|
Let us present a simple occupancy problem.
We have a roulette wheel with only 3 numbers.
How many times will we have to spin it to have a 50% chance (or better) that all three numbers will appear?
Usually in these problems, we are asked to think that we are dropping "r" balls randomly into "n" boxes.
Each trial ends when all the boxes have been filled (or occupied).
(Now you know why these are called occupancy problems.)
Let's see how we can solve this one.
We must spin the wheel at least 3 times to get all 3 numbers, so let's solve the problem for just 3 spins of the wheel.
With 3 spins, there are 33 (or 27) ways that the numbers can appear.
| 1 1 1 | 1 1 2 | 1 1 3 |
| 1 2 1 | 1 2 2 | 1 2 3 |
| 1 3 1 | 1 3 2 | 1 3 3 |
| 2 1 1 | 2 1 2 | 2 1 3 |
| 2 2 1 | 2 2 2 | 2 2 1 |
| 2 3 1 | 2 3 2 | 2 3 1 |
| 3 1 1 | 3 1 2 | 3 1 3 |
| 3 2 1 | 3 2 2 | 3 2 3 |
| 3 3 1 | 3 3 2 | 3 3 3 |
That list of 27 numbers contains all the results of 3 spins of a three-numbered roulette wheel.
We can search that list to see how many of those 27 numbers contain 1 2 and 3 in any order.
It turns out that there are six such occurrences:
| 1 2 3 | 1 3 2 | 2 1 3 |
| 2 3 1 | 3 1 2 | 3 2 1 |
We see there are 6 ways out of 27 that we can get all 3 numbers after three spins of the wheel.
Therefore, the probability of getting all 3 numbers after 3 spins of our three-numbered roulette wheel is:
6 ÷ 27 = .22222222...
Click to go to Page 2
|
Return To Home Page
Copyright © 1999 -
1728 Software Systems
|
|