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Find the probability of a faulty device when knowing the probability of the cause
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(Note: the problem is translated from Japanese so there might be some
mistakes)
A machine includes three devices: A1, A2, A3. This machine can have
these kind of problems: Overheated (B1), Violent motion (B2) and Stop
with no reason in working mode (B3). These problem can be caused only
when one or some of 3 devices A1, A2, A3 be broken. The probabilities
of that each devices is broken led to each problem are shown as below.
beginarrayc hline & B1 & B2 & B3 \ hline A1 & 0.20 &
0.30 & 0.50\ hline A2 & 0.40 & 0.50 & 0.10\ hline A3 & 0.60 & 0.10 & 0.30\ endarray
Pr(A1), Pr(A2), Pr(A3) are respectively the probability that each
devices are broken. Given that Pr(A3) = 2Pr(A1), Pr(A2) = 1.5Pr(A1),
Pr(A1) + Pr(A2) + Pr(A3) = 1.
When the probability of B3 to occur is 0.23, find the probability that A3 is broken.
Find the probability that B1 occurs.
Find the probability that at least one of B1, B2, B3 occurs. B1, B2, B3 are independent.
For number 1, I don't know that they ask for Pr(A3) or not, but if it's Pr(A3), isn't it can be calculated through the given condition, like below?
Pr(A1) + Pr(A2) + Pr(A3) = Pr(A1) + 1.5Pr(A1) + 2Pr(A1) = 4.5Pr(A1) = 1
Then
Pr(A1) = 0.(2), Pr(A3) = 2Pr(A1) = 0.(4)
So why it's related to the number 0.23?
For number 2, I did it but I'm not sure, as below:
Pr(B1) = Pr(B1|A1)*Pr(A1) + Pr(B1|A2)*Pr(A2) + Pr(B1|A3)*Pr(A3)
With according to the table Pr(B1|A1) is 0.2, Pr(B1|A2) is 0.40... (If I'm not mistaken), and Pr(A1), Pr(A2), Pr(A3) can be found like in question 1.
For number 3, with Pr(B1), Pr(B2), Pr(B3) found in the same way as question 2, we calculate Pr(B1) + Pr(B2) + Pr(B3) - Pr(B1)Pr(B2) - Pr(B2)Pr(B3) - Pr(B3)Pr(B1) + Pr(B1)Pr(B2)Pr(B3). Is it true? I'm not really confident.
probability statistics
asked 2 days ago
nghia95
274
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up vote
1
down vote
favorite
(Note: the problem is translated from Japanese so there might be some
mistakes)
A machine includes three devices: A1, A2, A3. This machine can have
these kind of problems: Overheated (B1), Violent motion (B2) and Stop
with no reason in working mode (B3). These problem can be caused only
when one or some of 3 devices A1, A2, A3 be broken. The probabilities
of that each devices is broken led to each problem are shown as below.
beginarrayc hline & B1 & B2 & B3 \ hline A1 & 0.20 &
0.30 & 0.50\ hline A2 & 0.40 & 0.50 & 0.10\ hline A3 & 0.60 & 0.10 & 0.30\ endarray
Pr(A1), Pr(A2), Pr(A3) are respectively the probability that each
devices are broken. Given that Pr(A3) = 2Pr(A1), Pr(A2) = 1.5Pr(A1),
Pr(A1) + Pr(A2) + Pr(A3) = 1.
When the probability of B3 to occur is 0.23, find the probability that A3 is broken.
Find the probability that B1 occurs.
Find the probability that at least one of B1, B2, B3 occurs. B1, B2, B3 are independent.
For number 1, I don't know that they ask for Pr(A3) or not, but if it's Pr(A3), isn't it can be calculated through the given condition, like below?
Pr(A1) + Pr(A2) + Pr(A3) = Pr(A1) + 1.5Pr(A1) + 2Pr(A1) = 4.5Pr(A1) = 1
Then
Pr(A1) = 0.(2), Pr(A3) = 2Pr(A1) = 0.(4)
So why it's related to the number 0.23?
For number 2, I did it but I'm not sure, as below:
Pr(B1) = Pr(B1|A1)*Pr(A1) + Pr(B1|A2)*Pr(A2) + Pr(B1|A3)*Pr(A3)
With according to the table Pr(B1|A1) is 0.2, Pr(B1|A2) is 0.40... (If I'm not mistaken), and Pr(A1), Pr(A2), Pr(A3) can be found like in question 1.
For number 3, with Pr(B1), Pr(B2), Pr(B3) found in the same way as question 2, we calculate Pr(B1) + Pr(B2) + Pr(B3) - Pr(B1)Pr(B2) - Pr(B2)Pr(B3) - Pr(B3)Pr(B1) + Pr(B1)Pr(B2)Pr(B3). Is it true? I'm not really confident.
probability statistics
asked 2 days ago
nghia95
274
Sorry it's my mistake. Already edited.
– nghia95
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago
add a comment |Â
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1
down vote
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up vote
1
down vote
favorite
(Note: the problem is translated from Japanese so there might be some
mistakes)
A machine includes three devices: A1, A2, A3. This machine can have
these kind of problems: Overheated (B1), Violent motion (B2) and Stop
with no reason in working mode (B3). These problem can be caused only
when one or some of 3 devices A1, A2, A3 be broken. The probabilities
of that each devices is broken led to each problem are shown as below.
beginarrayc hline & B1 & B2 & B3 \ hline A1 & 0.20 &
0.30 & 0.50\ hline A2 & 0.40 & 0.50 & 0.10\ hline A3 & 0.60 & 0.10 & 0.30\ endarray
Pr(A1), Pr(A2), Pr(A3) are respectively the probability that each
devices are broken. Given that Pr(A3) = 2Pr(A1), Pr(A2) = 1.5Pr(A1),
Pr(A1) + Pr(A2) + Pr(A3) = 1.
When the probability of B3 to occur is 0.23, find the probability that A3 is broken.
Find the probability that B1 occurs.
Find the probability that at least one of B1, B2, B3 occurs. B1, B2, B3 are independent.
For number 1, I don't know that they ask for Pr(A3) or not, but if it's Pr(A3), isn't it can be calculated through the given condition, like below?
Pr(A1) + Pr(A2) + Pr(A3) = Pr(A1) + 1.5Pr(A1) + 2Pr(A1) = 4.5Pr(A1) = 1
Then
Pr(A1) = 0.(2), Pr(A3) = 2Pr(A1) = 0.(4)
So why it's related to the number 0.23?
For number 2, I did it but I'm not sure, as below:
Pr(B1) = Pr(B1|A1)*Pr(A1) + Pr(B1|A2)*Pr(A2) + Pr(B1|A3)*Pr(A3)
With according to the table Pr(B1|A1) is 0.2, Pr(B1|A2) is 0.40... (If I'm not mistaken), and Pr(A1), Pr(A2), Pr(A3) can be found like in question 1.
For number 3, with Pr(B1), Pr(B2), Pr(B3) found in the same way as question 2, we calculate Pr(B1) + Pr(B2) + Pr(B3) - Pr(B1)Pr(B2) - Pr(B2)Pr(B3) - Pr(B3)Pr(B1) + Pr(B1)Pr(B2)Pr(B3). Is it true? I'm not really confident.
probability statistics
asked 2 days ago
nghia95
274
(Note: the problem is translated from Japanese so there might be some
mistakes)
A machine includes three devices: A1, A2, A3. This machine can have
these kind of problems: Overheated (B1), Violent motion (B2) and Stop
with no reason in working mode (B3). These problem can be caused only
when one or some of 3 devices A1, A2, A3 be broken. The probabilities
of that each devices is broken led to each problem are shown as below.
beginarrayc hline & B1 & B2 & B3 \ hline A1 & 0.20 &
0.30 & 0.50\ hline A2 & 0.40 & 0.50 & 0.10\ hline A3 & 0.60 & 0.10 & 0.30\ endarray
Pr(A1), Pr(A2), Pr(A3) are respectively the probability that each
devices are broken. Given that Pr(A3) = 2Pr(A1), Pr(A2) = 1.5Pr(A1),
Pr(A1) + Pr(A2) + Pr(A3) = 1.
When the probability of B3 to occur is 0.23, find the probability that A3 is broken.
Find the probability that B1 occurs.
Find the probability that at least one of B1, B2, B3 occurs. B1, B2, B3 are independent.
For number 1, I don't know that they ask for Pr(A3) or not, but if it's Pr(A3), isn't it can be calculated through the given condition, like below?
Pr(A1) + Pr(A2) + Pr(A3) = Pr(A1) + 1.5Pr(A1) + 2Pr(A1) = 4.5Pr(A1) = 1
Then
Pr(A1) = 0.(2), Pr(A3) = 2Pr(A1) = 0.(4)
So why it's related to the number 0.23?
For number 2, I did it but I'm not sure, as below:
Pr(B1) = Pr(B1|A1)*Pr(A1) + Pr(B1|A2)*Pr(A2) + Pr(B1|A3)*Pr(A3)
With according to the table Pr(B1|A1) is 0.2, Pr(B1|A2) is 0.40... (If I'm not mistaken), and Pr(A1), Pr(A2), Pr(A3) can be found like in question 1.
For number 3, with Pr(B1), Pr(B2), Pr(B3) found in the same way as question 2, we calculate Pr(B1) + Pr(B2) + Pr(B3) - Pr(B1)Pr(B2) - Pr(B2)Pr(B3) - Pr(B3)Pr(B1) + Pr(B1)Pr(B2)Pr(B3). Is it true? I'm not really confident.
probability statistics
asked 2 days ago
nghia95
274
edited 2 days ago
asked 2 days ago
nghia95
274
asked 2 days ago
nghia95
274
asked 2 days ago
nghia95
274
274
Sorry it's my mistake. Already edited.
– nghia95
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago
add a comment |Â
Sorry it's my mistake. Already edited.
– nghia95
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago
Sorry it's my mistake. Already edited.
– nghia95
2 days ago
Sorry it's my mistake. Already edited.
– nghia95
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago
add a comment |Â
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Sorry it's my mistake. Already edited.
– nghia95
2 days ago
In problems like 3, where you are asked to find the probability that at least one of several events occurs, it is frequently easier to find the probability that none of the events occurs, and then subtract from 1. You might compare the result with your approach above.
– awkward
2 days ago