# Variation on Birthday Problem

1 vote
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So, I have read the birthday paradox problem, and now I came across below question:

Assuming the following: there are no leap years, all years have $n = 365$ days and that people’s birthdays
are uniformly distributed across the $n$ days of the year.

(i) How many people must be there in a room before the probability that someone else has the same birthday as you do is at least $0.5$?

(ii) How many people must there be before the probability that at least two people have a birthday on 11th November is greater than $0.5$?

I think the answer to both of the parts must be $23$ people, as we are interested generally in only finding at least $2$ sets of people from $m$ people who have same birthday and then giving this probability.

Any corrections/suggestions/improvements on this is really welcomed. Please let me know whether I am correct?

Well, for both parts, let's say I have birthday on 11th Nov. Now if I want all people to have distinct birthdays other than that of mine so they can choose days from 364.

So, required probability model is :

$1-\frac{^{364}P_n}{365^n}$

For $n=23$, this works out to be 0.53 and Yes it seems to me I am done.

Please correct me If I am wrong.
0
$(i)$ $0.5<\binom{n}{1}\left ( \frac{1}{365}.\left ( \frac{364}{365} \right )^{n-1} \right )$

$(ii)$ $0.5<1-[\binom{n}{0}\left ( \frac{364}{365} \right )^{n} +\binom{n}{1}\left (\frac{364}{365}\right )^{n-1}]$
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@srestha

Can you explain your approach to the first part.? What have you selected.

Is $\binom{n}{2}\ast \frac{1}{365} > \frac{1}{2}$ incorrect?

0

@imShreyas

How are u getting this?

0

@srestha
There are n elements and m slots. Here, m = 365, Select any 2 people out of n people such that the probability of that they have same birthday is gretater than 1/2. probability of selecting the same date for 2 people I took 1/m.

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