Question

In a hall full of 90 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

Why the answer is A: 75

This makes sense because there's a 1/365 chance that the second person has the same birthday as the first person. We can re-word this and say that the probability that one pair of people have the same birthday is 364/365.

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Answer 2

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

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

Since, we have 90 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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90th person can have birthday on any one day out of the 365 days.

So, Total possible outcomes in sample space S are

$\rm \: n(S) = {(365)}^{90}$

Now, Let's 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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90th person can have birthday on any one day out of the 276 days.

So, Total favourable outcomes are

$\rm \: n(A) = 365 \times 364 \times 363 \times 362 \times \cdots \cdots \times 276$

can be further rewritten as

$\rm \:= \dfrac{365 \times 364 \times 363 \times 362 \times \cdots \cdots \times 276 \times 275!}{275!}$

$\rm \:= \dfrac{365 \times 364 \times 363 \times 362 \times \cdots \cdots \times 276 \times 275!}{(365 - 90)!}$

$\rm \: = \: ^{365}P_{90}$

$\red{\rm\implies \:\boxed{ \rm{ \:n(A) = \: ^{365}P_{90} \: }}}$

So, the probability of the event A that no two people have same birthday is

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

$\red{\rm\implies \:\rm \: P(A) = \dfrac{^{365}P_{90}}{ {(365)}^{90} } \: }$

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

$\rm \: = \: P(A')$

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

$\rm \: = \: 1 \: - \: \dfrac{^{365}P_{90}}{ {(365)}^{90} }$