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Let's say I have a collection of objects. Each object has a set of predictive attributes where the attribute may be true or false. Each attribute when true predicts "success" with a known probability, and these probabilities are not necessarily the same for different attributes. What is the probability of success for an object with more than one true attribute? For example, if being male has a 25% chance of success, and being under 35 has a 50% chance of success, what are the chances of success for a male under 35, assuming the two attributes are independent in terms of success?

Second question. Let's say I know the z-value for attribute A, and the z-value for attribute B. What would I expect the z-value to be for (A and B), assuming A and B are independent?

Thanks.

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    $\begingroup$ In general, it's not possible to answer from the information given. $\endgroup$ – Glen_b -Reinstate Monica Sep 19 '13 at 5:36
  • $\begingroup$ Ordinarily we can guess what you might mean by a "z-value" but it is strange to see it applied to binary attributes and even stranger to see its use in the second question, which is difficult to interpret. So that we can be sure we understand what you're asking, could you please explain what you mean by this term? $\endgroup$ – whuber Sep 19 '13 at 7:22
  • $\begingroup$ Is the question still unclear, or is it not possible to answer? Any help would be greatly appreciated. $\endgroup$ – Tyro Sep 20 '13 at 23:52
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If two events are independent, then $P(A|B)=P(A)$. If on the other hand, two independent drawings are made from two urns to determine whether you find at least one red ball, then the probability becomes $1-(1-(P(\not R_A))*(1-P(\not R_B))$ if $R_A$ corresponds to "drawing a red ball from urn A".

It is not straightforward to link this with yout sentence: "assuming the two attributes are independent in terms of success".

Given the numbers you have shown the first interpretation is not easy t apply as there are three "events": some one is over 35 (A), someone is male (B) and someone is successful (C). Independen could mean: $P(C|A) = P(C|A \land B)$ and $P(C|B) = P(C|A \land B)$. But this only works if $P(C|A)=P(C|B)$, which is not the case in the example.

The second interpretation would be equivalent to saying: You basically have two indepndent shots at being successful, one is based on your age, one on yor gender (this makes of course not much sense conceptually or empirically, but would correspond to the run case). In this case I would calculate: $P(C)= $1-(1-(P(C|A))(1-P(C|B))=1-((1-0.25)(1-0.5))=1-0.375=0.625.

The answer by Hotaka uses yet another idea of independence that uses a sequential process: only for those candidates that are not successful based on the first criterion, is the chance for being successful based on the second criterion the specified probability. I would not consider this to be a valid case of independence.

Regarding the second question, I am a bit lost as to how binary attributes are linked to z-values here...

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1st question: Think about it this way: consider one factor at a time. First, 25% of males will succeed. So no need to worry about them anymore, let's worry about the 75% that are left behind. As long as the factors are independent, 50% of them will succeed. So 75% x 50% = 32.5% will succeed. Finally, add 32.5 and 25 and you get 57.5% success rate.

2nd question is interesting. Imagine a scatterplot of z scores for A on the y-axis and B on the x-axis. They are independent, so basically you'd see a circle with a lot of dots in the middle and fewer as you go outward. The z score of A and B combined is the distance between that specific dot of interest and the very center of the circle, which is 0. So let's go back to the Pythagorean theorem: z of A and B = ((z of A)^2 + (z of B)^2)^.5

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    $\begingroup$ I would ask you to reflect on what it means for two predictors of a third value to be "independent" and consider explaining that to your readers. In particular, it is strongly counter-intuitive that combining two "independent" predictors would actually lower the chances predicted by each predictor separately. I suspect you may have interpreted this question in a radically different way than I read it. $\endgroup$ – whuber Sep 19 '13 at 7:20
  • $\begingroup$ Thanks for the catch. i misread the question. I changed my response. how is it now? $\endgroup$ – Hotaka Sep 19 '13 at 7:40
  • $\begingroup$ It's very late here and my thinking is fuzzy, but it strikes me there might still be something missing. I'm imagining (for instance--this is not the only possibility) a log-linear or Poisson model in which the log odds of success are posited to be a linear combination of the two binary predictors. In such a model I believe the answer to question one would differ from your new answer. This suggests we should take to heart the comment left by @glen_b. I don't know what to make of the second question so I have left a comment to the original proposer (OP) asking for clarification. $\endgroup$ – whuber Sep 19 '13 at 7:45
  • $\begingroup$ By the way, I see this is your first day here: welcome to our site! $\endgroup$ – whuber Sep 19 '13 at 7:45

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