help with this problem appreciated

Not a very difficult one, once you understand what it's all about.

Look at it this way: say each student is born in a separate month -- one in January, one in February... and so on. You get that 12 students could have birthdays in separate months. But you have 27 students, which is obviously greater than 12. This means that at least two students share months of birth, since 27/12 = 2 remainder 3. That covers 24 students: you've concluded that at least two students have the same month of birth. But since you have three students left, then they also have to be distributed in the 12 months of the year. This means that three months of the year will have at least 3 students, which makes the probability you're looking for 1.

Keep in mind that the case outlined above is a worse-case scenario, when all the students are spread as fairly as possible all over the calendar, so to speak. You could also have a case when all students are born in the same month, or 26 students are born in January and one student is born is February. No matter what, the result will be the same: at least 3 students will have to share one month of birth.

DanaJ wrote:Not a very difficult one, once you understand what it's all about.

Look at it this way: say each student is born in a separate month -- one in January, one in February... and so on. You get that 12 students could have birthdays in separate months. But you have 27 students, which is obviously greater than 12. This means that at least two students share months of birth, since 27/12 = 2 remainder 3. That covers 24 students: you've concluded that at least two students have the same month of birth. But since you have three students left, then they also have to be distributed in the 12 months of the year. This means that three months of the year will have at least 3 students, which makes the probability you're looking for 1.

Keep in mind that the case outlined above is a worse-case scenario, when all the students are spread as fairly as possible all over the calendar, so to speak. You could also have a case when all students are born in the same month, or 26 students are born in January and one student is born is February. No matter what, the result will be the same: at least 3 students will have to share one month of birth.

Cool approach DanaJ.. Nice stuff..

DanaJ wrote:Not a very difficult one, once you understand what it's all about.

Look at it this way: say each student is born in a separate month -- one in January, one in February... and so on. You get that 12 students could have birthdays in separate months. But you have 27 students, which is obviously greater than 12. This means that at least two students share months of birth, since 27/12 = 2 remainder 3. That covers 24 students: you've concluded that at least two students have the same month of birth. But since you have three students left, then they also have to be distributed in the 12 months of the year. This means that three months of the year will have at least 3 students, which makes the probability you're looking for 1.

Keep in mind that the case outlined above is a worse-case scenario, when all the students are spread as fairly as possible all over the calendar, so to speak. You could also have a case when all students are born in the same month, or 26 students are born in January and one student is born is February. No matter what, the result will be the same: at least 3 students will have to share one month of birth.

Indeed great explanation thank you so much thought about this one whole day:) was trgyin to calculate it with formulas etc:) now it's all clear.