Question

What is the probability that at least two out of n people have the same birthday (assume 365 days in a year and that all days are equally likely)?

Answer 1

No. of days = 365
Day of birth = 1
P(both have same birthday) = 1/365

Answer 2

Answer:

The required probabilty is:  $1- \frac{365\times364\times363\times...(365-n+1)}{365^n}}$

Step-by-step explanation:

As per the data given in the question,

We know,

In a group of $n$ people they can have birthdays in =Q=$365^{n}$ ways

For no two people to have birthdays on same day:

The first person can have b'day in any one of the 365 days

So, second can have b'day in any one of the 364 days

And, third can have b'day in any one of the 363 days

Hence, for $n$ people. No. of ways are =P= $365\times364\times363\times...(365-n+1)$

Therefore, the probability that no two birthdays coincide is given by = $\frac{P}{Q}$

$\frac{P}{Q}=\frac{365\times364\times363\times...(365-n+1)}{365^n}}$

Hence,

Probability that at least two person will have the same birthday  = 1 - (the probability that no two birthdays coincide)

$= 1 - \frac{P}{Q} \\=1- \frac{365\times364\times363\times...(365-n+1)}{365^n}}$

So,

The required probabilty is:  $1- \frac{365\times364\times363\times...(365-n+1)}{365^n}}$

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