# Thread: Probability Inclusion Exclusion

1. ## Probability Inclusion Exclusion

Hi,
Assume hat every time you buy a box of Corn Flakes, you receive one of the pictures of the k players of the Florida Marlins. Over a period of time you buy n >= k boxes of Corn Flakes. Let Ej, j=1,2,3,...,k, denote the event you do not get the jth player's picture.

What is the probability of Ej?
What is the probability of Ej intersect Ei (i is just another player)?

Thanks,

I think once i have those I can use the inclusion-exclusion principle to answer what is the even you do not get at least one picture.

2. I'm pretty sure that that
P(Ei) = ? / k^n

The sample space is k*k*k*k n-times.
But i don't see how to count not getting the jth player.

3. yes k^n different orderings off the boxes.

4. I'm not sure but does P(Ei) = (k-1)! / k^n make any sense?

5. There are k! ways of getting the jth player. So there are (n-k)! ways of not getting the jth player.

So I think P(E_i) = (n-k)!/k^n

6. Yeah i think that makes sense actually.
So the probability of not getting the ith or jth player would be
P(Ei int Ej) = (n-k-1)!/k^n?

7. yes that is correct I believe.

8. How would I get P(E1 union E2 union E3 union ... union En)? THis is the inclusion-exclusion principle.

9. I think that you both need to rethink this problem.
Suppose you toss a die ten times. What is the probability of not getting a two?
In the problem if k=6 then what is the probability of not getting player b in ten boxes?

Toss a die ten times; what is the probability of getting neither a two nor a three?
If k=6 then in buying 10 boxes what is the probability of getting neither player b nor player c?

Toss a die ten times; what is the probability of getting none of 2, 3 or 4?
If k=6 then in buying 10 boxes what is the probability of getting none of player b, player c or player d?

10. So are the answers (5/6)^10, (4/6)^10 and (3/6)^10 for each set of questions?

11. yeah I was right before..before I deleted it. Over thought it.

12. $P(Ei) = (\frac{k-1}{k})^n$
$P(Ei int Ej) = (\frac{k-2}{k})^n$
$P(Ei int Ej int Er) = (\frac{k-3}{k})^n$

$P(E1 union E2 union ... union Ek) = (\frac{k-1}{k})^n * (n choose 1) + (\frac{k-2}{k})^n * (n choose 2)+...$

Does that make any sense?
THen i have to estimate the solution of an infinite series?