Question

What is the probability that in a group of ‘N’ people , at least two of them have the same birthday.

Answer 1

Hint: At first , excluding leap years there are 365 different birthdays possible , so in a group of N people the person can have birthdays in ${\left( {365} \right)^N}$different ways. This can be denoted by n . Next now we can calculate the number of different ways that no people have the same birthday. We get that the number of different ways is .$m = 365*364*363*...*(365 - N + 1)$.

Complete step-by-step answer:
Step 1 :
We are asked to find the probability that in a group of ‘N’ people , at least two of them have the same birthday.
Firstly , let's find the probability that no two persons have the same birthday and subtract it from 1 as the total probability of a success event is 1 .
Step 2:
Excluding leap years there are 365 different birthdays possible
So , any person can have any one of the 365 days of the year as a birthday
Same way the second person may also have any one of the 365 days of the year as a birthday and so on.
Hence in a group of N people , there are ${\left( {365} \right)^N}$possible combination of birthdays
So now let n = ${\left( {365} \right)^N}$
Step 3 :
Now assuming that no two people have their birthday on the same day
The first person can have any one of the 365 days as his birthday
So the second person will have his birthday in any one of the 364 days
And the third person will have his birthday in any one of the 363 days and so on
From this , we can get that the Nth person may have his birthday in any one of the ( 365 – N +1 ) days
The number of ways that all N people can have different birthdays is then
$m = 365*364*363*...*(365 - N + 1)$
Therefore , the probability that no two birthdays coincide is given by $\dfrac{m}{n}$
$ \Rightarrow \dfrac{m}{n} = \dfrac{{365*364*363*...*(365 - N + 1)}}{{{{\left( {365} \right)}^N}}}$
Step 4 :
Probability that at least two person will have the same birthday = 1 – (probability that no two birthdays coincide)
$
   \Rightarrow 1 - \dfrac{{365*364*363*...*(365 - N + 1)}}{{{{\left( {365} \right)}^N}}} \\
    \\
$
The above expression gives the probability that at least two people will have the same birthday.


Note: The assumptions that a year has 365 days and that all days are equally likely to be the birthday of a random individual are false, because one year in four has 366 days and because birth dates are not distributed uniformly throughout the year. Moreover, if one attempts to apply this result to an actual group of individuals, it is necessary to ask what it means for these to be “randomly selected.” It would naturally be unreasonable to apply it to a group known to contain twins.