Probability of a device having anomaly type A

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$A$ = component has anomaly A



$B$ = component has anomaly B.



$P(B)=0.09$



$P(A|B) = 0.5$



$P(A|B^c) = 0.01$



I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$



The next question is to figure out what $P(A)$ is.



I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$










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    $A$ = component has anomaly A



    $B$ = component has anomaly B.



    $P(B)=0.09$



    $P(A|B) = 0.5$



    $P(A|B^c) = 0.01$



    I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$



    The next question is to figure out what $P(A)$ is.



    I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$










    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $A$ = component has anomaly A



      $B$ = component has anomaly B.



      $P(B)=0.09$



      $P(A|B) = 0.5$



      $P(A|B^c) = 0.01$



      I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$



      The next question is to figure out what $P(A)$ is.



      I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$










      share|cite|improve this question













      $A$ = component has anomaly A



      $B$ = component has anomaly B.



      $P(B)=0.09$



      $P(A|B) = 0.5$



      $P(A|B^c) = 0.01$



      I figured out that the probability of device having both anomalies is equal to $P(B cap A) = P(B)*P(A|B)=0.045$



      The next question is to figure out what $P(A)$ is.



      I tried $P(A cap B) = P(A)*P(B|A)$ but I only know $P(A cap B)$ in this equation. I dont see how to apply what I know about $P(A|B^c)$







      probability statistics conditional-probability






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      asked Sep 2 at 10:46









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          We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$






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            accepted










            We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$






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              accepted










              We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$






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                down vote



                accepted







                up vote
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                down vote



                accepted






                We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$






                share|cite|improve this answer












                We have $mathbbP(B^c) = 1 - mathbbP(B) = 0.91$ , so you also know $mathbbP(A cap B^c) = 0.0091$ and $$ mathbbP(A) = mathbbP(A cap B) + mathbbP(A cap B^c) = 0.0541 $$







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                answered Sep 2 at 10:49









                LucaMac

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