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?
Saturday, November 24, 2007
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3 Comments:
According to me the answer is 3.
The method applied to by me is like this, please confirm the same.
To raise the probability above 50% what we can do is find a complement case.
Now, probability of a person to be born in a non-leap year is 3/4 (Almost, it is slightly lesses, but 3/4 works out fine for the sum).
Now,
for randomly selected people, these are independent events and hence the probability can be multiplied, now we are to find the number of people such that the probability goes below 0.5
Hence,
Num = (3/4)*(3/4)... n times such that n is the smallest number to give Num < 0.5
For n = 3,
Num = (3/4)*(3/4)*(3/4) = 0.421875
Hence, probability of atleast one of them to be born in a leap year becomes (1 - 0.421875) i.e. > 0.50.
good one and real good way of solving..
;) same as i did it...
another 1 for you
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