In a group of 3 people, the probability that at least two will have the same birthday is (ignoring leap year)

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364 × 363 × 362 3652

1 − 3643652

1 − 364 × 363 3652

364 × 363 3652

answer is C.

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Detailed Solution

Given, a group of 3 people.
We know that for this question, the number of ways that each individual can celebrate his or her birthday is 365 ways.
Mention the number of ways, as mentioned in the group, 3 people can have a birthday in

(365) 3

days.
So, the number of ways in which all the 3 birthdays will be celebrated is

^{365} P_{3}^{}

ways
Hence, we can write that,
P(A)=1-P(A)
=1-

\frac{^{365} P_{3}^{}}{{(365)}^{3}}

=1-

\frac{(365)!}{{(365)}^{3} (365 - 3)!}

This gives us the answer as

1 − 364 × 363 3652

.
In a group of 3 people, the probability that at least two will have the same birthday is

1 − 364 × 363 3652

.
Hence, the correct option is 3.

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