Question

This question is also known as the “Birthday Problem.”

In a room full of 50 people, what is the probability that at least two people have the same birthday? Assume that all birthdays are equally likely (uniform distribution) and there are 365 days in the year.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that E be the event that no two people have the same birthday.

Since, we have 50 people in a room.

So,

Ist person can have birthday on any one day out of the 365 days.

2nd person can have birthday on any one day out of the 365 days.

3rd person can have birthday on any one day out of the 365 days.

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50th person can have birthday on any one day out of the 365 days.

So, Total possible outcomes in sample space S are

$\rm \: n(S) = {(365)}^{50}$

Now, we have to find out the probability that no two persons have the same birthday

So,

Ist person can have birthday on any one day out of the 365 days.

2nd person can have birthday on any one day out of the remaining 364 days.

3rd person can have birthday on any one day out of the remaining 363 days.

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50th person can have birthday on any one day out of the 316 days.

So, Total favourable outcomes are

$\rm \: n(E) = 365 \times 364 \times 363 \times 362 \times \cdots \cdots \times 316$

can be rewritten as

$\rm \: n(E) = \dfrac{365 \times 364 \times 363 \times 362 \times \cdots \times 316 \times 315!}{315!}$

$\rm \: n(E) = \dfrac{365 \times 364 \times 363 \times 362 \times \cdots \times 316 \times 315!}{(365 - 50)!}$

$\rm\implies \:n(E) \: = \: ^{365}P_{50}$

So, Probability of the event E that no two people have the same birthday is

$\rm \: P(E) = \dfrac{n(E)}{n(S)}$

$\rm \: \rm\implies \:\boxed{ \rm{ \:P(E) = \frac{\: ^{365}P_{50}}{ {(365)}^{50} } \: }}$

So, required probability that atleast two people have same birthday is

$\rm \: P(E')$

$\rm \: = \: 1 - P(E)$

$\rm \: = \: 1 - \dfrac{\: ^{365}P_{50} \: }{ \: \: {(365)}^{50} \: \: }$