# dependent probability question?

• Nov 30th 2013, 06:29 PM
random512
dependent probability question?
Hello, I had a question about how to calculate dependent probability. Calculating independent probability is easy, simply multiply the probability of each event: P1 * P2 * P3.

Therefore, if 3 different resources are working on 3 independent and unrelated tasks, and each resource has an 80% probability of completing his task, then the total probability of all resources completing all tasks, successfully, would be .80 * .80 * .80 = 51.2%

However, I need to figure out the probability formula, for a more advanced scenario. Consider a scenario, where 1 resource is working on 3 tasks in sequence, and the probability of the resource completing each individual task is 80%.

In this scenario, the first task would be independent, but the 2nd task would be dependent on completion of the 1st task, and the 3rd task would be dependent on completion of the 1st and 2nd task. So can you tell me the formula I would use to calculate total probability in this scenario?
• Nov 30th 2013, 07:30 PM
romsek
Re: dependent probability question?
Let's look at failure rather than success. Then the overall probability of success is 1 - probability of failure.

You can fail at step 1, step 2, or step 3.

Pr[fail at step 1] = 1 - .8 = .2

Pr[fail at step 2] = Pr[fail at step 2 | success at step 1]

Now assuming failure at step 2 is independent of success at step 1 we get

Pr[fail at step 2] = Pr[fail at step 2] Pr[success at step 1] = .2 * .8 = .16

Pr[fail at step 3] = Pr[fail at step 3 | success at 1 and 2] = .2 * .8 * .8 = .128

The total probabily of failure is then the sum of these (.2 + .16 + .128) = .488

So Pr[Success] = 1 - .488 = .512

This is identical to the non sequenced case. Why? Because in order to succeed all 3 steps have to be succeeded at in either case.
Where this methodology becomes very useful is when there are multiple paths to success as well as failure.
• Nov 30th 2013, 07:45 PM
Melody2
Re: dependent probability question?
My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.

Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.
• Nov 30th 2013, 07:51 PM
romsek
Re: dependent probability question?
Quote:

Originally Posted by Melody2
My probability abilities are not advanced but i can get a surprising number of problems correct by using probability trees and other basic probability techniques.
When I learn more advanced probability theories they are more likely to make sense because I can 'see' that the formula is logical.

Learn to use trees effectively and the answer to these problems becomes much easier to discover. (You don't always need to draw the whole tree)
That's what I think anyway.

In industry probability trees are used EXTENSIVELY.
• Nov 30th 2013, 11:09 PM
random512
Re: dependent probability question?
romsec - thanks for your feedback. However, I think the total probability should be lower, if 1 resource works on 3 tasks in sequence. Here's why: If the resource fails at task 1, then he will never get to task 2 or task 3. If the resource fails at task 2, then he will never get to task 3.

On the other hand, if 3 different resources work on these 3 different tasks independently, then this approach will completely sidestep the dependencies/risks that I described with the single resource above, so it seems like the realistic probability of success should be greater.

It seems like probability theory should have a formula that factors in the dependencies I described above. Can you provide an updated probability formula based on the additional context I provided above?
• Nov 30th 2013, 11:22 PM
romsek
Re: dependent probability question?
I tell you what. You look at my derivation and tell me specifically what bits you think are incorrect. I took into account all the dependencies you put in by requiring serial operation.
• Dec 1st 2013, 12:16 AM
random512
Re: dependent probability question?
@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?

I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:

Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%

Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???

Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.

Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?
• Dec 1st 2013, 01:21 AM
romsek
Re: dependent probability question?
Quote:

Originally Posted by random512
@romsek - I'm only a beginner at statistics. Maybe there's some type of statistics concept that we're not accounting for here?

I was under the impression that the total probability for independent events was calculated differently than the total probability for dependent events, and the total probability for dependent events would be lower. Maybe it will help if I simplify my example?:

Scenario 1: 3 independent events, each event has a probability of 80%: .80 * .80 * .80 = 51.2%

Scenario 2: 3 events with sequential dependency, each distinct event has a probability of 80%: ???

Maybe part of the problem here, is that when I say each event has a probability of 80%, that implies that the dependency relationship has no impact, compared to scenario 1. I was assuming that the total probability formula for dependent events would have a built-in function to lower the total probability.

Does my simplified example change anything? Or do you still assert that the total probability for these scenarios should be identical?

In both cases there is a single outcome that denotes success of the entire venture. All 3 tasks must be completed successfully. There is only a single way this can be achieved and that is that all 3 tasks are done successfully.

The probability that your 3 independent folks each succeeding is .8. The probability that all 3 independently succeed is (.8)^3 = .512

Task 1 succeeds with pr. .8. Only then does it proceed to Task 2. The probability that we get to Task 2 is thus .8. We don't care what happens if we fail at Task 1, if we do we fail.

We have arrived at Task 2 with pr .8. Now we have .8 pr of completing Task 2. There's no other dependency of our success on Task 1, it got here, that's all we need to know. So our pr of success of Task 2 is again .8 but also times the pr. that it got here at all, which we know to be .8. So the overall pr of completing Task 2 is (.8)^2 = .64

Now we make it to Task 3 with pr .64. Once again there is no dependency on how well things went at task 1 or 2, the thing got here and that's all we know and we have pr .8 of success at Task 3. So the overall pr of completing Task 3 is .8 times the pr it got here at all .64, or (.8)^3 = .512.

There are all sorts of dependencies you could put between your tasks that would show the sort of conditional probability effects you are thinking of. But you haven't done that here.
• Dec 1st 2013, 02:17 AM
random512
Re: dependent probability question?
@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.

3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:

1 - (.50 * .50 * .50) = 87.5%

So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.
• Dec 1st 2013, 02:25 AM
romsek
Re: dependent probability question?
If you want to set up an example demonstrating what you have in mind consider this.

Let's say we're making toy dolls since it's the holidays. We have to cut the parts, assemble the parts, paint the doll.
We have 3 elves and they have different abilities regarding making these dolls. For elf A his pr success at those tasks is say {0.85, .6, .4}
elf B could be {0.6, 0.9, 0.6}, and elf C could be {0.5, 0.5, 0.8}

You could have each elf build the entire doll at their station but it's pretty clear that there's a better strategy and that's the pair the elf's task with their best ability.

That's where your sequential operation comes in since you have to build the dolls in a sequence.

If you really want to get fancy you can have the probability of success at a station depend on the quality that the previous station produced. It's harder to assemble poorly cut parts. It's harder to paint a poorly assembled doll, and incorporate all that in to further show how important your strategy of matching worker and task is.

All this will certainly show the effect of sequential vs parallel operation.
• Dec 1st 2013, 02:28 AM
romsek
Re: dependent probability question?
Quote:

Originally Posted by random512
@Romsek - Thanks for your knowledge and patience. I think I figured out what I was missing: total probability of overrun.

3 independent events can execute concurrently. Therefore, if each event has an overrun probability of 50% then the total probability of overrun can be kept to 50%. However, if the 3 events must execute in sequence, then the probability of overrun is:

1 - (.50 * .50 * .50) = 87.5%

So that's how the concept of total probability can be used to prove the cost/impact of dependent events, as opposed to independent events.

ok I'm glad you figured it out.
• Dec 1st 2013, 09:21 AM
random512
Re: dependent probability question?
One more related question. What's the proper way to describe the probability multiplication factor? For example:

* the probability of a single risk event occurring is 50%
* the probability of at least 1/3 risk events occurring is:
1 - ( .50 * .50 * .50) = 87.5%

Is it valid to say that additional risk factors create an "exponential" probability that at least 1 risk event will occur? "Exponential" is a good word, because it's descriptive, and powerful. However, I want to make sure that I use the word appropriately. So can I describe this multiplication effect as "exponential" or should I choose a different word?