoN]

So would I need to do the same for if one of the tickets is a winner? What about if two of the five are winning tickets? How would I do that? I looked at the back for an answer and it said that to get at least one of the prizes, it's 17/38. I don't know if that's including "WW" or not with "WL" and "LW".

What I was thinking was that if you one the first prize, you would have 4 tickets afterwards, so I thought it would be something like 5*1/10, then when you're going for the next prize, you'd have 4 tickets left, so, 5 * 1/10 * 4 * 1/19. It this right?

 

[AlonzoN]

Picking upon on reply #6, I would do it this way: [imath]\dfrac{\dbinom{18}{5}}{\dbinom{20}{5}}[/imath] SEE HERE
Use that link to see that the probability of no prizes in five tickets is about 55.2%.
You can try other combinations.

How do i try other combinations?