Probability question

This is a discussion on Probability question within the Psychology, Risk & Money Management forums, part of the Methods category; You have to make a yes/no decision. To help you, you have two independent decision factors A and B with ...

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Old Apr 13, 2007, 11:41am   #1
 
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Probability question

You have to make a yes/no decision.
To help you, you have two independent decision factors A and B with probabilities p(A) and p(B) of being correct. The probability of making the correct choice is p(A) if you use decision factor A alone, and p(B) using B alone.
What is the probability of being correct using both A and B?
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Old Apr 13, 2007, 11:55am   #2
 
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p(A)*p(B)

pA(0,1) and pB(0,1)
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Old Apr 13, 2007, 12:12pm   #3
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Your question is not phrased as I would expect and would benefit from some clarification to be sure that I am reading it correctly. As I understand the question you are asking:

"If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Assuming this to be the case and there is only 1 event but there are 2 unrelated decision factors it is additive: P(A)+P(B).

It would be P(A)*P(B) if we were considering the probability of 2 independent events occurring each with probabilities P(A) and P(B) but, judging by your question it is the former and not the latter case in which you are interested.
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Old Apr 13, 2007, 12:30pm   #4
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At a guess

1 - [(1-P(A))*(1-P(B)]

Good question.
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Old Apr 13, 2007, 12:33pm   #5
 
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WinstonSmith started this thread "If I have a single event and can choose between 2 decision factors (A and B) to determine the probability of the event, where A and B have probabilities of being correct of P(A) and P(B), what is the total probability of making the correct decision if I use both factors".

Yes that's correct. Let's give an example: You have 2 decision factors A and B to decide whether to enter a trade. P(A)=0.7 and P(B)=0.6. You propose to use B as a filter to be used if A is satisfied. Are you better or worse off?
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Old Apr 13, 2007, 12:41pm   #6
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1-[(1-0.7)*(1-0.6)] = 0.88

T4 would say it improves your chances to use both.
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Old Apr 13, 2007, 12:52pm   #7
 
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Quote:
Originally Posted by TWI
1-[(1-0.7)*(1-0.6)] = 0.88

T4 would say it improves your chances to use both.
Thanks TWI. Could you explain your reasoning?
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Old Apr 13, 2007, 1:07pm   #8
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You have to look at th chance of being incorrect rather than chance of being correct.
so 1-P(A) and 1-P(B)
P(A) and P(B) are independent so the probability of both occurring is the multiple.
Now you have the chance of being wrong when they are combined.
Simply subtract this from 1 to get the chance of being correct.
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