A cancer test is 90 percent positive when cancer is present. It gives a false positive in 10 percent of the tests when the cancer is not present. If 2 percent of the population has this cancer what is the probability that someone has cancer given that the test is positive?I multiplied the 90 by 10 divided by 90 times 10 plus 2.

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Keely Fernandez · Accepted Answer

Do you know Bayes Theorem?If not you can get a feeling for it using an &quot;expected average tree diagram&quot;.Imagine 1000 patients.2% have cancer so you expect to have a split:- 980 cancer free- 20 with cancerLook at the 980 cancer free. We have 10% false positives (test indicates cancer when there is none); so you expect this splits:- 98 test positive cancer but are cancer free- 882 test negative for cancer but are cancer freeLook at the 20 with cancer. We have 90% true positives (test indicates cancer when there is cancer); so you expect this splits:- 18 test positive cancer and have cancer- 2 test negative for cancer and have cancerTherefore 98+18=116 test positive for cancer and of these only 18 have cancer.So the probability of having cancer, given a positive test, is small:   18  /  116  =  9  /  58  ≈  15.52  %.Bayes Theorem works as follows:  P  [  cancer | positive  ]  =                P      [      cancer and positive      ]                  P      [      positive      ]        .Note                          P        [        positive        ]                            =              P        [        (positive | cancer) or (positive | no cancer)        ]                                          =              P        [        positive | cancer        ]        +              P        [        positive | no cancer        ]        .              ,and so  P  [  cancer | positive  ]  =            0.02      (      0.9      )              0.02      (      0.9      )      +      (      0.98      )      (      0.1      )        =      9    58    .

prirodnogbk · Accepted Answer

Let A be the event that someone has cancer.  P  (  A  )  =  0.02Let B be the event that the cancer test returns positive. There are four cases to consider:- Test positive, cancer present:   P  (  B  |  A  )  =  0.9- Test negative, cancer present:   P  (      B    c    |  A  )  =  0.1- Test positive, cancer not present:   P  (  B  |      A    c    )  =  0.1- Test negative, cancer not present:   P  (      B    c    |      A    c    )  =  0.9The two probabilities are the same here, we will need   P  (  B  |  A  )  ,  P  (  B  |      A    c    ), which are given in the question.We need to find   P  (  A  |  B  ).Bayes&#039; Theorem tells us that:  P  (  A  |  B  )  =            P      (      B          |      A      )      P      (      A      )              P      (      B      )      and conditional probabilities tell us:                    P        (        B        )                            =        P        (        B              |        A        )        P        (        A        )        +        P        (        B              |                  A          c                )        P        (                  A          c                )                                          =        0.9        ×        0.02        +        0.1        ×        0.98                                          =        0.018        +        0.098                                          =        0.116            So   P  (  B  |  A  )  =            0.018      0.116        ≈  0.15517

A cancer test is 90 percent positive when cancer is present. It gives a false positive in 10 percent

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