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If ten people each have a ten percent chance of winning a prize. What is the probability that at least one of them wins the prize?

Background :

there are 100 prizes to be won, and 1000 people with their names in the hat to win the prize.

Therefore, one person has a ten percent chance of winning... If you can place your name in the hat ten times... what are your odds of winning. (assuming your ten tickets are included within the 1000 total)

How did you figure this out?

Thanks!

Daniel

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    Find the probability that no one wins, and subtract from 1.2012-06-15
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    It is important to say whether these events are independent. Otherwise the question may be understood in the way that 10 people are competing and one of them must win.2012-06-15
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    Depends on the situation. If there are $10$ people and one is picked at random to get the prize, the answer is $1$. But if you assume **independence** $\dots$.2012-06-15
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    Daniel: So do you have 10 people or 1000 people?2012-06-15
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    there are 990 total people registered within the competition, and you have the remaining 10 "tickets"2012-06-15
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    Please clarify in your question (1) how many people are we talking about (2) how many tickets does each person have (3) how many prizes.2012-06-15
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    each person has one ticket, except for one individual who has ten tickets. Total number of tickets is 1000. There are 100 prizes to be won. Chosen randomly2012-06-15
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    So there are 991 people?2012-06-15
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    technically yes... the real question is if I hold 10 of the 1000 tickets... What are my odds of winning at least one of the 100 prizes2012-06-15
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    @Daniel: In that case it seems Martin's answer is what you are looking for.2012-06-15
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    10% is the chance that any one of the 1000 wins a prize because there are 100 prizes that everyone has a chance to win (assuming 1000 people with 1 ticket each). But how are the numbers drawn to determine prize winners? (1) Sample with replacement in which case the draws are independent but the probability of winning each time decreases because there are fewer prizes to win or (2) Sample without replacement in which case the successive draws are dependent since the person once drawn no longer has a chance of selection.2012-06-15

2 Answers 2

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Now for the edited version of question the probability is $$p=1-\frac{\binom{990}{100}}{\binom{1000}{100}}=1-\frac{990\cdot989\cdots891}{1000\cdot999\cdots901}=1-\frac{900\cdot899\cdots891}{1000\cdot999\cdots991}.$$ The binomial coefficient $\binom{990}{100}$ is the number of possible ways to choose 10 tickets out of 990, that are not yours. The number $\binom{1000}{100}$ is the number of all possibilities.

According to WolframAlpha this is approximately 65%.

(This is the probability that you win at least one of the 100 prizes.)

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    Thank you! Now what about if I only had 3 tickets?2012-06-15
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    If you had 997 out of 10000 tickets - try to use 997 instead of 990.2012-06-15
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    @MartinSleziak You mean that what you calculated is the probability that a person with 10 tickets will win at least one of the one hundred prizes. Then wouldn't the other 990 ticket holders have about a 6.5% chance of winning since you have 10 times their chance?2012-06-15
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    @MartinSleziak There is a typo in your formula. You wrote it correctly the first time. But then you repeated in and put 991 where you meant to have 901.2012-06-15
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    @MichaelChernick The second formula has only 10 terms in numerator and 10 terms in denominator. It gives the same result, see [here](http://www.wolframalpha.com/input/?i=1-%28900*899*898*897*896*895*894*893*892*891%29%2F%281000*999*998*997*996*995*994*993*992*991%29).2012-06-15
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    @MartinSleziak Oh I see there were many terms in numerator and denominator that cancelled from the first formula going to the second. But you meant to say "997 out of 1000 tickets" above and not "997 out of 10000 tickets."2012-06-15
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If you have only three tickets the same argument Martin gave would apply but with 997 replacing 990 in the formula. With 10 tickets and you having a 65% chance to win at least 1 prize the other 990 would each have a 6.5% chance of winning. Say the answer for 3 tickets is X% (clearly X is a lot less than 65) then the other 997 would have an X/3% chance to win at least one prize.