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There are $W \,,X\,, Y\,, \& Z$ possible activities a nurse can carry out whilst on a shift. Whilst observing n nurses, $W \,,X\,, Y\,, \& Z$ happened $w$% $x$% $y$% and $z$% of the time respectively. However since w+x+y+z sums to more than 100% then some nurses do more than one task at a time. We know that the probability of doing: 1 activity =63.8%, 2 activities =30%, 3 activities =6% 4 activities= 0.2%

Is it possible to calculate the probability of doing activity X? or is more information needed?

Thank you, and please ask for clarification is this is unclear. Best regards.

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Isn't it x%? Then it can't be estimated from the data since there is no information about which activities are taking place, only the number of activities; by symmetry any argument that P(X)=p would apply just as well to P(W)=p. – Dan Brumleve Jan 24 '12 at 9:12
I'm not sure to understand completely what you mean by "w% x% y% and z% of the time". Is a "time" a whole shift? Do you mean than W, X, Y and Z occurred exactly w, x, y and z times, and since w+x+y+z > n, then we know that a nurse had to do more than one task during her shift? – Charles Jan 24 '12 at 9:14
@DanBrumleve x% is the percentage of times a nurse carried out that particular activity with a patient. – HCAI Jan 24 '12 at 9:31
@CharlesMorisset time refers to each time a nurse visits a patient. So every time an activity/ies (be it 1, 2, 3 or 4) was recorded – HCAI Jan 24 '12 at 9:31
@user1134241, I'm not sure I understand, but please comment on my answer if I am misinterpreting the problem. – Dan Brumleve Jan 24 '12 at 9:34
up vote 1 down vote accepted

It can't be estimated from the data since there is no information about which activities are taking place. For example, it may be that X alone is active 64% of the time, Y & Z alone are active 30% of the time, Y & Z & W alone are active 25% of the time, and all four are active 1% of the time, in which case X is active 65% of the time (64% + 1%), because these events are mutually exclusive. But if we swap e.g. X and W we get a different answer. So more information is needed.

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Thank you for your answer. So from the data isn't there any way of inferring one type of activity (say X) is more likely to be happening given that only 1 activity is happening: from $P(X|1\, \text{type})$? What extra data would we need? – HCAI Jan 24 '12 at 9:38
If we had all 16 mutually-exclusive probabilities for every subset of $\{W,X,Y,Z\}$, that would be enough, because we could take the sum over all containing $X$. But maybe less is needed? The 8 subsets containing X would be enough, but you have only given 4 data points. – Dan Brumleve Jan 24 '12 at 9:41
I see. So since $x+y+z+w>100$ is there anything plausible that can be deduced from this data? Cheers – HCAI Jan 24 '12 at 10:36
Based on that inequality, only that the probability of two activities is positive. But you already stated that it is 30%. I'm still not sure if I understand your problem correctly so please take my answer with a grain of salt. – Dan Brumleve Jan 24 '12 at 10:38
Ok let's forget figures and that inequality (because I'm dubious of it's validity). If a nurse is seeing to a patient and someone records that she carried out 2 different activities, what is the probability that they are: X & Y? – HCAI Jan 24 '12 at 10:47

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