Hello !
I've found that for calculation probabelities that all boxes are filled I have to deal with Biiiiiiiiiiiig numbers. How I calculate:
For example, let's take 5 types of chocolates and 3 boxes.
All possible distinguished variants for placement chocolates is:
Box1 Box2 Box3
I I III
I II II
Let's take first placemets. All possible variants of box permutation is 3!/2! (as two boxes will have equally number of chocolates). Probabilities such one placement would be 1/(3^5). Permutation of chocolates themselves inside boxes will be 5!/3! (as one of boxes will have 3 chocolates). Well, common probabilities placement of first type is 1/(3^5)*(3!/2!)*(5!/3!)=60/243
Calculate probabilities for second type placement like for first one.
All possible variants of box permutation is 3!/2!. Probabilities such one placement would be 1/(3^5). Permutation of chocolates themselves inside boxes will be 5!/(2!*2!). Well, common probabilities placement of second type is 1/(3^5)*(3!/2!)*(5!/(2!*2!))=90/243
Common probabilities for all placement where all boxes are filled:
60/243+90/243=150/243
So probabilities at least one boxes is empty is (1-150/243).
I show easy example. But when quantity of chocolates and boxes grows, numbers become too large.
Should I use long ariphmetic ?
![:(](./images/smilies/icon_frown.gif)