SOLUTION: This is known as the Birthday Problem. (a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (For simplic

Algebra.Com
Question 1051953: This is known as the Birthday Problem.
(a) Consider a class with 30 students. Compute the probability that at least two of them
have their birthdays on the same day. (For simplicity, ignore the leap year).
(b) How many students should be in class in order to have this probability above 0.5?

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

The complement event would be for all 30 students to have
all different birthdays.  There are 365P30 ways to assign
them all different birthdays and 36530 ways to assign
then any birthdays.  So the probability of all 30 having
different birthdays is







So the probability that at least 2 students have the same birthday is
that number subtracted from 1, which is:



or about a 71% probability that at least 2 students among the 30
have the same birthday.

Edwin

RELATED QUESTIONS

Suppose that three students are selected at random from your class. What is the... (answered by swincher4391)
What is the probability that there are at least two people with the same birthday in a... (answered by Shin123,math_helper)
Please help me solve this problem The STAT 230 class has 34 students. What is the... (answered by dabanfield)
Calculate the probability that in a class of 40 students, at least 2 will share the same... (answered by rothauserc)
What is the probability that at least one out of 40 students in a class will have a... (answered by stanbon)
I'd appreciate some help with this problem. It is too complex for me to even begin to... (answered by stanbon)
In a room full of 30 people, what is the probability that at least two people have the... (answered by vleith)
Determine the probability that in a class of 8 students, at least two students have the... (answered by stanbon)
Suppose that you are in a class of 33 students and it is assumed that approximately 17%... (answered by stanbon)