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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, 17:25
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Question Stats:

16% (03:56) correct 83% (01:16) wrong based on 1 sessions
Tough and tricky 3:

How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5
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Re: Tough and tricky 3: leap year [#permalink] New post 11 Oct 2009, 17:51
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, 09:39
Set of Tough & tricky questions is now combined in one thread: tough-tricky-set-of-problms-85211.html

You can continue discussions and see the solutions there.
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Re: Tough and tricky 3: leap year [#permalink] New post 03 May 2011, 23:54
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, 10:18
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, 11:58
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, 11:58
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