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Tough and tricky 3: leap year

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Tough and tricky 3: leap year [#permalink] New post 11 Oct 2009, 16:25
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A
B
C
D
E

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(N/A)

Question Stats:

38% (02:08) correct 63% (01:16) wrong based on 16 sessions
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Re: Tough and tricky 3: leap year [#permalink] New post 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

Therefore the answer is 3.
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Re: Tough and tricky 3: leap year [#permalink] New post 12 Oct 2009, 08:39
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Re: Tough and tricky 3: leap year [#permalink] New post 03 May 2011, 22:54
3 it is.Has to be an odd number of people. 1 and 2 are ruled out straightaway.
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Re: Tough and tricky 3: leap year [#permalink] New post 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%
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Re: Tough and tricky 3: leap year [#permalink] New post 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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Re: Tough and tricky 3: leap year   [#permalink] 14 May 2011, 10:58
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