^
i think my link in post #144 should lead you to the answer
^
i think my link in post #144 should lead you to the answer
willy
something you have singularly failed to understand.It's basic statistics.
first, select 1000 people and test them.
now, how many do we expect to have the disease? 1/1000 of 1000 people is 1.
only 1 of 1000 test subjects actually has the disease, the other 999 do not.
we also know that 5% of those tested but who do not have the disease will actually test positive.
and as there are 999 disease-free people, so we would expect 5% of those, i.e. about 50, to test positive who do not have the disease.
now back to the original question.
there are 51 people who test positive in our example (the one unfortunate person who actually has the disease, plus the 50 people who tested positive but don’t).
only one of these people has the disease, so thats 1 in 51 or 2% give or take.
geddit?
Cheers Tax
I did provide a link from the Washington facility of education but in true TD style that was ignored as well
Your reasoning is. About as sound as this old chestnut
So you want a day off? Let’s take a look at what you are asking for!
There are 365 days this year.
There are 52 weeks per year in which you already have 2 days off per week, leaving 261 days available for work.
Since you spend 16 hours each day away from work, you have used up 170 days, leaving only 91 days available.
You spend 30 minutes each day on coffee break. That accounts for 23 days each year, leaving only 68 days available.
With a one hour lunch period each day, you have used up another 46 days, leaving only 22 days available for work.
You normally spend 2 days per year on sick leave. This leaves you only 20 days available for work.
We are off for 5 holidays per year, so your available working time is down to 15 days.
We generously give you 14 days vacation per year which leaves only one day available for work and I’ll be damned if you’re going to take that day off!
willy can only count up to 5, the number of repo comments per page. beyond that he is lost in the humid mists of his confusion.
what difference would that make to the question or the answer?
If I read anymore shit about this teaser, I am packing up my toys and taking them all home.
So still no answer on how many were false negatives nor where you mention 49 late in the debate but not in the OP
Did you go to the link I provided?
From the link:
A test to detect this disease has a false positive rate of 5%.I.e 0% false negativesAssume that the test diagnoses correctly every person who has the disease.
That makes it 95% accurate but only stating 95% accurate does not provide the same information
^^
try 0 does that work for you.
Yes but the wording does not say that as Troy indicates
In terms of machine learning and pattern classification, the labels of a set of random observations can be divided into 2 or more classes. Each observation is called an instance and the class it belongs to is the label. The Bayes error rate of the data distribution is the probability an instance is misclassified by a classifier that knows the true class probabilities given the predictors.
For a multiclass classifier, the expected prediction error may be calculated as follows:^{[3]}
{\displaystyle EPE=E_{x}[\sum _{k=1}^{K}L(C_{k},{\hat {C}}(x))P(C_{k}|x)]}where x is the instance, {\displaystyle E[]} the expectation value, C_{k} is a class into which an instance is classified, P(C_{k}|x) is the conditional probability of label k for instance x, and L() is the 0-1 loss function:
{\displaystyle L(x,y)=1-\delta _{x,y}={\begin{cases}0&{\text{if }}x=y\\1&{\text{if }}x\neq y\end{cases}},}where {\displaystyle \delta _{x,y}} is the Kronecker delta.
When the learner knows the conditional probability, then one solution is:
{\displaystyle {\hat {C}}_{B}(x)=\arg \max _{k\in \{1...K\}}P(C_{k}|X=x)}This solution is known as the Bayes classifier.
The corresponding expected Prediction Error is called the Bayes error rate:
{\displaystyle BE=E_{x}[\sum _{k=1}^{K}L(C_{k},{\hat {C}}_{B}(x))P(C_{k}|x)]=E_{x}[\sum _{k=1,\ C_{k}\neq {\hat {C}}_{B}(x)}^{K}P(C_{k}|x)]=E_{x}[1-P({\hat {C}}_{B}(x)|x)]},where the sum can be omitted in the last step due to considering the counter event. By the definition of the Bayes classifier, it maximizes {\displaystyle P({\hat {C}}_{B}(x)|x)} and, therefore, minimizes the Bayes error BE.
The Bayes error is non-zero if the classification labels are not deterministic, i.e., there is a non-zero probability of a given instance belonging to more than one class.^{[citation needed]}. In a regression context with squared error, the Bayes error is equal to the noise variance.^{[3]}
Just a small observation, both Willy and David gave wrong answers early in the thread #104 and #106 and then suddenly needed more information. Funnily they did not need this info prior to getting the answer wrong.
It was the Reichsmark , still have some, between the RentenMarks and the much lamented Deutschmark which gave way to plastic ECB mafia debt on German working class to support Graeco Latin criminals profligates and assorted hangers on
Of course during the last few Months of the conflict a pack of American camels could get you a lot of action
^
We could test 10,000,100,000 1,000,000 the result would be the same. I don’t think you quite get it.
That 9 dots one was depressing because I think I would have got it 5 or 10 years ago
With the medical test with 95% accuracy (or 5% false positive), if 1000 people are tested then 50 will test positive but only 1 will actually have the disease so the probability that you have the disease if you test positive is 1/50 or 2%
I read somewhere that 80% of doctors get that one wrong so if your doctor tells you you are positive then you still only have 2.5% chance of being positive
The difference is that 95% accurate could also mean 0% false positives and 5% false negatives, which means a positive result indicates you are 100% likely to have the disease.
The answer assumes that 95% accurate is based on 5% false positives and 0% false negatives, assumptions that are supplied in the link but not in the question posed.
Apologies for being pedantic, I didn't bother answering the question originally, I am just explaining what david44 is talking about
Information that kceman thinks he is clever in not providing.
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