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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.