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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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    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

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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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    @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.