What is the Probability of Winning in a Raffle Ticket Draw with Replacement?

[neden]

Hi,

This is the question:
Players are awarded $1000 dollars in a contest. Each player draws a ticket from a bowl of 100 raffle-tickets. Once a winning ticket is drawn the draw is over. For every ticket drawn it is with replacement. What is the probability one of the first 25 customers is the winner?

My thinking
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I don't really have too much confidence in my understanding of probability but from what I can see this is how it looks like:

1) We know that we only need one person to win.
2) Secondly, there's some sort of combination of winners that can arise from the first 25 customers (although I am not exactly sure how the combination looks)

I have two ways of approaching this:

1) Solving it according to my own logic and thinking.
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25*(1/100)(99/100)*<10 choose 1>

25 because 25 customers,
1/100 is the chance of getting the ticket
99/100 is the chance of failure
<10 choose 1> because you can have a combination

2) It is a simple plug in and solve question
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Thus I deduced that using the Binomial Distribution formula would solve the answer.

P(event) = <n (choose) k> (success)^k * (failure)^(n-k)
where n is 100, k = 1, success= 1/100, failure = 1 - success.
(... something tells me this is completely not the right answer since, I never even accounted for the 10 customers anywhere ... confused!)

Nevertheless, is this question as simply plugging in and finding the answer or no? (and to be honest, I don't completely understand how the logic behind binomial distribution formula works).

 

[tiny-tim]

hi neden!

Players are awarded $1000 dollars in a contest. Each player draws a ticket from a bowl of 100 raffle-tickets. Once a winning ticket is drawn the draw is over. For every ticket drawn it is with replacement. What is the probability one of the first 25 customers is the winner?.

oooh, you're making this far too complicated …

what is the probability none of the first 25 customers is the winner? ​

 

[neden]

Basically you are saying 1 - P(none of the first 25 customers is the winner)

Ok...

So the answer is: 1 - (24/25 * <100 choose 1>) ??

 

[tiny-tim]

(¹⁰⁰C₁ is just 100)

Nooo … what is the probability of tossing a coin 25 times and getting no heads?

 

[neden]

(1/2) ^ 25, which is the same probability of tossing a coin 25 times and getting no tails; since both are have equiprobable outcomes.

 

[tiny-tim]

Yup!

Now use the same method on the raffle tickets.

 

[neden]

none of the first 25 customers is the winner: (99/100)^24*(1/100) ?

Edit: Wait... that doesn't make sense ... I'm confused!

 

[tiny-tim]

none of the first 25 customers is the winner: (99/100)^24*(1/100) ?

(try using the X² icon just above the Reply box )

No, that's the probability of the 25th customer winning.

 

[neden]

Ok for the coin question there's half the chance of getting heads in 25 coin tosses, which is why I said (1/2)²⁵

In this case, I don't even know whether to start with 1/25 or 1/100, they both make sense; 1 winner out of the 25 customers or 1 winning ticket out of the 100 tickets. But which one?

 

[tiny-tim]

The question asks for the probability that one of the first 25 customers is the winner.

So that's 1 - the probability that none of the first 25 customers is the winner.

If none is the winner, why do you want to use 1/100 ?

Use 99/100.

(and I'm off to bed :zzz: … see you tomorrow)

 

[neden]

Ok, so (99/100)²⁴ is the full solution? Well I guess I'll wait for the answer tomorrow.
I'm sorry but I'm completely new to probability.

Edit: (99/100)²⁵ because total trials is 25, we don't care if 24 non-winners and 1 winner, we want the probability as a whole, right?

 

[tiny-tim]

hi neden!

(just got up :zzz: …)

Edit: (99/100)²⁵ because total trials is 25, we don't care if 24 non-winners and 1 winner, we want the probability as a whole, right?

s'right!

(well, 1 minus that, of course! )

 

[neden]

Thank you.