QTRA - I'm sorry i don't agree with it!

[Amelanchier]

I'm currently working out what "affirming the consequent" might mean:001_smile:

 

Let me google that for you

 

[HCR]

Interesting! Or is it that 0.8 trees will fail? Now that's confusing

 

Yes, 100% chance that 0.8 of a tree will fail.

 

I'm not sure how that would stack up in court though...if you think about it, using this system you can 'pass' 10,000 trees as 'safe', i.e. each tree has minimum of a 1:10,000 chance of failing. If one tree (or fewer) fails from the 10,000, you have met your criteria, based on empirical evidence.

 

Now, what if that one tree had a serious discoverable defect? What if it killed someone?

 

[Amelanchier]

Hang on though. To make the numbers easier, let's imagine that daltontrees has to survey 10,000 trees at a risk threshold of 1/10,000. Each tree would be assessed against that threshold so wouldn't the the population risk be 0.0001x0.0001x0.0001... etc 10,000 times... Which is too small a number for my calculator to show me?!?

[HCR]

Hmm. That would give you the probability of all the trees failing. Assuming DaltonTrees knows what he's on about (and I'm sure he does), as you rightly predict that would be a very small number.

 

Successive events: multiply the probabilities

Independent events: add them

 

i.e. 0.0001 + 0.0001......etc up to n=10,000 would give you a probability of 1.0

[Amelanchier]

Don't you only add independent event probabilities if the events are mutually exclusive? I.e., if one tree failing prevents the others from failing?

 

Seems odd that your would end up with one tree (that had been individually assessed as having a prob. of 0.0001 likelihood of failure) that then was certain to fail when the population was examined?!?

[HCR]

Effectively the events are mutually exclusive, as we are looking for the chances of one (and only one) tree failing.

 

However, I know it's not that simple and I don't know the proper answer.

 

Imagine a die with 10,000 sides. If you threw it 10,000 times would you get each number represented once? Highly unlikely. It would be exactly the same odds as any number not being repeated.

[Amelanchier]

However, I know it's not that simple and I don't know the proper answer.

 

Me neither - probability can be entirely unintuitive!

 

Imagine a die with 10,000 sides. If you threw it 10,000 times would you get each number represented once? Highly unlikely. It would be exactly the same odds as any number not being repeated.

 

But in that instance I would think that you would multiply each distinct event? That's going to bug me all day now...

[RobArb]

I still can't get past this figure of 1 in 10000 as being an accessible risk, based on a book from 1987 do you not think things have changed since then?

 

In Dalton trees case it had been specified that the trees (8000) must pass this threshold, very black and white... What happened to the grey areas if it wasn't objective?

 

Sent from my Galaxy S2

[RobArb]

What proportion of the 8000 trees do you think would "fail" this test compared to using a different system?

 

Sent from my Galaxy S2

[HCR]

It's based on his subjective assessment as a competent person.

 

If it goes horribly wrong (i.e. a tree that passed the test subsequently goes on to fall over and kill someone) a number of points will be examined.

 

1. was an assessment of the tree carried out?

2. was the assessment an adequate one?

3. was the assessor competent to carry out that assessment?

4. was the tree defective in a manner that a reasonably competent tree inspector would have discovered?

 

Assuming the answer to the first 3 questions is yes, the answer to the third one comes down to what the jurors/court believes on the basis of the evidence presented to them. I doubt the 1:10,000 thing would be hugely relevant to be honest, but I'm open to other opinions.