Mary is certainly one of medical doctor Brown's patients. She has carried out a domestic pregnancy take a look at which has given a high quality end result. what's the possibility that the pregnancy take a look at used by doctor Brown in his surgical procedure will say Mary is pregnant given that the home check became nice?Home pregnancy test is 85% accurateDoctor Brown pregnancy test is 95% accurate20 females are pregnant and 80 females are not

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Mary is certainly one of medical doctor Brown&#039;s patients. She has carried out a domestic pregnancy take a look at which has given a high quality end result. what&#039;s the possibility that the pregnancy take a look at used by doctor Brown in his surgical procedure will say Mary is pregnant given that the home check became nice?Home pregnancy test is 85% accurateDoctor Brown pregnancy test is 95% accurate20 females are pregnant and 80 females are not

Teagan Raymond · Accepted Answer

Strictly speaking, you don&#039;t have enough information to solve this. Here&#039;s how I&#039;d approach it.Let X be the random variable which is 1 if the home test is positive and 0 otherwise. Similarly let   Y  =  1 if Dr. Brown&#039;s test is positive, and let P=1 if Mary is actually pregnant. We want the conditional probability   P  r  (  Y  =  1  ∣  X  =  1  ). Write                          P        r        (        Y        =        1        ∣        X        =        1        )                            =                            P        r        (        X        =        1        ∧        Y        =        1        )                  /                P        r        (        X        =        1        )                            =                                                  P            r            (            X            =            1            ∧            Y            =            1            ∣            P            =            1            )            P            r            (            P            =            1            )            +            P            r            (            X            =            1            ∧            Y            =            1            ∣            P            =            0            )            P            r            (            P            =            0            )                                P            r            (            X            =            1            ∣            P            =            1            )            P            r            (            P            =            1            )            +            P            r            (            X            =            1            ∣            P            =            0            )            P            r            (            P            =            0            )            .                              If we&#039;re willing to assume X and Y are independent given P, we can write   P  r  (  X  =  1  ∧  Y  =  1  ∣  P  =  1  )  =  P  r  (  X  =  1  ∣  P  =  1  )    P  r  (  Y  =  1  ∣  P  =  1  )  . This may be a reasonable assumption if Dr. Brown&#039;s test works on different principles than the home test, but it might not. Given that assumption, this gives an answer of            (      0.85      )      (      0.95      )      (      0.2      )      +      (      0.15      )      (      0.05      )      (      0.8      )              0.85      )      (      0.2      )      +      (      0.15      )      (      0.8      )        ≈  0.56

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