Problem Solving

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?

ok. here comes a hard nut.

let me read it again.

2 is it.

i m not sure whether i hv grasped the question completely.


can u tell me?

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.

now that was really tough one

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...

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