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Maths question - Help! watch

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    A hospital uses an test that gives the following results:

    For a patient who really has cancer, the test is positive 97% of the time.

    For a patient who does not have cancer, the test is positive 2% of the time (‘false positive’).

    Let’s say that 1% of all patients have cancer.

    If Mr Smith has a test, and the result comes back negative, what is the chance that he really is cancer-free?
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    (Original post by Bigbosshead)
    A hospital uses an test that gives the following results:

    For a patient who really has cancer, the test is positive 97% of the time.

    For a patient who does not have cancer, the test is positive 2% of the time (‘false positive’).

    Let’s say that 1% of all patients have cancer.

    If Mr Smith has a test, and the result comes back negative, what is the chance that he really is cancer-free?
    Ugh my stats is rusty.

    This is one of those "What is the probability of A given B"

     \mathrm{P(A \ | \ B)} = \dfrac {P(A \cap B)}{P(B)}

    Let A be the probability of not having cancer. Let B be the probability of having a negative test.

    To get  \ P(A \cap B) you just multiply the probability of not getting cancer by the probability of getting a negative test from that.
    It's helpful to draw a probability tree diagram for this.

    The probability of not getting cancer is 0.99 (99%). If you don't have cancer, the probability of getting a negative result is 0.98 (1-0.02).
    Spoiler:
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    Therefore  \ P(A \cap B)  = 0.99 \times 0.98  = 0.9702
    To get  \ P(B) you need to find the ways to get a negative test. We already found way which is to not get cancer and to get a negative test. The other way is to have cancer and then get a negative test.
    Again a probability tree diagram is helpful.
    Spoiler:
    Show
    Therefore  \ P(B)  = (0.99 \times 0.98) + (0.01 \times 0.03) = 0.9705
    Now plug it into the equation.
 
 
 
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