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Re: Tough and tricky 3: leap year [#permalink]
11 Oct 2009, 16:51

The answer is C:3.

Essentially, you need to find the number of people so that the probability of all of them not being from a leap year is LESS THAN 50%. The probability of any one person not being born on a leap year is approximately 3/4.

So you are essentially solving for:

(\frac{3}{4})^n < \frac{1}{2}

n = 1: 3/4 < 1/2? NO n = 2: (3/4)(3/4) = 9 / 16 < 1/2? NO n = 3 (3/4)(3/4)(3/4) = 27 / 64 < 1/2? YES

Re: Tough and tricky 3: leap year [#permalink]
14 May 2011, 09:18

why should there be an odd number of people ? it is clear... if there is 1 person the probability is 1/4 , if there are 2 , the probability is 1/4 +1/4 = 0.5 or 50%. if there are 3, p(L) > 50% _________________

What is of supreme importance in war is to attack the enemy's strategy.

Re: Tough and tricky 3: leap year [#permalink]
14 May 2011, 10:58

garimavyas wrote:

why should there be an odd number of people ? it is clear... if there is 1 person the probability is 1/4 , if there are 2 , the probability is 1/4 +1/4 = 0.5 or 50%. if there are 3, p(L) > 50%

for two people - how can u add 1/4+1/4? may I know your logic _________________

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