Probability that at least 1 machine breaks down












1












$begingroup$


In the manufactoring plant of a company two large production machines are used. Machine A breaks down with probability of 0.01, whereas a breakdown of machine B occurs with probability 0.005. Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1. What is the probability that the production stops, i.e. at least one of the production machines breaks down?



Basically, my solution so far is:

C - event of interest where at least one production machine breaks down (=1-C¬)

C¬ - complement of C, aka production doesn't stop as no machines break down



P(A) = 0.01 ==> P(A¬) = 1-0.01=0.99

P(B) = 0.005 ==> P(B¬) = 1-0.005 = 0.995

P(C¬)=P(A¬)*P(B¬)=0.98505 chance that no machine will break down



==> 1 - 0.98505 = 0.01495 chance that at least 1 machine will break down and production will stop



However, I feel like I'm missing something because this solutions seems just too easy and I don't know what is P(BlA)=0.1 given for (from: "Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1")



Any insight will be much appreciated. Thank you in advance!






probability statistics







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  • 1




    $begingroup$
    Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:25












  • $begingroup$
    This approach looks fine to me.
    $endgroup$
    – Warren Hill
    Dec 11 '18 at 16:53










  • $begingroup$
    We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
    $endgroup$
    – Sagnik
    Dec 11 '18 at 17:18


















1












$begingroup$


In the manufactoring plant of a company two large production machines are used. Machine A breaks down with probability of 0.01, whereas a breakdown of machine B occurs with probability 0.005. Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1. What is the probability that the production stops, i.e. at least one of the production machines breaks down?



Basically, my solution so far is:

C - event of interest where at least one production machine breaks down (=1-C¬)

C¬ - complement of C, aka production doesn't stop as no machines break down



P(A) = 0.01 ==> P(A¬) = 1-0.01=0.99

P(B) = 0.005 ==> P(B¬) = 1-0.005 = 0.995

P(C¬)=P(A¬)*P(B¬)=0.98505 chance that no machine will break down



==> 1 - 0.98505 = 0.01495 chance that at least 1 machine will break down and production will stop



However, I feel like I'm missing something because this solutions seems just too easy and I don't know what is P(BlA)=0.1 given for (from: "Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1")



Any insight will be much appreciated. Thank you in advance!






probability statistics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:25












  • $begingroup$
    This approach looks fine to me.
    $endgroup$
    – Warren Hill
    Dec 11 '18 at 16:53










  • $begingroup$
    We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
    $endgroup$
    – Sagnik
    Dec 11 '18 at 17:18
















1












1








1





$begingroup$


In the manufactoring plant of a company two large production machines are used. Machine A breaks down with probability of 0.01, whereas a breakdown of machine B occurs with probability 0.005. Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1. What is the probability that the production stops, i.e. at least one of the production machines breaks down?



Basically, my solution so far is:

C - event of interest where at least one production machine breaks down (=1-C¬)

C¬ - complement of C, aka production doesn't stop as no machines break down



P(A) = 0.01 ==> P(A¬) = 1-0.01=0.99

P(B) = 0.005 ==> P(B¬) = 1-0.005 = 0.995

P(C¬)=P(A¬)*P(B¬)=0.98505 chance that no machine will break down



==> 1 - 0.98505 = 0.01495 chance that at least 1 machine will break down and production will stop



However, I feel like I'm missing something because this solutions seems just too easy and I don't know what is P(BlA)=0.1 given for (from: "Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1")



Any insight will be much appreciated. Thank you in advance!






probability statistics







share|cite|improve this question









$endgroup$




In the manufactoring plant of a company two large production machines are used. Machine A breaks down with probability of 0.01, whereas a breakdown of machine B occurs with probability 0.005. Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1. What is the probability that the production stops, i.e. at least one of the production machines breaks down?



Basically, my solution so far is:

C - event of interest where at least one production machine breaks down (=1-C¬)

C¬ - complement of C, aka production doesn't stop as no machines break down



P(A) = 0.01 ==> P(A¬) = 1-0.01=0.99

P(B) = 0.005 ==> P(B¬) = 1-0.005 = 0.995

P(C¬)=P(A¬)*P(B¬)=0.98505 chance that no machine will break down



==> 1 - 0.98505 = 0.01495 chance that at least 1 machine will break down and production will stop



However, I feel like I'm missing something because this solutions seems just too easy and I don't know what is P(BlA)=0.1 given for (from: "Further, if machine A breaks down the probability of a breakdown of machine B is equal to 0.1")



Any insight will be much appreciated. Thank you in advance!






probability statistics




probability statistics






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asked Dec 11 '18 at 16:22









VRTVRT

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626








  • 1




    $begingroup$
    Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:25












  • $begingroup$
    This approach looks fine to me.
    $endgroup$
    – Warren Hill
    Dec 11 '18 at 16:53










  • $begingroup$
    We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
    $endgroup$
    – Sagnik
    Dec 11 '18 at 17:18
















  • 1




    $begingroup$
    Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
    $endgroup$
    – Felix Marin
    Dec 11 '18 at 16:25












  • $begingroup$
    This approach looks fine to me.
    $endgroup$
    – Warren Hill
    Dec 11 '18 at 16:53










  • $begingroup$
    We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
    $endgroup$
    – Sagnik
    Dec 11 '18 at 17:18










1




1




$begingroup$
Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
$endgroup$
– Felix Marin
Dec 11 '18 at 16:25






$begingroup$
Please, read the MSE $texttt{MathJax}$ Basic Tutorial and Quick Reference. Thanks.
$endgroup$
– Felix Marin
Dec 11 '18 at 16:25














$begingroup$
This approach looks fine to me.
$endgroup$
– Warren Hill
Dec 11 '18 at 16:53




$begingroup$
This approach looks fine to me.
$endgroup$
– Warren Hill
Dec 11 '18 at 16:53












$begingroup$
We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
$endgroup$
– Sagnik
Dec 11 '18 at 17:18






$begingroup$
We have P(B) << P(A). So the final probability will depend majorly on P(A). So we can see that the final answer = 1 - (P(A) + P(B)) which is approximately equal to P(A) = 0.01
$endgroup$
– Sagnik
Dec 11 '18 at 17:18












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$begingroup$

You can use the addition rule:
$$begin{align}P(Acup B)&=P(A)+P(B)-P(Acap B)=\
&=0.01+0.005-P(A)cdot P(B|A)=\
&=0.01+0.005-0.01cdot 0.1=\
&=0.014.end{align}$$
Also, note:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$






share|cite|improve this answer









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    $begingroup$

    You can use the addition rule:
    $$begin{align}P(Acup B)&=P(A)+P(B)-P(Acap B)=\
    &=0.01+0.005-P(A)cdot P(B|A)=\
    &=0.01+0.005-0.01cdot 0.1=\
    &=0.014.end{align}$$
    Also, note:
    $$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can use the addition rule:
      $$begin{align}P(Acup B)&=P(A)+P(B)-P(Acap B)=\
      &=0.01+0.005-P(A)cdot P(B|A)=\
      &=0.01+0.005-0.01cdot 0.1=\
      &=0.014.end{align}$$
      Also, note:
      $$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can use the addition rule:
        $$begin{align}P(Acup B)&=P(A)+P(B)-P(Acap B)=\
        &=0.01+0.005-P(A)cdot P(B|A)=\
        &=0.01+0.005-0.01cdot 0.1=\
        &=0.014.end{align}$$
        Also, note:
        $$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$






        share|cite|improve this answer









        $endgroup$



        You can use the addition rule:
        $$begin{align}P(Acup B)&=P(A)+P(B)-P(Acap B)=\
        &=0.01+0.005-P(A)cdot P(B|A)=\
        &=0.01+0.005-0.01cdot 0.1=\
        &=0.014.end{align}$$
        Also, note:
        $$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 18:03









        farruhotafarruhota

        20.2k2738




        20.2k2738






























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