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A math class has 25 students. Assuming that all of the students were born in the first
half of the year—between January 1st and June 30th— what is the probability that at
least two students have the same birthday? Assume that nobody was born on leap day,
so there are 181 possible birthdays.

### 1 comment

$\frac{_{24}^{181}\textrm{C}._{1}^{24}\textrm{C}+_{23}^{181}\textrm{C}._{2}^{23}\textrm{C}+_{22}^{181}\textrm{C}._{3}^{22}\textrm{C}+.................}{_{25}^{181}\textrm{C}}$

## 1 Answer

P( at least 2 students sharing the same birthday)

= 1 - P( no 2 students having the same birthday)

= 1 - P( all students having distinct birthdays)

So now, P( all students having distinct birthdays) =>

The first student can have any birthday out of 181 possible days => 181/181

If the second student is not supposed to have the same birthday as first, then he is left with only (181 - 1) choices => 180/181

Similarly, if the third student is not supposed to share a birthday with either of the previous two students, then he is left with (181 - 2) choices  => 179/181

Similarly, we can find the individual probabilities for a distinct birthday for each student as:

181/181, 180/181, 179/181, 178/181, ............................., 158/181, 157/181         ( for 25 terms)

=> P( all students having distinct birthdays) =

181/181 * 180/181 * 179/181 * 178/181 * .............................* 158/181 * 157/181

= (181!/(181 - 25)!)/(181^25) = 0.1759

Thus, P( at least 2 students sharing the same birthday) = 1 - 0.1759 = 0.8241

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