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Decision Tables
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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.
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.
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Join Date: Mar 2008
Posts: 19
Reputation:
Solved Threads: 2
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?
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?
Steve
www.pcresolver.es
www.pcresolver.es
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