Question

In: Statistics and Probability

PROBLEM 6. Suppose that the birthdays of different people in a group of n people are...

PROBLEM 6. Suppose that the birthdays of different people in a group of n people are independent, each equally likely to be on the 365 possible days. (Pretend there's no such thing as a leap day.)What's the smallest n so that it's more likely than not that someone among the n people has the same birthday as you? (You're not part of this group.)

Solutions

Expert Solution

solution :

There are n people in the gathering whose birthday are similarly likely and autonomous

let An is an occasion that no individual has indistinguishable birthday from yours

P(A)= 1-P(Ac)

=1 - P(different birthday)

= 1 - {365364363.........(365-n+1)}/365n

presently we need to locate the littlest estimation of n. By Birthday Paradox the littlest estimation of n is 23 which isn't same as this issue.

assume n=2 i.e there is two individuals and none of their birthday same as yours

at that point Q2 =1 - (364/365)2 ~ .28

on the off chance that n=3 and none has same birthday as yours

Q3=1 - (365/365)(364/365)(363/365) ~ .55

et cetera...

on the off chance that there is n people and none has same birthday as yours

Qn= 1-(356/365)(364/365).....(365-n+1/365)

on the off chance that n is substantial then calculations end up harder so we utilize basic analytics

1 - x < e-x

there is different strategy to discover the estimation of n and one is

All together for the likelihood of no less than one other individual to share your birthday to surpass half, we require n sufficiently huge that

1 - (364/365) n-1 0.5

so we can discover the esteem n for which n > 253

n=253 is the most modest number for which your birthday will be same as others.


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