What is the cumulative probability? - page 5
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1-(1-P(A))*(1-P(B)) (no guarantee)
A little bit of abstraction, I think it makes more sense.
A The chance of getting sick from an open window is 0.5
B, the probability of getting sick from wet feet is 0.5.
The probability of getting sick if we have both A and B is 1 - the probability of not getting sick, i.e. 1 - (1 - P(A))*(1 - P(B)). = 0.75
Everything is correct.
I have my doubts about something else... How can bulls and bears opinions be independent???
Conclusion -- I think the solution to the problem is meaningless, because the conditions are incorrect and it can only be solved by determining the relationship between A and B.
It is the same as trying to calculate the probability from the outcomes of individual experts in an expert system, if all the experts have the same input.
I need to know a reliable prediction of the probability of a certain price occurring when I determine that probability with an Up-trend indicator on one side and a Down-trend indicator on the other. What will be the final probability?
Simpler: A bullish indicator tells you: the price will be in the area of interest with probability P1. And a bearish indicator tells you: the price will appear in the zone with the probability P2. How do you determine the final probability?
finally a problem statement:)
and the solution:
up: P1*(1-P2) and down: P2*(1-P1)
although: with what probability do the indicators give the right recommendations?
finally, a problem statement:)
and a solution:
upwards: P1*(1-P2) and correspondingly downwards: P2*(1-P1)
Brilliant! Let me remind you that up + down gives 100%
Make up your mind...
A little bit of abstraction, I think it makes more sense.
A The chance of getting sick from an open window is 0.5
B, the probability of getting sick from wet feet is 0.5.
The probability of getting sick if we have both A and B is 1 - the probability of not getting sick, i.e. 1 - (1 - P(A))*(1 - P(B)). = 0.75
Everything is correct.
I have doubts about something else... How can the opinion of bulls and bears be independent???
Conclusion -- I think the problem is meaningless because the conditions are incorrect and it can only be solved by determining the relationship between A and B.
It is the same as trying to calculate the probability according to the outcomes of individual experts of an expert system, if all the experts have the same input.
Not correct. Where did you get 1 for the probability of getting sick? What if the probability of getting sick from an open window is 0.7 and from wet feet is 0.8?
finally, a problem statement:)
and the solution:
up: P1*(1-P2) and down: P2*(1-P1)
>> although: with what probability do the indicators give the right recommendations?
Not up and down. It is the probability of price in a particular zone in terms of the two different indicators determining that probability, with a slight difference.
That's great! Now, let me remind you that up + down gives 100%.
>> Decide further...
Unfortunately wrong. The event space I have is as follows (if of course we are talking about independent events):
P1*(1-P2)+(1-P1)*P2+(1-P1)*(1-P2)+P1*P2
in numbers:
0.4*(1-0.2)+(1-0.4)*0.2+(1-0.4)*(1-0.2)+0.4*0.2=1
How about you? :)
Not up and down. It is the probability of price in a particular zone in terms of two differently polarised indicators which determine that probability with a slight difference.
I think you got what you wanted....
I think you got what you wanted....
>> Where?
Not correct. Where did you get 1 for the probability of getting sick? What if the probability of getting sick from an open window is 0.7 and from wet feet is 0.8?
Not like that. 1 minus the probability of getting sick. The answer is 0.94 probability of getting sick.
Unfortunately, this is wrong. My event space is as follows (if we are talking about independent events, of course):
P1*(1-P2)+(1-P1)*P2+(1-P1)*(1-P2)+P1*P1
in numbers:
0.4*(1-0.2)+(1-0.4)*0.2+(1-0.4)*(1-0.2)+0.4*0.2=1
How about you? :)
I can do the math too. Where did the last 2 summands come from???
I quote again:
finally, the problem statement:)
and solution:
up: P1*(1-P2) and down: P2*(1-P1).
although: with what probability do the indicators give the right recommendations?
we get the system
up P1*(1-P2)
down P2*(1-P1)
up + down -- complete group of events whose sum of probabilities is 1
we get --
P1*(1-P2) + P2*(1-P1) == 1
Waiting for an explanation.