# Help with probability question.

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. – David Mitra Jun 15 '12 at 17:36
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. – Martin Sleziak Jun 15 '12 at 17:42
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 $\dots$. – André Nicolas Jun 15 '12 at 17:42
Daniel: So do you have 10 people or 1000 people? – Thomas Jun 15 '12 at 17:45
there are 990 total people registered within the competition, and you have the remaining 10 "tickets" – Daniel Jun 15 '12 at 17:47

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? – Daniel Jun 15 '12 at 17:56
If you had 997 out of 10000 tickets - try to use 997 instead of 990. – Martin Sleziak Jun 15 '12 at 18:05
@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? – Michael Chernick Jun 15 '12 at 18:08
@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. – Michael Chernick Jun 15 '12 at 18:16
@MichaelChernick The second formula has only 10 terms in numerator and 10 terms in denominator. It gives the same result, see here. – Martin Sleziak Jun 15 '12 at 18:35

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.

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