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The prevalence of breast cancer in women over 40 in country X is estimated to be $0.8\%$ (i.e., $8$ in every $1,000$ women in that age group).

Mammograms test for the presence of breast cancer. A positive result indicates that the disease is present. A negative result indicates that it is not.

The sensitivity of a mammogram test for breast cancer is estimated to be $90\%$. This is the probability that the mammogram will give a positive result when the person being tested does have breast cancer.

The false positive rate for the mammogram is $7.5\%$. This is the probability that the mammogram will give a positive test result when the person being tested does not have breast cancer.

(At this point, there are $816$ women who test positive. I've calculated that myself ;) ) All women who test positive ($816$) in the mammogram are referred for a further, different examination, which however has the same sensitivity and false positive rates as the first test.

What is the probability that a woman referred for this examination and testing positive again, actually does have breast cancer?

What is the probability that a woman referred for this second examination and testing negative this time, actually does not have breast cancer?

Thanks so much. I'm struggling.

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closed as off-topic by Johanna, heropup, Mark Fantini, user147263, Michael Albanese Mar 16 '15 at 1:20

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  • $\begingroup$ Please edit your title to make it more descriptive. $\endgroup$ – Johanna Mar 15 '15 at 20:48
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Ah, the classical cancer detection problem. Let's use a large sample for this one. Say, 1,000,000 women.

Out of these 1,000,000 women :

  • 8,000 will have cancer
  • 7,200 (90% of 8,000) of them will get a positive test result
  • 800 of them will have a negative result (a false negative)
  • 992,000 will NOT have cancer
  • 74,400 (7.5% of 992,000) of them will get a (false) positive test result
  • 917,600 will get a negative result

So, out of the 81,600 (7,200+74,400) women who get a positive result, only 7,200 of them actually have cancer. Meaning that if your test is positive you have a 9.12% probability to have cancer.

Now, if we gather these 81,600 women :

  • From the 7,200 having cancer :
  • 6,480 will get a second positive result
  • 720 will get a negative one
  • From the 74,400 NOT having cancer :
  • 5,580 will get a second (false) positive result
  • 68,820 will get a negative one

To conclude, out of the 12,060 women with a double positive result, 6,480 (53,7%) will actually have cancer. Conversely, out of the 69,540 women having a positive, then negative result, 68,820 (99,0%) actually do NOT have cancer.

A tree to visualize all of this may be useful !

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  • $\begingroup$ A tiny correction: $7.5\%$ of $992,\!000$ is $744,\!000$ rather than $774,\!000$ $\endgroup$ – Henry Mar 15 '15 at 21:00
  • $\begingroup$ There are actually $74,400$ ($7.5\%$ of $992,000$) women who will get a false positive result on the first test. While I corrected the initial number, you will have to correct the others. $\endgroup$ – N. F. Taussig Mar 15 '15 at 23:32
  • $\begingroup$ Numbers corrected. Thanks! $\endgroup$ – jrmyp Mar 16 '15 at 9:59
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You seem to have started with $10000$ women over $40$. It might have been easier if you had multiplied this by $10$ or $100$, but let's take your approach.

  • an expected $80$ have breast cancer:

  • an expected $9920$ do not have breast cancer

Now you test them the first time:

  • an expected $72$ have breast cancer and test positive

  • an expected $744$ do not have breast cancer but test positive

making an expected total of $816$ testing positive the first time. Test those a second time:

  • an expected $64.8$ have breast cancer and test positive twice, while an expected $7.2$ have breast cancer and test positive then negative

  • an expected $55.8$ do not have breast cancer but test positive twice , while an expected $688.2$ do not have breast cancer and test positive then negative

So testing positive twice gives a probability of $\dfrac{64.8}{64.8+55.8} \approx 54\%$ of having breast cancer, while testing positive then negative gives a probability of $\dfrac{688.2}{7.2+688.2} \approx 99\%$ of not having breast cancer.

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