Originally Posted by: Gerry Knowles Lets just face it there are many ways of assessing risk, and one of those is by having a mathematical claculation which is conveniently placed on a risk matrix. The size of the matrix will change the accuracy but not necessarly the outcome. So in theory a 5x5 matrix will give a different outcome to a 10x10 matrix. However it is still down to a person deciding what number is allocated to the probability and the outcome so I may decide that a probability score for a process may be three and severity of the injury may be a four so the risk score is a twelve. This will probably give a medium rating (work to do to reduce the risk). The same process may see the scores doubled but in the end the process will still be a medium rating.
I guess I am saying that lets not get tied up with formula and their outcome. Our profession would command a much higher level of respect if we kept it simple and easy for everyone to understand.
Gerry is abolutly spot on and he already knew that;-) and does not need me to confim or ratify his assertion! You can skip to my point at the end of my post because this middle bit...... good luck if you read it.
The UK regulator does not require an assessment of "likelihood" of an outcome. BS EN IEC 31010:2019 allows for the interchangeable use of "Likelihood" and "Probability" as there are few direct equivalences in other languages. “Probability” is a mathematical term. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes
Discrete Random Variables
Suppose that you have a stochastic process that takes discrete values (e.g., outcomes of tossing a coin 10 times, number of customers who arrive at a store in 10 minutes etc). In such cases, we can calculate the probability of observing a particular set of outcomes by making suitable assumptions about the underlying stochastic process (e.g., probability of coin landing heads is pp and that coin tosses are independent).
Denote the observed outcomes by OO and the set of parameters that describe the stochastic process as θθ. Thus, when we speak of probability we want to calculate P(O|θ)P(O|θ). In other words, given specific values for θθ, P(O|θ)P(O|θ) is the probability that we would observe the outcomes represented by OO.
However, when we model a real life stochastic process, we often do not know θθ. We simply observe OO and the goal then is to arrive at an estimate for θθ that would be a plausible choice given the observed outcomes OO. We know that given a value of θθ the probability of observing OO is P(O|θ)P(O|θ). Thus, a 'natural' estimation process is to choose that value of θθ that would maximize the probability that we would actually observe OO. In other words, we find the parameter values θθ that maximize the following function:
L(θ|O)=P(O|θ)L(θ|O)=P(O|θ)
L(θ|O)L(θ|O) is called the likelihood function. Notice that by definition the likelihood function is conditioned on the observed OO and that it is a function of the unknown parameters θθ.
Continuous Random Variables
In the continuous case the situation is similar with one important difference. We can no longer talk about the probability that we observed OO given θθ because in the continuous case P(O|θ)=0P(O|θ)=0. Without getting into technicalities, the basic idea is as follows:
Denote the probability density function (pdf) associated with the outcomes OO as: f(O|θ)f(O|θ). Thus, in the continuous case we estimate θθ given observed outcomes OO by maximizing the following function:
L(θ|O)=f(O|θ)L(θ|O)=f(O|θ)
In this situation, we cannot technically assert that we are finding the parameter value that maximizes the probability that we observe OO as we maximize the PDF associated with the observed outcomes OO.
The point I am trying to make, let’s not bet on outcomes for workers. The "incident" is binary outcome (1/0) and they will either get hurt or not. If we imagine they will get hurt how do we control the "energy" that will cause the damage. We focus to much on prevention and rely on the preventative measures, yet people still get hurt because there is so much variability in the systems they work in however, we don’t think that that bad outcome could happed because because we beleive the preventative measures we took would be enough to .... well prevent it. Sadly, our imagination lets us down. To the point being made by Gerry here we all perceive risk differently. If we assume that even with all the preventative measures in place the worker will still be hurt that forces us to think about controls that separate them from the harmful “energy” in the system so that that “incident/failure” can still happen but there is no loss or damage because the “system” can recover and the worker can move on.
Yes most of this was a cut and paste job from the interweb and yes I may have been trying to add to the confusion (except that last paragraph) I hope it didn't make your eyes bleed.
IMHO, this thread is a non-conversation (yet I still posted twice). Lets stop making bets on workers safety!