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Mathematics 15 Online
OpenStudy (baseballer2014):

In a room full of 30 people, what is the probability that at least 2 of them have the same birthday?

OpenStudy (baseballer2014):

@ganeshie8 @dan815

ganeshie8 (ganeshie8):

day as in `monday, tuesday,... ` ?

OpenStudy (baseballer2014):

as in day of the month

OpenStudy (dan815):

30 people where atleast 2 out of the 365 numbers common

OpenStudy (baseballer2014):

what dan said

OpenStudy (dan815):

hold i gota move my laptop first its over heating

ganeshie8 (ganeshie8):

so 1/1/1990 and 1/1/2000 are considered same birth days ?

OpenStudy (dan815):

yep

OpenStudy (anonymous):

I can tell that it's more than 50% because I only take 23 people for that to happen

OpenStudy (anonymous):

if you search birthday paradox you see how it's calculated. The idea is just to do 1 - P(no one share a same birthday)

OpenStudy (dan815):

it looks like a cool question to think about

OpenStudy (baseballer2014):

the goal is to come to the answer without the internet

OpenStudy (dan815):

so you're saying find the prob that none of them share brithdays

OpenStudy (anonymous):

yes, it's just 1 - (365/365) * (364/365) * (363/364) * .. all the way to the 30th person

OpenStudy (anonymous):

typo 1 - (365/365) * (364/365) * (363/365) * .. all the way to the 30th person

ganeshie8 (ganeshie8):

\[\sum \limits_{k=2}^{30}\binom{30}{k} \left(\dfrac{1}{365}\right)^k \left(\dfrac{364}{365}\right)^{30-k}\]

OpenStudy (dan815):

i was thinking it was like that too but is there a closed form for this ganeshie

OpenStudy (dan815):

i feel like theres a nice function or something that gives the sum of that

ganeshie8 (ganeshie8):

its a binomial distribution right ? p = 1/365 n = 30 k >= 2

OpenStudy (dan815):

yeah

OpenStudy (dan815):

true okay i get it!

OpenStudy (dan815):

we can integrate the normal distribution curve

ganeshie8 (ganeshie8):

yes thats the only simple way i guess ! we can approximate it by converting it to normal curve

OpenStudy (dan815):

ok cool

OpenStudy (dan815):

@geerky42 are you @usukidoll?

OpenStudy (baseballer2014):

@dan815 you could also go back to a permutation by selecting 30 birthdays out of the 365 and assign them to the people which is 365p30 and then divide it by the total possible ways of assigning a birthday to someone?

OpenStudy (baseballer2014):

that of course is the compliment event where everyone has a different birthday but we can subtract it from 1?

OpenStudy (anonymous):

if no one share a same birthday, the complement of that is at least two people share a same birthday. So we subtract from 1. If fact, the formula is Pr = 1 - (365Pn) / 365^n in the case of 30, Pr = 70.6%

ganeshie8 (ganeshie8):

Yup ^

OpenStudy (baseballer2014):

yes exactly i am curious as to how @ganeshie8 and @dan815 were going to solve it

ganeshie8 (ganeshie8):

Forget it, its not a direct binomial distribution - need to tweak some more >.<

geerky42 (geerky42):

No, I am not? @dan815

OpenStudy (usukidoll):

@dan815 I AM OVER HERE s t h u

OpenStudy (dan815):

@usukidoll :O

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