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Difficult Math Question #3

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Master | Next Rank: 500 Posts
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by 800guy » Mon Aug 28, 2006 7:26 pm
OA will come after some people answer...

How many randomly assembled people 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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by aim-wsc » Wed Aug 30, 2006 5:02 am
ok. here comes a hard nut.

let me read it again.

2 is it.
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by aim-wsc » Wed Aug 30, 2006 10:46 pm
i m not sure whether i hv grasped the question completely.


can u tell me?
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by 800guy » Thu Aug 31, 2006 6:16 pm
OA:

Soln: Prob. of a randomly selected person to have NOT been born in a leap yr = 3/4
Take 2 people, probability that none of them was born in a leap = 3/4*3/4 = 9/16. The probability at least one born in leap = 1- 9/16 = 7/16 < 0.5
Take 3 people, probability that none born in leap year = 3/4*3/4*3/4 = 27/64.
The probability that at least one born = 1 - 27/64 = 37/64 > 0.5
Thus min 3 people are needed.
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by aim-wsc » Thu Aug 31, 2006 7:34 pm
now that was really tough one :x

nice work there :)
800guy.
but why arent you involved in this?

if you find math difficult then why dont you share your doubts & problems with us...may be ppl could provide better explanation.

for example in this given: i had a problem.
dear members,
when i read the question i conclude that it is impossible to get the probablity more than 50%.,,,, since the probalility of getting a person borned in leap year is 1/4 ie 25%....so it never could go beyond that.

but reading the solution i got the right method to approach such nuts...
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by narfa17 » Wed Oct 06, 2010 5:36 am
Another way of solving:

1. P(a) for one person = 1/4

2. P(a) for two persons P(A+B) = 1/4 + 1/4 - 1/4*1/4 = 8/16 - 1/16 = 7/16 < 0,5 ( A+B - AB, cause A or B, two independent nonexclusive)

3. P(a) for 3 persons = P(A+B+C) = 7/16 +1/4 - 7/16*1/4 = 44/64 - 7/64 = 37/64 > 0,5

Answer: 3 persons
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