Mixing
Find a value for a specific conditional probability
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So I'm given an assignment in probability that states the following:
One system is made out of 2 modules. The first module has an error with a probability of 0.4 and the second one with a probability of 0.2 independent from the first module. If there is an error only in the first module(there isn't an error in the second one) the system will not function with a probability of 0.6. For the second module that probability is 0.7. If there is an error in the two modules the system will not function with a probability of 0.8.
-Lets assume that the system is functioning. Find the probability that there isn't an error in both of the modules.
So judging by the assignment here's what I came up with:
A - The first module has an error
B - The second module has an error
M - The system is not functioning
$P(A) = 0.4$ $P(B) = 0.2$ $P(A`) = 0.6$ $P(B`) = 0.8$
$P(M|AB') = 0.6$ $P(M|A`B) = 0.7$ $P(M|AB)=0.8$
Using the law of total probability I've managed to calculate P ( M ) = 0.34 and P( M` ) = 0.66. Now for the actual assignment I need to find:
$ P(A`B`|M`) = frac{P(A`)P(B`)P(M`|A`B`)}{P(M`)}$ $(Using$ $Bayes'$ $ formula)$
However I have no idea how to find $P(M`|A`B`)$.
Any ideas?
probability conditional-probability
asked Jan 28 at 19:38
David MasonDavid Mason
587
$endgroup$
add a comment |
$begingroup$
So I'm given an assignment in probability that states the following:
One system is made out of 2 modules. The first module has an error with a probability of 0.4 and the second one with a probability of 0.2 independent from the first module. If there is an error only in the first module(there isn't an error in the second one) the system will not function with a probability of 0.6. For the second module that probability is 0.7. If there is an error in the two modules the system will not function with a probability of 0.8.
-Lets assume that the system is functioning. Find the probability that there isn't an error in both of the modules.
So judging by the assignment here's what I came up with:
A - The first module has an error
B - The second module has an error
M - The system is not functioning
$P(A) = 0.4$ $P(B) = 0.2$ $P(A`) = 0.6$ $P(B`) = 0.8$
$P(M|AB') = 0.6$ $P(M|A`B) = 0.7$ $P(M|AB)=0.8$
Using the law of total probability I've managed to calculate P ( M ) = 0.34 and P( M` ) = 0.66. Now for the actual assignment I need to find:
$ P(A`B`|M`) = frac{P(A`)P(B`)P(M`|A`B`)}{P(M`)}$ $(Using$ $Bayes'$ $ formula)$
However I have no idea how to find $P(M`|A`B`)$.
Any ideas?
probability conditional-probability
asked Jan 28 at 19:38
David MasonDavid Mason
587
$endgroup$
$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52
add a comment |
$begingroup$
So I'm given an assignment in probability that states the following:
One system is made out of 2 modules. The first module has an error with a probability of 0.4 and the second one with a probability of 0.2 independent from the first module. If there is an error only in the first module(there isn't an error in the second one) the system will not function with a probability of 0.6. For the second module that probability is 0.7. If there is an error in the two modules the system will not function with a probability of 0.8.
-Lets assume that the system is functioning. Find the probability that there isn't an error in both of the modules.
So judging by the assignment here's what I came up with:
A - The first module has an error
B - The second module has an error
M - The system is not functioning
$P(A) = 0.4$ $P(B) = 0.2$ $P(A`) = 0.6$ $P(B`) = 0.8$
$P(M|AB') = 0.6$ $P(M|A`B) = 0.7$ $P(M|AB)=0.8$
Using the law of total probability I've managed to calculate P ( M ) = 0.34 and P( M` ) = 0.66. Now for the actual assignment I need to find:
$ P(A`B`|M`) = frac{P(A`)P(B`)P(M`|A`B`)}{P(M`)}$ $(Using$ $Bayes'$ $ formula)$
However I have no idea how to find $P(M`|A`B`)$.
Any ideas?
probability conditional-probability
asked Jan 28 at 19:38
David MasonDavid Mason
587
$endgroup$
So I'm given an assignment in probability that states the following:
One system is made out of 2 modules. The first module has an error with a probability of 0.4 and the second one with a probability of 0.2 independent from the first module. If there is an error only in the first module(there isn't an error in the second one) the system will not function with a probability of 0.6. For the second module that probability is 0.7. If there is an error in the two modules the system will not function with a probability of 0.8.
-Lets assume that the system is functioning. Find the probability that there isn't an error in both of the modules.
So judging by the assignment here's what I came up with:
A - The first module has an error
B - The second module has an error
M - The system is not functioning
$P(A) = 0.4$ $P(B) = 0.2$ $P(A`) = 0.6$ $P(B`) = 0.8$
$P(M|AB') = 0.6$ $P(M|A`B) = 0.7$ $P(M|AB)=0.8$
Using the law of total probability I've managed to calculate P ( M ) = 0.34 and P( M` ) = 0.66. Now for the actual assignment I need to find:
$ P(A`B`|M`) = frac{P(A`)P(B`)P(M`|A`B`)}{P(M`)}$ $(Using$ $Bayes'$ $ formula)$
However I have no idea how to find $P(M`|A`B`)$.
Any ideas?
probability conditional-probability
probability conditional-probability
asked Jan 28 at 19:38
David MasonDavid Mason
587
asked Jan 28 at 19:38
David MasonDavid Mason
587
edited Jan 28 at 19:55
David Mason
asked Jan 28 at 19:38
David MasonDavid Mason
587
asked Jan 28 at 19:38
David MasonDavid Mason
587
asked Jan 28 at 19:38
David MasonDavid Mason
587
587
$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52
add a comment |
$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52
$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52
$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52
add a comment |
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$begingroup$
The problem doesn't say, so far as I can see. However, you're looking for the probability that the system functions if neither module is in error, and I suspect you are supposed to assume that is $1.$ If I were doing the assignment, that's what I would assume, and I would include a note to that effect.
$endgroup$
– saulspatz
Jan 28 at 19:52