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ps-leap year

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ps-leap year [#permalink] New post 30 May 2005, 14:54
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How many randomly assembled ppl do u need to have a better than 50% prob. that at least 1 of them was born in a leap year ?
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Re: ps-leap year [#permalink] New post 30 May 2005, 15:14
christoph wrote:
How many randomly assembled ppl do u need to have a better than 50% prob. that at least 1 of them was born in a leap year ?


prob of being born in a leap year is 1/4 and not being born in a leap year is 3/4
(note: if considering days of birth, than probabilities will be slightly different because leap year has 366 days, so prob of being born on a leap year is 366/(3*365+366) = approx 1/4 )

prob(none out of n people is born in a leap year) is (3/4)^n

so

prob(at least one of them is born in a leap year) = 1-(3/4)^n >1/2 => (3/4)^n < 1/2

minimum number of people is 3 (n=3).
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Re: ps-leap year [#permalink] New post 30 May 2005, 16:46
sparky wrote:
christoph wrote:
How many randomly assembled ppl do u need to have a better than 50% prob. that at least 1 of them was born in a leap year ?


prob of being born in a leap year is 1/4 and not being born in a leap year is 3/4
(note: if considering days of birth, than probabilities will be slightly different because leap year has 366 days, so prob of being born on a leap year is 366/(3*365+366) = approx 1/4 )

prob(none out of n people is born in a leap year) is (3/4)^n

so

prob(at least one of them is born in a leap year) = 1-(3/4)^n >1/2 => (3/4)^n < 1/2

minimum number of people is 3 (n=3).


Good one sparky :)

I couldnt get it :(
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 [#permalink] New post 30 May 2005, 23:14
Provided I solved it right, this is a standard setup for probability-number of people problems. They all use binominal distribution. I can see this as a potential GMAT problem.
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 [#permalink] New post 31 May 2005, 04:58
Sparky: That was a good way to solve this problem....
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 [#permalink] New post 31 May 2005, 17:21
heh... and now my simple way since I dont understand binomial distributions.

Leap years comes around once in every 4 years (1/4)

So some random person has a 1/4 chance of being born in a leap year.

2 people have (1/4 + 1/4) = 1/2 chance

3 peple have (1/4 + 1/4 +1/4) = 3/4 chance > 1/2

n= 3

??
  [#permalink] 31 May 2005, 17:21
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