Probability that at least 1 machine breaks down
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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
asked Dec 11 '18 at 16:22
VRTVRT
957
$endgroup$
add a comment |
$begingroup$
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
asked Dec 11 '18 at 16:22
VRTVRT
957
$endgroup$
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
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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
add a comment |
$begingroup$
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
asked Dec 11 '18 at 16:22
VRTVRT
957
$endgroup$
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
asked Dec 11 '18 at 16:22
VRTVRT
957
asked Dec 11 '18 at 16:22
VRTVRT
957
asked Dec 11 '18 at 16:22
VRTVRT
957
asked Dec 11 '18 at 16:22
VRTVRT
957
asked Dec 11 '18 at 16:22
VRTVRT
957
957
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
add a comment |
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
add a comment |
1 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:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
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add a comment |
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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:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
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:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
$endgroup$
add a comment |
$begingroup$
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:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
$endgroup$
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:
$$P(Acap B)=P(A)cdot P(B|A)=P(B)cdot P(A|B).$$
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
answered Dec 11 '18 at 18:03
farruhotafarruhota
20.5k2739
20.5k2739
add a comment |
add a comment |
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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