1. probability

there are 30 days in june. 5 students have their birthdays in june. the birthdays are independent of each other and all dates are equally likely. whaat is the probability that at least 2 of the students have the same birthdays .

i am not sure really what to do. I tried this. total combinations of birthdays is 30 nCr 5 which is 142506.

2. Re: probability

The easiest solution is to find the total number of ways that no two people have the same birthday. $\dbinom{30}{5}$ is the number of ways to get a collection of five days. You need to multiply that by the number of ways to arrange the five days among the students (because students are unique). Then, divide by the total number of ways that five people can have a June birthday $30^5$. So, the probability that no two students have the same birthday is given by:

$\dfrac{30\cdot 29\cdot 28\cdot 27\cdot 26}{30^5}$

The probability that at least two people have the same birthday is one minus that:

$1- \dfrac{30\cdot 29\cdot 28 \cdot 27 \cdot 26}{30^5} \approx 29.63\%$

3. Re: probability

I am not sure where your answer (30*29*28*27*26)/30^5 is coming from ?
30nCr5 =142506.
I also do not understand what this is " multiply that by the number of ways to arrange the five days among the students (because students are unique)"

4. Re: probability

How I would do it: put the students in a line. The first student has some day in June as his birthday. The probability of that is 30/30= 1. If none have the same birthday, the second student must have one of the other 29 days as his birthday. There are 29 ways that can happen so the probability that will happen is 29/30. The third student must have one of the other 28 days as birthday, the fourth 27, and the fifth 26. That is where the "(30*29*28*27*26)/30^5= (30/30)(29/30)(28/30)(27/30)(26/30) comes from.

But that is for that particular order of the students in line. There are 5!= 5*4*3*2*1= 120 different orders of five people.

5. Re: probability

Thanks I get it now. The answer at the back of the book is
1111/3750 so I do not see why you have to take the order into account?

6. Re: probability

Originally Posted by markosheehan
Thanks I get it now. The answer at the back of the book is
1111/3750 so I do not see why you have to take the order into account?
Regardless of their order in line, they take their birthday with them. You do not want a collection of five birthdays in any order. You have five students with distinct birthdays. Order matters when they are distinct.