Decision Tables

[andy4919]

I have a little problem with decision tables. I do under stand that for every 2 conditions there should be 4 possible outcomes or true/false condidtions.

I have this telephone problem where the telephone company charges .10 a minute for all calls outside the callers area code (2 outcomes) and charges .13 cents a minute for all other calls (1 outcome?) The phone company wants a report showing the callers area code and number, the area code called and number, number of minutes and the charge per minute.

I only have 3 outcomes and there are supposed to be 4. I am totally stuck, does anyone know a little about desicion tables.

[PCResolver]

Hello Andy,

well, it has been some time since I used decision trees.

The thing with decision 'trees' is that each branch (where one option spits into 2 or more branches) should add up to 1. That is, the sum of the probabilities of the branches should add up to 1.

For instance, a coin toss yields 2 branches with a probability of 0.5 each.

A dice roll would have a probability of 1/6 for each branch.

These represent equal probability but the real strength is in 'biased' probabilities. For instance a loaded dice could be
6: 0.5
5: 0.1
4: 0.1
3: 0.1
2: 0.1
1: 0.1

So, in summary, you don't need to have 4 branches, you can have just the 2 that you need. For instance if 75% of the calls are outside the area code:

outside: .75
inside: .25

Does this help?

[andy4919]

I do nto have the book with me but I thought if you had 2 Conditions you needed 4 possible outcomes, if you had 3 you need 8, and so on.....