Mixing
Probability about functioning of two watches
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If factory A produces $1$ faulty watch in $100$ watches and factory B produces $1$ faulty watch in $200$. You are given two watches, you know that one is form factory A and one is form factory B. You don't know which is from which factory. Then
a) What is the probability that the second watch works?
b) If the first watch works, then what is the probability that the second one will be working?
My approach:
a) Say $A_1$ and $A_2$ be the events that the an watch is form factory A or factory B respectively. Let $W$ denotes the event that the second watch is working. Of course $ P(A_1)=P(A_2)=0.5$.
Then $P(W)= P(W/A_1)P(A_1)+P(W/A_2)P(A_2) = frac99100(0.5)+frac199200(0.5) =0.9925$
b) Say $W_1$ and $W_2$ be two events such that first and second watches are working respectively. So $W_1 cap W_2$ denotes both the watches are working. Then $P(W_1 cap W_2$)= $(frac99100)(frac199200)$.
Hence $P(W_2/W_1)= fracP(W_1 cap W_2)P(W_1)$= $frac(frac99100)(frac199200)0.9925 approx 0.992494$.
Am I correct?
probability probability-theory
asked Jul 21 at 16:16
Arnab Chowdhury
1489
 |Â
show 8 more comments
up vote
1
down vote
favorite
If factory A produces $1$ faulty watch in $100$ watches and factory B produces $1$ faulty watch in $200$. You are given two watches, you know that one is form factory A and one is form factory B. You don't know which is from which factory. Then
a) What is the probability that the second watch works?
b) If the first watch works, then what is the probability that the second one will be working?
My approach:
a) Say $A_1$ and $A_2$ be the events that the an watch is form factory A or factory B respectively. Let $W$ denotes the event that the second watch is working. Of course $ P(A_1)=P(A_2)=0.5$.
Then $P(W)= P(W/A_1)P(A_1)+P(W/A_2)P(A_2) = frac99100(0.5)+frac199200(0.5) =0.9925$
b) Say $W_1$ and $W_2$ be two events such that first and second watches are working respectively. So $W_1 cap W_2$ denotes both the watches are working. Then $P(W_1 cap W_2$)= $(frac99100)(frac199200)$.
Hence $P(W_2/W_1)= fracP(W_1 cap W_2)P(W_1)$= $frac(frac99100)(frac199200)0.9925 approx 0.992494$.
Am I correct?
probability probability-theory
asked Jul 21 at 16:16
Arnab Chowdhury
1489
1
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
3
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
2
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09
 |Â
show 8 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If factory A produces $1$ faulty watch in $100$ watches and factory B produces $1$ faulty watch in $200$. You are given two watches, you know that one is form factory A and one is form factory B. You don't know which is from which factory. Then
a) What is the probability that the second watch works?
b) If the first watch works, then what is the probability that the second one will be working?
My approach:
a) Say $A_1$ and $A_2$ be the events that the an watch is form factory A or factory B respectively. Let $W$ denotes the event that the second watch is working. Of course $ P(A_1)=P(A_2)=0.5$.
Then $P(W)= P(W/A_1)P(A_1)+P(W/A_2)P(A_2) = frac99100(0.5)+frac199200(0.5) =0.9925$
b) Say $W_1$ and $W_2$ be two events such that first and second watches are working respectively. So $W_1 cap W_2$ denotes both the watches are working. Then $P(W_1 cap W_2$)= $(frac99100)(frac199200)$.
Hence $P(W_2/W_1)= fracP(W_1 cap W_2)P(W_1)$= $frac(frac99100)(frac199200)0.9925 approx 0.992494$.
Am I correct?
probability probability-theory
asked Jul 21 at 16:16
Arnab Chowdhury
1489
If factory A produces $1$ faulty watch in $100$ watches and factory B produces $1$ faulty watch in $200$. You are given two watches, you know that one is form factory A and one is form factory B. You don't know which is from which factory. Then
a) What is the probability that the second watch works?
b) If the first watch works, then what is the probability that the second one will be working?
My approach:
a) Say $A_1$ and $A_2$ be the events that the an watch is form factory A or factory B respectively. Let $W$ denotes the event that the second watch is working. Of course $ P(A_1)=P(A_2)=0.5$.
Then $P(W)= P(W/A_1)P(A_1)+P(W/A_2)P(A_2) = frac99100(0.5)+frac199200(0.5) =0.9925$
b) Say $W_1$ and $W_2$ be two events such that first and second watches are working respectively. So $W_1 cap W_2$ denotes both the watches are working. Then $P(W_1 cap W_2$)= $(frac99100)(frac199200)$.
Hence $P(W_2/W_1)= fracP(W_1 cap W_2)P(W_1)$= $frac(frac99100)(frac199200)0.9925 approx 0.992494$.
Am I correct?
probability probability-theory
asked Jul 21 at 16:16
Arnab Chowdhury
1489
edited Jul 21 at 16:22
asked Jul 21 at 16:16
Arnab Chowdhury
1489
asked Jul 21 at 16:16
Arnab Chowdhury
1489
asked Jul 21 at 16:16
Arnab Chowdhury
1489
1489
1
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
3
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
2
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09
 |Â
show 8 more comments
1
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
3
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
2
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09
1
1
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
3
3
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
2
2
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09
 |Â
show 8 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $frac 12$) probability to each.
Thus the probability that the second one is working is $$frac 12times p_A+frac 12 times p_B=frac p_A+p_B2$$
Note: this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working. That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches. We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $frac 12times (p_A+p_B)$. Of that, the probability that it comes from $A$ explains $frac 12times p_A$. Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=frac p_Ap_A+p_B$$
Similarly $$P_1(B)=frac p_Bp_A+p_B$$
Note as a sanity check that if $p_A=p_B$ then we just get $frac 12$ again as nothing has happened to break the symmetry.
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$frac p_Bp_A+p_Btimes p_A+frac p_Ap_A+p_Btimes p_B=frac 2p_Ap_Bp_A+p_B$$
Using your numbers we get $boxed 0.992493703$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=psi$ makes the answer $psi$, as it should.
answered Jul 21 at 16:55
lulu
35.2k14172
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $frac 12$) probability to each.
Thus the probability that the second one is working is $$frac 12times p_A+frac 12 times p_B=frac p_A+p_B2$$
Note: this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working. That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches. We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $frac 12times (p_A+p_B)$. Of that, the probability that it comes from $A$ explains $frac 12times p_A$. Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=frac p_Ap_A+p_B$$
Similarly $$P_1(B)=frac p_Bp_A+p_B$$
Note as a sanity check that if $p_A=p_B$ then we just get $frac 12$ again as nothing has happened to break the symmetry.
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$frac p_Bp_A+p_Btimes p_A+frac p_Ap_A+p_Btimes p_B=frac 2p_Ap_Bp_A+p_B$$
Using your numbers we get $boxed 0.992493703$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=psi$ makes the answer $psi$, as it should.
answered Jul 21 at 16:55
lulu
35.2k14172
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
add a comment |Â
up vote
1
down vote
As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $frac 12$) probability to each.
Thus the probability that the second one is working is $$frac 12times p_A+frac 12 times p_B=frac p_A+p_B2$$
Note: this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working. That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches. We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $frac 12times (p_A+p_B)$. Of that, the probability that it comes from $A$ explains $frac 12times p_A$. Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=frac p_Ap_A+p_B$$
Similarly $$P_1(B)=frac p_Bp_A+p_B$$
Note as a sanity check that if $p_A=p_B$ then we just get $frac 12$ again as nothing has happened to break the symmetry.
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$frac p_Bp_A+p_Btimes p_A+frac p_Ap_A+p_Btimes p_B=frac 2p_Ap_Bp_A+p_B$$
Using your numbers we get $boxed 0.992493703$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=psi$ makes the answer $psi$, as it should.
answered Jul 21 at 16:55
lulu
35.2k14172
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $frac 12$) probability to each.
Thus the probability that the second one is working is $$frac 12times p_A+frac 12 times p_B=frac p_A+p_B2$$
Note: this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working. That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches. We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $frac 12times (p_A+p_B)$. Of that, the probability that it comes from $A$ explains $frac 12times p_A$. Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=frac p_Ap_A+p_B$$
Similarly $$P_1(B)=frac p_Bp_A+p_B$$
Note as a sanity check that if $p_A=p_B$ then we just get $frac 12$ again as nothing has happened to break the symmetry.
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$frac p_Bp_A+p_Btimes p_A+frac p_Ap_A+p_Btimes p_B=frac 2p_Ap_Bp_A+p_B$$
Using your numbers we get $boxed 0.992493703$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=psi$ makes the answer $psi$, as it should.
answered Jul 21 at 16:55
lulu
35.2k14172
As there is some disagreement in the comments, I'll post some notes.
To make the problem more general, suppose that a watch from factory $A$ (resp $B$) is working with probability $P_A$ (resp. $p_B$).
At the start we have no idea which watch comes from which factory, so our prior must be to assign equal probability (i.e. $frac 12$) probability to each.
Thus the probability that the second one is working is $$frac 12times p_A+frac 12 times p_B=frac p_A+p_B2$$
Note: this is entirely consistent with the argument given by the OP.
Now assume that we have seen that $W_1$ is working. That is (weak) evidence for the claim that $W_1$ comes from $B$ since $B$ is more likely to make working watches. We need to use Bayes to re-estimate the probability that $W_1$ comes from $A$ or $B$.
As we saw in the first computation, the total probability that it is defective is $frac 12times (p_A+p_B)$. Of that, the probability that it comes from $A$ explains $frac 12times p_A$. Thus the re-estimated probability that it comes from $A$ is $$P_1(A)=frac p_Ap_A+p_B$$
Similarly $$P_1(B)=frac p_Bp_A+p_B$$
Note as a sanity check that if $p_A=p_B$ then we just get $frac 12$ again as nothing has happened to break the symmetry.
Of course $P_2(A)=P_1(B), P_2(B)=P_1(A)$.
Now we can use the revised probabilities to conclude that, given that $W_1$ is working, the probability that the second is working is $$frac p_Bp_A+p_Btimes p_A+frac p_Ap_A+p_Btimes p_B=frac 2p_Ap_Bp_A+p_B$$
Using your numbers we get $boxed 0.992493703$ which is slightly less than the $.9925$ for part $A$, reflective of the fact that the evidence that $W_1$ came from $B$ was quite weak.
As a (crude) sanity check, note that taking $p_A=p_B=psi$ makes the answer $psi$, as it should.
answered Jul 21 at 16:55
lulu
35.2k14172
edited Jul 21 at 17:01
answered Jul 21 at 16:55
lulu
35.2k14172
answered Jul 21 at 16:55
lulu
35.2k14172
answered Jul 21 at 16:55
lulu
35.2k14172
35.2k14172
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
add a comment |Â
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
Thats exactly how I proceed
– Arnab Chowdhury
Jul 21 at 17:07
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
That may well be, I had trouble following your argument. That's why I wrote it out in detail.
– lulu
Jul 21 at 17:17
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
Actually I should have written the intermediate steps, I apologise for that...
– Arnab Chowdhury
Jul 21 at 17:19
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
No apologies needed. I had the sense (possibly incorrect) that there was some confusion in the comments, so I thought I'd just write it all out in detail. Maybe too much detail!
– lulu
Jul 21 at 17:20
add a comment |Â
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1
This is not clear. What do you mean by "the second watch"? What process are you using to sample the watches?
– lulu
Jul 21 at 16:17
sorry, I missed a line. Let me edit this one please
– Arnab Chowdhury
Jul 21 at 16:18
3
Possible duplicate of Probability questions of watch problem
– Parcly Taxel
Jul 21 at 16:19
Please you see that this question is slightly different... @ParclyTaxel
– Arnab Chowdhury
Jul 21 at 16:26
2
In the other question, it was given that both the watches come from the same factory... @joriki
– Arnab Chowdhury
Jul 21 at 17:09