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Math Help - probability of illnes

  1. #1
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    probability of illnes

    A= the paitent is ill: P(A)=0,001
    A*= the paitent is not ill
    B= the diagnos shows ill
    B*=the diagnos shows not ill

    given: p(B|A)= 0,8
    p(B*|A*)= 0,9

    Question: What the probability that the paitent is ill and the diagnos shows ill?

    started with: P(A|B)= P(A∩B)/P(B)= (P(A)xP(B|A))/P(B)

    .....
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  2. #2
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    Hello, sf1903!

    \begin{array}{ccc}A &= & \text{paitent is ill} \\<br />
A^* & = & \text{patient is not ill} \\<br />
B & = & \text{test shows ill} \\<br />
B^* & = & \text{test shows not ill} \end{array} \qquad P(A) \:= \:0.001\;{\color{blue}[1]}


    Given: . \begin{array}{cccc}P(B|A) &=& 0.8 & {\color{blue}[2]}\\<br />
P(B^*|A^*) &= & 0.9\end{array}

    What the probability that the paitent is ill and the test shows ill?
    They are asking for: . P(A \wedge B)


    Using Bayes' Theorem on [2]: . P(B|A) \:=\:\frac{P(A \wedge B)}{P(A)} \:=\:0.8

    Substitute [1]: . \frac{P(A \wedge B)}{0.001} \:=\:0.8

    Therefore: . P(A \wedge B) \;=\;(0.001)(0.8) \;=\;0.0008

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  3. #3
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    Cool

    Quote Originally Posted by sf1903 View Post
    A= the paitent is ill: P(A)=0,001
    A*= the paitent is not ill
    B= the diagnos shows ill
    B*=the diagnos shows not ill

    given: p(B|A)= 0,8
    p(B*|A*)= 0,9

    Question: What the probability that the paitent is ill and the diagnos shows ill?

    started with: P(A|B)= P(A∩B)/P(B)= (P(A)xP(B|A))/P(B)

    .....
    b) how many % of the patients are diagnosed wrong?

    P(B*
    |S)=0,2
    P(B
    | S*)=0,1


    ???
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  4. #4
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    Quote Originally Posted by sf1903 View Post
    b) how many % of the patients are diagnosed wrong?

    P(B*|S)=0,2
    P(B| S*)=0,1


    ???
    Using the notation you introduced in your first post (what's S and S* meant to be?): Pr(B | A') = 0.1 and Pr(B' | A) = 0.2.

    So the answer is 0.1 + 0.2 = 0.3
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