math
posted by Mackenzie
suppose a group of n people is randomly selected. For each value v of n, find the probability that everyone in the group has a differnet birthda;y. (assume no one is born on february 29 of a leap year, so there are 365 equally likely burthdays possible.)

Count Iblis
Denote the probability by P(n). Clearly:
P(n) = 0 for all n > 365.
For n smaller than 365, we can argue as follows. Since everyone's birthday is assumed to be random, selecting n people and comparing the birthdays is equivalent to randomly selecting a string of n integers between 1 and 365.
The total number of ways one can select a string of n integers between 1 and 365 is:
365^n
These will include strings containing integers that are all different and strings in which some integers occure more than once. All these strings are equally likely, so they all have a probability of 1/365^n.
The total number of strings containing numbers that are all different is:
365!/(365  n)!
you can also write this as:
365*(365  1)*(365  2)*...*(365  n +1)
For the first integer you have 365 possibilities, for the next integer there are 365  1 possibilites left etc. etc.
Since each of these strings has a probability of 1/365^n, the total probability of selecting a string in which non of the integers are repeated is:
P(n) = 365!/(365  n)! 365^(n) 
Damon
If there are two people, what is the probability that the second will have a different birthday?
That would be 364/365
Now if there are three people, what is the probability that the third has a different birthday?
That would be 363/365
so the probability of all three having different birthdays is
364 * 363/365^2
etc
so I get for n people
p no coincidence = { 364!/(365n)!} /365^(n1)
so for example for 26 people:
p no coincidence = (364!/339!)/365^25
Do this by doing on our calculator
364/365*363/365*362/365 etc to 340/365
I got about .403
Which means that in a class with 26 people, there is about a 60% probability that at least two will have the same birthday. 
Damon
Whew.
It turns out the Count and I gave you the same equation, but he did it much more neatly.
he wrote
P(n) = 365!/(365  n)! 365^(n)
I wrote
p no coincidence = { 364!/(365n)!} /365^(n1)
which can be written
(365!/365) /(365n)! / 365^(n1)
which is
(365! / (365n)!) / 365^n
Respond to this Question
Similar Questions

math
My study group is lost on this one also.. Suppose you are managing 25 employees, and you need to form three teams to work on three different projects. Assume that all employees will work on exactly one team. Also, each employee has … 
statistics
Data shows that 88% of the people in a certain population are righthanded. A group of 7 people from this population are selected at random. (a) What would be the expected value for the number of righthanded people in the group? 
math
a political discussion group consists of five Democrats and six Republicans. Four people are selected to attend a conference. A.in how many ways can four people be selected from this group of eleven? 
Statistics
The home states of a certain group of people are distributed as follows: 53 percent are from MISSOURI, 21 percent are from KANSAS, and 26 percent are from IOWA. (No one in the group had a home state other than one of these three.) … 
STATISTICS
From a group of 12 people, you randomly select 3 of them. Find the probability that at least one of them is among the 3 oldest people in the group 
STATISTICS
From a group of 12 people, you randomly select 3 of them. Find the probability that at least one of them is among the 3 oldest people in the group 
statistics
Suppose that in a large population the proportion of people that have a certain disease is 0.01. Use the Poisson approximation to find the probability that at least four people will have the disease in a group of 400 randomly selected … 
Math
200 elementary school students were surveyed to find their blood groups. Of those, 55 were of blood group A, 85 of blood group B, and 21 of blood group O. What is the probability that a randomly chosen student has either blood group … 
Statistics
The home states of a certain group of people are distributed as follows: 54 percent are from MISSOURI, 27 percent are from KANSAS, and 19 percent are from IOWA. (No one in the group had a home state other than one of these three.) … 
Statistics
A youth group has 7 boys and 5 girls. Experiment is select 5 people at random from the group. Let X be the number of boys in the selected 5 people. (a) Find the probability of exactly 1 boy, i.e., P([X = 1]). (b) Find the probability …