0

I got a question, that i could need some help with. Below you see the picutres

The following is given: A patient can have cancer, or not. There is a test, that has two outcomes: Positive, or negative. The prior prob. over all population, that a person has cancer is 0.01. The test returns positive outcome with prob. of 0.98, where the person has indeed cancer and negative with 0.95 where the person has NOT cancer. So far, so good.

1) I should summarize the above in terms on probabilities. That is my solution: P(C) = 0.01; P(not C) = 0.99 P(TP|C) = 0.98; P(TN|not C) = 0.95

2) A patient gets tested, and the test is positive. Do we tell this patient, he has cancer? I calculated P(C|TP) and got a solution value = 0.01 being 1%. Which means for me, we don't tell him, as there is only 1% chance, that he has cancer.

3) The patient gets testet again! Both tests are independent to each other. And i should calculate the "posterior probability" of cancer and not cancer following the two tests.

And here, for 3) i am lost a bit. I would now calculate just P(not C|TP) as i did in 2). But is that actually correct? If not, what should i calculate instead and how?

Task itself Try of Task

And for reference, a similar excercise, we discussed. The solutions are said to be correct for the next 2 pictures. I was trying to apply the same methods. But for Task 3) i am unsure, how to solve it.

example Solution for example

user3556093
  • 41

1 Answers1

1

You can tell without any calculation that your answer to $2)$ is wrong because the posterior probability is equal to the prior probability, and the test is clearly not so bad as to be completely irrelevant.

For $3)$, you need to calculate $P(\text{no cancer}\mid\text{test $1$ positive}\cap\text{test $2$ positive})$, where e.g. $P(\text{test $1$ positive}\cap\text{test $2$ positive}\mid\text{cancer})=P(\text{TP}\mid\text{C})^2$.

joriki
  • 238,052
  • Can you show me, how you can tell? I dont get it. Also, we have probably to calculate something. Pls help. – user3556093 Jan 09 '20 at 12:40
  • If someone tells me, how to get pictures in here, i can show, what i did. – user3556093 Jan 09 '20 at 12:42
  • @user3556093: There's an image button above the editor window. But it would be better to write what you did, not least because images are not searchable. I posted a tutorial and reference for typesetting math on this site in a comment under the question. I didn't mean to imply that you shouldn't calculate anything; you certainly should. I was just saying that it doesn't require calculation that your answer to $2)$ is wrong. I already explained in the answer why. If you don't understand the explanation, please state which part you don't understand (instead of asking for a new explanation). – joriki Jan 09 '20 at 12:48
  • Yes, thanks. So, you said, my probabilities are the same. Is at least my P(C|TP) = 0.01 correct? That means, that P(C) = P(C|TP) right? Maybe i did calculate the wrong thing in a whole? I just dont understand, why you say, the test is good, if it only gets correct in 1% of all cases. Probability a patient has cancer given the test is positve. That is, what i calculated. And according to the excercise, i think that is, what i should do. "new patient => Test lab is positive => should we tell him, he has cancer?" And if the probability is only 1%, that he has cancer, if the test is correct ... – user3556093 Jan 09 '20 at 12:59
  • I got the pictures now. Thank you for your help with that :) Hope that clarifies a bit, on what i tried and where i struggle :) – user3556093 Jan 09 '20 at 13:14
  • For 3) i dont understand, why i have to calculate, what you mentioned above. Why isn't it just P(not Cancer|TP) ? The tests are independent. Test 1 and Test 2 that is. So why do i have to cut (∩) them? I bet you are right. I surely have no clue. But that is also, why i don't understand. – user3556093 Jan 09 '20 at 13:18
  • @user3556093: I didn't say that the test is good if it's only correct in $1%$ of cases – I said that that result must be wrong because it would mean that the test is completely irrelevant (since $1%$ is already the a priori probability of having cancer, without a positive test result), whereas it's clearly not completely irrelevant – not based on your erroneous result but based on the given data, which show that there's a high correlation between the test results and having cancer. – joriki Jan 09 '20 at 14:24
  • @user3556093: The error in your calculation is where you have $0.95\cdot0.99$ in the denominator. This should be $P(\text{TP}\mid\neg\text C)P(\neg\text C)=0.02\cdot0.99$. – joriki Jan 09 '20 at 14:26
  • @user3556093: As to your last question above: The probability for two independent events to both occur is the product of their probabilities; in fact, this is the definition of independence. – joriki Jan 09 '20 at 14:28
  • Ah, i think i found the error you were mentioning. I now get a result of 33% for P(C|TP). Which i still think is not very good. Or am i wrong again? I also still don't understand task 3 and why i have to calculate P(no cancer∣test 1 positive∩test 2 positive). Do i also have to calculate P(cancer∣test 1 positive∩test 2 positive)? It says so, or can i just do it with 1- P(no cancer∣test 1 positive∩test 2 positive)? – user3556093 Jan 12 '20 at 09:24
  • @user3556093: Yes, conditional probabilities are probabilities, $P(A\mid C)=1-P(\overline A\mid C)$ holds just like $P(A)=1-P(\overline A)$. About $33%$: I'm sorry, I made a mistake above. It's right that you need $P(\text{TP}\mid\neg\text C)P(\neg\text C)$ there, but $P(\text{TP}\mid\neg\text C)=1-P(\text{TN}\mid\neg\text C)=1-0.95=0.05$. I'd calculated $1-P(\text{TP}\mid C)=1-0.98=0.02$ by mistake, which is $P(\text{TN}\mid C)$, not $P(\text{TP}\mid\neg\text C)$. About task $3)$: Can you say more about what's causing you difficulties? It's really essentially just the multiplication rule. – joriki Jan 12 '20 at 11:40
  • So is the solution 33% correct? Or not? My current Solution is: (0.980.01/(0.980.01+0.02*0.99)) = 0,33 %. Or were the 1% correct? I am super confused now. For 3) I have difficulties to understand, what exactly i have to calculate, and why it has to be what you said. Seems to be a very basic thing, i just dont get. We never have done such an exercise, so i have no solution which i could try to understand and compare. Thus i am completely lost and do not understand anything, regarding 3) Nor what i have to do, neither what you suggest to do to obtain the correct solution. – user3556093 Jan 16 '20 at 09:39
  • Edit: 0.33, not 0.33 %. Why is P(TN|C) = 0.02? I thought it is 0.05. P(TN|¬C) = 0.95. So shouldn't P(TN|C) be 0.05? Same with P(TP|C) is 0.98. So P(TP|¬C) should be 0.02, or not? P(TN|C) was given in the text: (Test returns positve AND cancer is actually present is 98%. Test is negative in only 95% where disease is NOT present) This is in the text. So did i calculate P(TP|¬C) and P(TN|C) wrong? If so, what would be correct, and why? Again, thank you for your help :) – user3556093 Jan 16 '20 at 09:46