Finding the probability for non defective battery
$begingroup$
A car manufacturer purchases car batteries from two different suppliers A and B.
Suppose supplier A provides 60% of the batteries and supplier B provides the rest. If 6%
of all batteries from supplier A are defective and 4% of the batteries from supplier B are
defective. Determine the probability that a randomly selected battery is not defective.
Solution:
P(Selecting not a defective battery) =1- P(Selecting a defective battery) or P(Selecting a defective battery given A is defective) or P(selecting a defective battery given B is defective)
P(Selecting a defective battery) = (.6*.06)+(.4*.04) = 0.052
P(Selecting a defective battery given A is defective) = (.6*.06) = 0.036
p(Selecting a defective battery given B is defective) - (.4*.04) = 0.016
Therefore
P(Selective a not defective battery is ) = 1 - (0.052+0.036+0.016) = 0.9284
Please let me know if this is correct. Please provide you comment and advice on my solution . I am a bit confused whether it is correct or not.
probability self-study
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
A car manufacturer purchases car batteries from two different suppliers A and B.
Suppose supplier A provides 60% of the batteries and supplier B provides the rest. If 6%
of all batteries from supplier A are defective and 4% of the batteries from supplier B are
defective. Determine the probability that a randomly selected battery is not defective.
Solution:
P(Selecting not a defective battery) =1- P(Selecting a defective battery) or P(Selecting a defective battery given A is defective) or P(selecting a defective battery given B is defective)
P(Selecting a defective battery) = (.6*.06)+(.4*.04) = 0.052
P(Selecting a defective battery given A is defective) = (.6*.06) = 0.036
p(Selecting a defective battery given B is defective) - (.4*.04) = 0.016
Therefore
P(Selective a not defective battery is ) = 1 - (0.052+0.036+0.016) = 0.9284
Please let me know if this is correct. Please provide you comment and advice on my solution . I am a bit confused whether it is correct or not.
probability self-study
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A car manufacturer purchases car batteries from two different suppliers A and B.
Suppose supplier A provides 60% of the batteries and supplier B provides the rest. If 6%
of all batteries from supplier A are defective and 4% of the batteries from supplier B are
defective. Determine the probability that a randomly selected battery is not defective.
Solution:
P(Selecting not a defective battery) =1- P(Selecting a defective battery) or P(Selecting a defective battery given A is defective) or P(selecting a defective battery given B is defective)
P(Selecting a defective battery) = (.6*.06)+(.4*.04) = 0.052
P(Selecting a defective battery given A is defective) = (.6*.06) = 0.036
p(Selecting a defective battery given B is defective) - (.4*.04) = 0.016
Therefore
P(Selective a not defective battery is ) = 1 - (0.052+0.036+0.016) = 0.9284
Please let me know if this is correct. Please provide you comment and advice on my solution . I am a bit confused whether it is correct or not.
probability self-study
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A car manufacturer purchases car batteries from two different suppliers A and B.
Suppose supplier A provides 60% of the batteries and supplier B provides the rest. If 6%
of all batteries from supplier A are defective and 4% of the batteries from supplier B are
defective. Determine the probability that a randomly selected battery is not defective.
Solution:
P(Selecting not a defective battery) =1- P(Selecting a defective battery) or P(Selecting a defective battery given A is defective) or P(selecting a defective battery given B is defective)
P(Selecting a defective battery) = (.6*.06)+(.4*.04) = 0.052
P(Selecting a defective battery given A is defective) = (.6*.06) = 0.036
p(Selecting a defective battery given B is defective) - (.4*.04) = 0.016
Therefore
P(Selective a not defective battery is ) = 1 - (0.052+0.036+0.016) = 0.9284
Please let me know if this is correct. Please provide you comment and advice on my solution . I am a bit confused whether it is correct or not.
probability self-study
probability self-study
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
prajyal senguptaprajyal sengupta
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
prajyal senguptaprajyal sengupta
82
asked yesterday
prajyal senguptaprajyal sengupta
82
82
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
prajyal sengupta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
1 Answer
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$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(text{selecting a defective battery})=P(text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(text{selecting a defective battery})=P(text{Battery comes from A and is defective}) + P(text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(text{battery comes from A and battery is defective})=P(text{battery is defective} | text{battery comes from A})P(text{battery comes from A}) = 0.6*0.06=.036.$$
and
$$P(text{battery comes from B and battery is defective})=P(text{battery is defective} | text{battery comes from B})P(text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(text{selecting a non-defective battery})=1-0.054=0.946.$$
answered yesterday
dlnBdlnB
69311
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dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
|
show 1 more comment
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(text{selecting a defective battery})=P(text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(text{selecting a defective battery})=P(text{Battery comes from A and is defective}) + P(text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(text{battery comes from A and battery is defective})=P(text{battery is defective} | text{battery comes from A})P(text{battery comes from A}) = 0.6*0.06=.036.$$
and
$$P(text{battery comes from B and battery is defective})=P(text{battery is defective} | text{battery comes from B})P(text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(text{selecting a non-defective battery})=1-0.054=0.946.$$
answered yesterday
dlnBdlnB
69311
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dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
|
show 1 more comment
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(text{selecting a defective battery})=P(text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(text{selecting a defective battery})=P(text{Battery comes from A and is defective}) + P(text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(text{battery comes from A and battery is defective})=P(text{battery is defective} | text{battery comes from A})P(text{battery comes from A}) = 0.6*0.06=.036.$$
and
$$P(text{battery comes from B and battery is defective})=P(text{battery is defective} | text{battery comes from B})P(text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(text{selecting a non-defective battery})=1-0.054=0.946.$$
answered yesterday
dlnBdlnB
69311
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
|
show 1 more comment
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(text{selecting a defective battery})=P(text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(text{selecting a defective battery})=P(text{Battery comes from A and is defective}) + P(text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(text{battery comes from A and battery is defective})=P(text{battery is defective} | text{battery comes from A})P(text{battery comes from A}) = 0.6*0.06=.036.$$
and
$$P(text{battery comes from B and battery is defective})=P(text{battery is defective} | text{battery comes from B})P(text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(text{selecting a non-defective battery})=1-0.054=0.946.$$
answered yesterday
dlnBdlnB
69311
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(text{selecting a defective battery})=P(text{(battery comes from A and battery is defective) or (battery comes from B and battery is defective)}).$$
Since (battery comes from A and battery is defective) and (battery comes from B and battery is defective) are mutually exclusive events, and more specifically form a partition of the set of all batteries, it follows
$$P(text{selecting a defective battery})=P(text{Battery comes from A and is defective}) + P(text{battery comes from B and battery is defective}).$$
Bayes' rule tells us
$$P(text{battery comes from A and battery is defective})=P(text{battery is defective} | text{battery comes from A})P(text{battery comes from A}) = 0.6*0.06=.036.$$
and
$$P(text{battery comes from B and battery is defective})=P(text{battery is defective} | text{battery comes from B})P(text{battery comes from B})=0.4*0.04=0.016.$$
Therefore
$$P(text{selecting a defective battery})=0.036+0.016=0.054.$$
This gives
$$P(text{selecting a non-defective battery})=1-0.054=0.946.$$
answered yesterday
dlnBdlnB
69311
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
answered yesterday
dlnBdlnB
69311
New contributor
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answered yesterday
dlnBdlnB
69311
answered yesterday
dlnBdlnB
69311
69311
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New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
|
show 1 more comment
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
The only question I had was it is a mutually exclusive event because both A and B can supply battery to the manufacture at the same time. What I mean is 60% of the battery is coming from A and 40% coming from B. Let me know your thoughts
$endgroup$
– prajyal sengupta
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
$begingroup$
I edited my answer to be more clear. Writing $P(D)=P(A cap D)+P(B cap D)$ actually requires more than just mutual exclusivity. It requires $A$ and $B$ to make up a partition of the entire set of batteries, i.e. $P(A cap B)=0$ and $P(A cup B)=1$.
$endgroup$
– dlnB
yesterday
|
show 1 more comment
prajyal sengupta is a new contributor. Be nice, and check out our Code of Conduct.
prajyal sengupta is a new contributor. Be nice, and check out our Code of Conduct.
prajyal sengupta is a new contributor. Be nice, and check out our Code of Conduct.
prajyal sengupta is a new contributor. Be nice, and check out our Code of Conduct.
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