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*In a hospital, the probability that a patient has Disease M is $0.003$. A test is carried out for Disease M. This test has a probability p of giving a positive result if patient has Disease M and the same probability p of giving a negative result if patient does not have Disease M. A patient is given the test. Taking p to be $0.897$, find the probability that the result of the test is positive.

This problem can be very easily solved with the aid of a tree diagram. What I have a problem with is this:

To find the p(test positive), I simply take $0.003p + 0.997(1-p)$, and of course I'll then use $p = 0.897$. I know this in an extremely mechanistic manner and I feel like I don't truly understand it intuitively. What I'm finding now is p(patient has disease and test positive) + p(patient does not have disease and test positive). Why does this automatically give me p(test positive)? Why are they equivalent? Can somebody explain in a way which would allow me to understand this intuitively? Ridiculous as it may sound, why can't I just say, the probability of the test being positive is simply p + (1-p), i.e. disregarding whether the patient actually has the disease or not? I mean, the question asks for the probability of the test being positive - it didn't ask about the status of the patient!

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What I'm finding now is p(patient has disease and test positive) + p(patient does not have disease and test positive). Why does this automatically give me p(test positive)

Let $B$ the event a patient is diseased-$\overline B$ the converse event.

And $A$ is the event that the test is positive.

You are asking why

$=P(A\cap B)+P(A\cap \overline B)=P(A)$

To see that this is true you can draw a venn diagram:

enter image description here

1) I think you can identify $A \cap B$.

2) To get $A\cap \overline B$ you have to mark $A$ first. I have marked it with green lines. And $\overline B$ is the area marked with black lines. The intersection of $A$ and $\overline B$ is the area with green and black lines. You can say it is the area $A$ without the intersection of $A$ and $B$.

You have to sum up the areas of 1) and 2) to get $A$. I hope it is comprehensible.


Using the equation of conditional probability $P(A\cap B)+P(A\cap \overline B)$ is equal to

$P(B)\cdot P(A|B)+P(\overline B)\cdot P(A|\overline B)=P(A)$

with $P(B)=0.003, P(\overline B)=1-P(B)=1-0.003, P(A|B)=0.897$ and

$P(A|\overline B)=1-P(\overline A|\overline B)=1-0.897$

Thus $P(A)=0.003\cdot 0.897+(1-0.003)\cdot (1-0.897)$

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  • $\begingroup$ @Charlz97 You are welcome. Could you elaborate or reformulate your question ? I have to admit I don´t really know what do you mean or you referring to. $\endgroup$ – callculus Dec 2 '16 at 0:19
  • $\begingroup$ Oops sorry thought I could delete the whole comment thread. I worked out what I was trying to ask and yeah its kinda nonsensical. But thanks anyways! $\endgroup$ – Charlz97 Dec 2 '16 at 0:35
  • $\begingroup$ @Charlz97 It would be nice if you would accept my answer, especially because a troll has downvoted my answer. $\endgroup$ – callculus Dec 2 '16 at 0:37
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I'm not sure whether or not this will help you to understand, but I will give it a try...


First, split the given scenario into disjoint events:

  • The patient is sick and the test is positive
  • The patient is not sick and the test is positive
  • The patient is sick and the test is not positive
  • The patient is not sick and the test is not positive

Then, find the probabilities of the desired events:

  • The patient is sick $\color\red{\text{and}}$ the test is positive: $(0.003)\color\red\times(0.897)$
  • The patient is not sick $\color\red{\text{and}}$ the test is positive: $(1-0.003)\color\red\times(1-0.897)$

Finally, find the probability of the unified event:

  • The 1st event $\color\green{\text{or}}$ the 2nd event: $(0.003)\times(0.897)\color\green+(1-0.003)\times(1-0.897)$
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"[T]he probability that a patient has Disease M is $0.003$." This means that (on average) out of every $1000$ people, $3$ have Disease M and $997$ do not have Disease M.

"This test has a probability $p$ of giving a positive result if patient has Disease M and the same probability $p$ of giving a negative result if patient does not have Disease M. A patient is given the test. Taking $p$ to be $0.897$, ... ." This is actually two separate statements, one about people who have the disease and one about about people who don't:

  • Out of every $1000$ people who do have Disease M, on average $897$ will test positive and $103$ will test negative.
  • Out of every $1000$ people who do not have Disease M, on average $103$ will test positive and $897$ will test negative.

Now suppose we administer this test to one million people. The expected results are:

Of the people tested, $3000$ have the disease. Of those $3000$ people, $2691$ ($897$ out of every thousand) tested positive and $309$ ($103$ out of every thousand) tested negative.

Of the people tested, $997000$ do not have the disease. Of those $997000$ people, $102691$ ($103$ out of every thousand) tested positive and $894309$ ($897$ out of every thousand) tested negative.

That's four different groups of people, and if we knew exactly who had the disease and who did not, we could put each person's name on one of four lists according to which group they belonged to. But the question is asking us just to list the people who tested positive. There will be $2691$ ill-and-the-test-detected-it people on that list, and $102691$ healthy-but-tested-positive-anyway people, for a total of $105382$ people with positive tests.

In summary, given the numbers in the problem statement (probability $0.003$ to have the disease, $p=0.897$), we have

  • $1000000 \times 0.003 \times 0.897$ ill-and-the-test-detected-it people, and
  • $1000000 \times 0.997 \times (1 - 0.897)$ healthy-but-tested-positive-anyway people,

for a total of $1000000 \times (0.003 \times 0.897 + 0.997 \times (1 - 0.897))$ positive tests.

If we went instead with the formula $p + (1 - p)$, notice that $p + (1 - p) = 1$, which tells us that everyone tests positive!

To put it in much less formal terms, since the test has a much higher percentage of positive results ($89.7\%$) for sick people than for healthy people ($10.3\%$), shouldn't we intuitively believe that the fact of a person being sick or healthy has some effect on whether they test positive? And that if a lot more people in our population were sick, we'd have more positive results?

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  • $\begingroup$ If you contrive arbitrary formulas that aren't answering the original question, to make sense of them we might just have to contrive arbitrary variations on the question. For any patient, the test has only two possible outcomes, with probabilities $p$ and $1-p$, so $p + (1-p)$ is the probability that you get an outcome (positive or negative) when you give the test; which of course always happens. If you test two healthy people, the probability they get different test results is $p(1-p)$. The possible interpretations are limited only by your patience for contriving them. $\endgroup$ – David K Dec 2 '16 at 0:44

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