# Another prob and stat expected value of x

1. Jan 2, 2009

### Proggy99

1. The problem statement, all variables and given/known data
Let X be the number of different birthdays among four persons selected randomly. Find E(X).

2. Relevant equations

3. The attempt at a solution
I am not completely sure how to start this problem. I was thinking of trying to figure out the total number of combinations of four birthdays over 365 days broken into totals showing number of combinations with 1, 2, 3 and 4 unique birthdays. Then divide those totals by the grand total and multiply each by 1, 2, 3 and 4 respectively. But this seemed to blow up. Am I approaching this with the right method or is there a better way?

2. Jan 2, 2009

Could you be more specific about the random variable X or is this the whole text of the problem?

3. Jan 2, 2009

### Proggy99

heh, yeah, that is the entire text of the problem. I keep trying to work it out like this:
x4/y * 4 + x3/y * 3 + x2/y * 2 + x1/y * 1 = answer.
Where x4 = number of combinations where i have four different birthdays, ie none of the four are the same.
x3 = three different birthdays where two of the people have the same birthday and the other two have different.
x2 = two birthdays where 3 have the same or two pairs have the same.
x1 = all four have the same birthday
y = number of different combinations, in this case I believe it would be 365!/361!

I am still trying to work it out in this method.

4. Jan 2, 2009

Hm, the problem is that I don't see how we would define the domain of this random variable X. The domain would be the events, for example, one of it is "we have randomly selected 4 persons with 4 different birthday dates". The image would be the probabilities of specific selections (i.e. events). I'm not quite sure what you were trying to say in your previous post.

5. Jan 2, 2009

Gee, am I stupid.

Simply take {0, 1, 2, 4} for the domain of X, where the elements of the domain denote the number of persons with different birthdays. Now all you have to do is calculate the probability of each of these events, and your random variable X is completely defined.

6. Jan 2, 2009

### Proggy99

I think that is what I am trying to do with this:
x4/y * 4 + x3/y * 3 + x2/y * 2 + x1/y * 1 = answer

where x4/y is the probability that I will have four different birthdays multiplied by 4, the number of birthdays. I just have not figured out how to figure out x4/y

7. Jan 2, 2009

### Proggy99

so the way I am doing this is:
365^4 = number of ways that 4 birthdays can be distributed amongst 365 days
365!/(365-4)! = number of permutations of 4 different birthdays over 365 days
365!/(365-3)! = number of permutations of 3 different birthdays over 365 days
365!/(365-2)! = number of permutations of 2 different birthdays over 365 days
365!/(365-1)! = number of permutations of 1 different birthdays over 365 days
therefore,
365!/(365-4)! / 365^4 * 4 = 3.93
365!/(365-3)! / 365^4 * 3 = 0.01
365!/(365-2)! / 365^4 * 2 = 0.00
365!/(365-1)! / 365^4 * 1 = 0.00
3.93 + .01 + 0 + 0 = 3.94

The correct answer is 3.98, so not sure yet where I am going wrong.

8. Jan 2, 2009

Interesting, my answer is 3.94, although I did it differently.

But first of all, why do you multiply by 4, 3, 2, 1, and not by 0, 1, 2, 4, since this represents the number of different birthdays:

0 - no different bds, they are all the same
1 - 1 different one, 3 same
2 - 2 vs 2
4 - all different

Further on, for example, to calculate the probability of case 4, I used p(4) = 364/365 * 363/365 * 362/365 = ... and so on.

Perhaps I'm missing something here.

9. Jan 2, 2009

### Proggy99

I am reading the problem as asking for the probability of different birthdays.
If they all have the same birthday, then there is 1 birthday represented.
If three have the same and the fourth has a different birthday, or if two have one and two have another, then there are two different birthdays.
If two have the same but the other two have different, then that would be three different birthdays.
If they all have different birthdays, then that would be four different birthdays.
Thus I am looking at 1, 2, 3 or 4 represented birthdays.

I am in no way positive that I am approaching this correctly. I did double check and the book's answer is definitely 3.98.

10. Jan 2, 2009

OK, the 1 2 3 4 thing makes perfect sense now, and it seems a correct way of setting the problem. But I don't see why you calculated the numbers of permutations, since you need the probabilities of every of these cases.

11. Jan 2, 2009

### Proggy99

our formulas are the same.
Yours is
while mine is
which is the same as 365/365 * 364/365 * 363/365 * 362/365

Then I take the probability of having 4 different birthdays and multiply by 4, then the prob of 3 birthdays by 3, etc. But I still land on 3.94 and not 3.98

12. Jan 2, 2009

### carlodelmundo

Just curious: Is this an AP Stats Question? What course are you taking that's asking such a q?

13. Jan 2, 2009

### Proggy99

it is math 3355 Probability thru LSU's online independent study course offerings. It is the final course I have to take to finish out my math requirement for my 7-12 math teaching license. It comes after linear algebra, number theory and abstract algebra, though technically only Multidimensional Calculus is listed as a prereq.

14. Jan 5, 2009

### Proggy99

anyone care to tell me what I am doing wrong with this final answer? I get 3.94, but the book says 3.98.

The problem statement is:
Let X be the number of different birthdays among four persons selected randomly. Find E(X).

my last try at a solution:
365^4 = number of ways that 4 birthdays can be distributed amongst 365 days
365!/(365-4)! =permutations of 4 different birthdays in 365 days
365!/(365-3)! =permutations of 3 different birthdays in 365 days
365!/(365-2)! =permutations of 2 different birthdays in 365 days
365!/(365-1)! =permutations of 1 different birthdays in 365 days
365!/(365-4)! / 365^4 * 4 = 3.93
365!/(365-3)! / 365^4 * 3 = 0.01
365!/(365-2)! / 365^4 * 2 = 0.00
365!/(365-1)! / 365^4 * 1 = 0.00
E(X) = 3.93 + .01 + 0 + 0 = 3.94

Thanks for any help. Turning this in tonight as is if no one can give me any further tips.

15. Jan 5, 2009

### Sisplat

Perhaps the solution takes into account the leap year?

16. Jan 5, 2009

### ko12sd

Hmm... When I saw this I thought maybe it's a binomial problem, but I forgot the conditions since last year's AP exam, so I looked the up and here they are:
1: The number of observations n is fixed.
2: Each observation is independent.
3: Each observation represents one of two outcomes ("success" or "failure").
4: The probability of "success" p is the same for each outcome.
Where, n = 4 because there are 4 people, the observations are independent because one outcome doesn't effect the next, success = different birthdays & failure = same birthdays.
p = (364/365) And P(x=k) = nCk*p^k*(1-p)^(n-k) where nCk is n choose k.

x P(x) x*P(x)
1 ≈8.2E-8 ≈8.2E-8
2 ≈4.5E-5 ≈9.0E-5
3 ≈0.011 ≈0.033
4 ≈0.989 ≈3.956
E(x) = Sum(x*P(x)) ≈3.989 round to evens so E(x) = 3.98

Not sure if that is the right method, but that's how I've done it in the past. I'm new here, is it all right to show that work given that the answer is already known... I'll cover with spoiler just in case. Hopefully it looks okay. What do you think?

17. Jan 5, 2009

### ko12sd

okay, so the spoiler thing kinda screws it up a bit. The point is it came out right as a binomial. :)

18. Jan 7, 2009

### Proggy99

hmmmm, I replied to this a couple days ago and I do not know where that post went. Anyway, in summary, sisplat you are right in that I should have been doing 366 days instead of 365. I was thinking I could not take the factorial of 365 1/4 days, but I would not do that because there are 366 possible birth dates to consider. Using 366 instead of 365 causes my method to reach the correct answer.

However, ko12sd, your method is much more elegant. I originally said I would have to study your answer a bit but then I started on the next chapter and it is on binomials and your method became clear. So either method works, I just had not reached binomials in the book yet.