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Arjun
asked
in Probability
Jun 10, 2018

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1 vote

1

So for exactly 2 people to have birthday on same day we have to choose 2 person out of 11 => 11C2

Now for the first person selected probability of it’s birthday falling on any day = 1

and the probability of its birthday falling on same day = 1/365

For rest other case their birthday should fall on different days = (364/365)*(363/365)*..(356/365)

P(exactly 2) = 11C2 * 1 * (1/365) * (364/365) * (363/365) * ….(356/365) =0.1707 ?

Now for the first person selected probability of it’s birthday falling on any day = 1

and the probability of its birthday falling on same day = 1/365

For rest other case their birthday should fall on different days = (364/365)*(363/365)*..(356/365)

P(exactly 2) = 11C2 * 1 * (1/365) * (364/365) * (363/365) * ….(356/365) =0.1707 ?

0

5 votes

It is a famous probability problem called The Birthday Paradox.

It is used to compute that at least two people share the same birthday (ignore year) in some people in a room Or in a group of n people.

Now, I'm going to compute the probability in the following manner-

Probability of at least two people share the same birthday = $1- $ Probability of no people share the same birthday

Now, the first people can born in any day of the year (assuming it's a normal year not a leap year) = $\dfrac{365}{365}$

the second people can have $364$ possible birthdays, so the second people's probability is** **$\dfrac{364}{365}$

the third people can have $363$ possible birthdays, so the third people's probability is** **$\dfrac{363}{365}$

$\vdots $

$11^{th}$ number people can have $355$ (because $10^{th}$ number people have the possibility of $356$ birthdays) possible birthdays, & the probability will be $\dfrac{355}{365}$

∴ Probability of no two people have the same bday = $\dfrac{365}{365} \times \dfrac{364}{365} \times \dfrac{363}{365} \times \ldots \times \dfrac{355}{365}$

$\quad \quad \quad = \dfrac{365!}{(354! *(365^{11}))}$

∴ Probability of two people have same birthday $= 1-\dfrac{365!}{(354! *(365^{11}))} \\ = 0.1411 = 14.11\%$

0 votes

Q: In a group of 11 people what is the probability that **exactly **two people have birthday on same day?

We want exactly two people out of 11 people to share same birthday.

let’s ignore leap years.

First we need to chose 2 people out of 11 to share the same birthday.

That is **C(11,2) = 55**

No. of ways exactly two people have same birthday is

**55*365*364*….*356 = 55*365!/355!**

Probability is:

**(55*365!)/(355! * 365^11) = 0.132 = 13.2%**