GATE Suitability Test | Test 1 | Question: 5

Question

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

Comments

$0.1411$ or $14.11\%$

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https://www.khanacademy.org/math/precalculus/prob-comb/prob-combinatorics-precalc/v/birthday-probability-problem

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Is the answer 0.132?

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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 ?

Answer 1

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 $= 1-\dfrac{365!}{(354! *(365^{11}))}  \\ = 0.1411 = 14.11\%$

Comments

it's wrongly typed, please make the last term of the product series as 355

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anyway, you should consider 366 days but not 365 days.

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@Subarna there are 366 different birthday possible, right?

Answer 2

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%