Mixing
Conditional probability : what is $mathbb P(F)$ for $F:$“ power supply fluctuate”.
$begingroup$
Here's the statement:
Quality control in a factory detects that 8% of the produced items are class $B$ quality only (i.e. not flawless). With a probability of 4% the production line functions suboptimally due to pollution, and with 50% probability the material used has quality issues. If pollution occurs, the items are class $B$ with a probability of 10%. Technicians also find that if there are fluctuations in the power supply then there is 10% probability that the product is class $B$. Based on these data, calculate the probability that power supply fluctuates.
I really have difficulty understanding the statement. As I see the problem: Let $P$ be the event "pollution", $Q$ the event a quality defect, $F$ the event "power supply fluctuate", and $B$ the event "$B$ quality".
I see the problem as follows. (see picture below).
So we just have
$$x=mathbb P(F)=1-0.04-0.5=0.46?$$
But I don't use $mathbb P(B)$. So maybe it should be $$mathbb P(B)=mathbb P(P)mathbb P(Bmid P)+mathbb P(Q)mathbb P(Bmid Q)+mathbb P(F)mathbb P(Bmid F)$$
which is equivalent to $$x=frac{0.01}{0.1}=0.1,$$
could it be possible?
probability
edited Jan 17 at 19:39
Bernard
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
$endgroup$
add a comment |
$begingroup$
Here's the statement:
Quality control in a factory detects that 8% of the produced items are class $B$ quality only (i.e. not flawless). With a probability of 4% the production line functions suboptimally due to pollution, and with 50% probability the material used has quality issues. If pollution occurs, the items are class $B$ with a probability of 10%. Technicians also find that if there are fluctuations in the power supply then there is 10% probability that the product is class $B$. Based on these data, calculate the probability that power supply fluctuates.
I really have difficulty understanding the statement. As I see the problem: Let $P$ be the event "pollution", $Q$ the event a quality defect, $F$ the event "power supply fluctuate", and $B$ the event "$B$ quality".
I see the problem as follows. (see picture below).
So we just have
$$x=mathbb P(F)=1-0.04-0.5=0.46?$$
But I don't use $mathbb P(B)$. So maybe it should be $$mathbb P(B)=mathbb P(P)mathbb P(Bmid P)+mathbb P(Q)mathbb P(Bmid Q)+mathbb P(F)mathbb P(Bmid F)$$
which is equivalent to $$x=frac{0.01}{0.1}=0.1,$$
could it be possible?
probability
edited Jan 17 at 19:39
Bernard
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
$endgroup$
$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53
add a comment |
$begingroup$
Here's the statement:
Quality control in a factory detects that 8% of the produced items are class $B$ quality only (i.e. not flawless). With a probability of 4% the production line functions suboptimally due to pollution, and with 50% probability the material used has quality issues. If pollution occurs, the items are class $B$ with a probability of 10%. Technicians also find that if there are fluctuations in the power supply then there is 10% probability that the product is class $B$. Based on these data, calculate the probability that power supply fluctuates.
I really have difficulty understanding the statement. As I see the problem: Let $P$ be the event "pollution", $Q$ the event a quality defect, $F$ the event "power supply fluctuate", and $B$ the event "$B$ quality".
I see the problem as follows. (see picture below).
So we just have
$$x=mathbb P(F)=1-0.04-0.5=0.46?$$
But I don't use $mathbb P(B)$. So maybe it should be $$mathbb P(B)=mathbb P(P)mathbb P(Bmid P)+mathbb P(Q)mathbb P(Bmid Q)+mathbb P(F)mathbb P(Bmid F)$$
which is equivalent to $$x=frac{0.01}{0.1}=0.1,$$
could it be possible?
probability
edited Jan 17 at 19:39
Bernard
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
$endgroup$
Here's the statement:
Quality control in a factory detects that 8% of the produced items are class $B$ quality only (i.e. not flawless). With a probability of 4% the production line functions suboptimally due to pollution, and with 50% probability the material used has quality issues. If pollution occurs, the items are class $B$ with a probability of 10%. Technicians also find that if there are fluctuations in the power supply then there is 10% probability that the product is class $B$. Based on these data, calculate the probability that power supply fluctuates.
I really have difficulty understanding the statement. As I see the problem: Let $P$ be the event "pollution", $Q$ the event a quality defect, $F$ the event "power supply fluctuate", and $B$ the event "$B$ quality".
I see the problem as follows. (see picture below).
So we just have
$$x=mathbb P(F)=1-0.04-0.5=0.46?$$
But I don't use $mathbb P(B)$. So maybe it should be $$mathbb P(B)=mathbb P(P)mathbb P(Bmid P)+mathbb P(Q)mathbb P(Bmid Q)+mathbb P(F)mathbb P(Bmid F)$$
which is equivalent to $$x=frac{0.01}{0.1}=0.1,$$
could it be possible?
probability
probability
edited Jan 17 at 19:39
Bernard
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
edited Jan 17 at 19:39
Bernard
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
edited Jan 17 at 19:39
Bernard
121k740116
edited Jan 17 at 19:39
Bernard
121k740116
edited Jan 17 at 19:39
Bernard
121k740116
121k740116
asked Jan 17 at 17:54
user623855user623855
1457
asked Jan 17 at 17:54
user623855user623855
1457
asked Jan 17 at 17:54
user623855user623855
1457
1457
$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53
add a comment |
$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53
$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53
$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).
Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.
Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.
Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$
Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$
Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$
Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$
Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$
Thus we get the (somewhat unpleasant) linear equation:
$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$
And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$
Note: the above is badly error prone so I strongly advise checking each stage carefully.
A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$
Which again implies $$Phi=boxed {frac {110}{931}}$$
answered Jan 17 at 18:52
lulululu
42.4k25080
$endgroup$
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
|
show 5 more comments
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1 Answer
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$begingroup$
I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).
Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.
Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.
Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$
Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$
Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$
Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$
Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$
Thus we get the (somewhat unpleasant) linear equation:
$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$
And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$
Note: the above is badly error prone so I strongly advise checking each stage carefully.
A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$
Which again implies $$Phi=boxed {frac {110}{931}}$$
answered Jan 17 at 18:52
lulululu
42.4k25080
$endgroup$
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
|
show 5 more comments
$begingroup$
I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).
Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.
Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.
Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$
Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$
Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$
Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$
Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$
Thus we get the (somewhat unpleasant) linear equation:
$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$
And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$
Note: the above is badly error prone so I strongly advise checking each stage carefully.
A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$
Which again implies $$Phi=boxed {frac {110}{931}}$$
answered Jan 17 at 18:52
lulululu
42.4k25080
$endgroup$
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
|
show 5 more comments
$begingroup$
I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).
Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.
Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.
Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$
Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$
Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$
Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$
Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$
Thus we get the (somewhat unpleasant) linear equation:
$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$
And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$
Note: the above is badly error prone so I strongly advise checking each stage carefully.
A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$
Which again implies $$Phi=boxed {frac {110}{931}}$$
answered Jan 17 at 18:52
lulululu
42.4k25080
$endgroup$
I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).
Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.
Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.
Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$
Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$
Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$
Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$
Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$
Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$
Thus we get the (somewhat unpleasant) linear equation:
$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$
And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$
Note: the above is badly error prone so I strongly advise checking each stage carefully.
A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$
Which again implies $$Phi=boxed {frac {110}{931}}$$
answered Jan 17 at 18:52
lulululu
42.4k25080
edited Jan 17 at 20:20
answered Jan 17 at 18:52
lulululu
42.4k25080
answered Jan 17 at 18:52
lulululu
42.4k25080
answered Jan 17 at 18:52
lulululu
42.4k25080
42.4k25080
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
|
show 5 more comments
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
Thank you for your answer. You have used $Phi$ twice : for the fluctuation and for not be in $B$... is it a mistake ? Also, I don't really understand why my second method is not working, could you give me a reason ? (except that it's not working ^^)
$endgroup$
– user623855
Jan 17 at 19:16
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
I only use $Phi$ to denote the fluctuation probability. Assuming independence, I just use $Phi$ to get the known probability, $.92$, that a random item is not in Class $B$. I couldn't follow your method, I'll look again.
$endgroup$
– lulu
Jan 17 at 19:20
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Sorry, I don't get it, you mean $Phi=mathbb P(Fmid B^c)$ ?
$endgroup$
– user623855
Jan 17 at 19:22
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
Ok, the problem with your second method is that you don't address overlap between the external factors. If you assume that the factors are mutually exclusive you can find a simple solution along your lines, but that assumption seems badly unrealistic.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
$begingroup$
$Phi$ is the probability that fluctuations occur. It says nothing about Class $B$.
$endgroup$
– lulu
Jan 17 at 19:23
|
show 5 more comments
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$begingroup$
I've posted something below. A couple of caveats: first, I made some assumptions (spelled out at the start) which you might or might not have intended. second, the computation (while logically straight forward) is arithmetically unpleasant and error prone. I advise checking it carefully, line by line.
$endgroup$
– lulu
Jan 17 at 18:53