Finding the probability for non defective batteryHow do I optimize this scenario to achieve the lowest score?Need help finding the probabilityFinding probability values for logitProbability that only one component is defective?Finding given probabilityThe Defective Lock ProblemWhat is the highest possible ratio to be almost certain that someone is randomly selecting answers on a test?Finding probabilityPosterior mean computation of “Monty Hall Poblem”Finding the probability
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Finding the probability for non defective battery
How do I optimize this scenario to achieve the lowest score?Need help finding the probabilityFinding probability values for logitProbability that only one component is defective?Finding given probabilityThe Defective Lock ProblemWhat is the highest possible ratio to be almost certain that someone is randomly selecting answers on a test?Finding probabilityPosterior mean computation of “Monty Hall Poblem”Finding the probability
$begingroup$
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 Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
$endgroup$
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 Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
$endgroup$
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 Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
$endgroup$
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 Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
asked Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
asked Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
asked Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
asked Mar 16 at 4:04
prajyal senguptaprajyal sengupta
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(textselecting a defective battery)=P(text(battery comes from A and battery is defective) or (battery comes from B and battery is defective)).$$
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
$$P(textselecting a defective battery)=P(textBattery comes from A and is defective) + P(textbattery comes from B and battery is defective).$$
Bayes' rule tells us
$$P(textbattery comes from A and battery is defective)=P(textbattery is defective | textbattery comes from A)P(textbattery comes from A) = 0.6*0.06=.036.$$
and
$$P(textbattery comes from B and battery is defective)=P(textbattery is defective | textbattery comes from B)P(textbattery comes from B)=0.4*0.04=0.016.$$
Therefore
$$P(textselecting a defective battery)=0.036+0.016=0.054.$$
This gives
$$P(textselecting a non-defective battery)=1-0.054=0.946.$$
answered Mar 16 at 4:17
dlnBdlnB
93712
$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(textselecting a defective battery)=P(text(battery comes from A and battery is defective) or (battery comes from B and battery is defective)).$$
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
$$P(textselecting a defective battery)=P(textBattery comes from A and is defective) + P(textbattery comes from B and battery is defective).$$
Bayes' rule tells us
$$P(textbattery comes from A and battery is defective)=P(textbattery is defective | textbattery comes from A)P(textbattery comes from A) = 0.6*0.06=.036.$$
and
$$P(textbattery comes from B and battery is defective)=P(textbattery is defective | textbattery comes from B)P(textbattery comes from B)=0.4*0.04=0.016.$$
Therefore
$$P(textselecting a defective battery)=0.036+0.016=0.054.$$
This gives
$$P(textselecting a non-defective battery)=1-0.054=0.946.$$
answered Mar 16 at 4:17
dlnBdlnB
93712
$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
|
show 1 more comment
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(textselecting a defective battery)=P(text(battery comes from A and battery is defective) or (battery comes from B and battery is defective)).$$
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
$$P(textselecting a defective battery)=P(textBattery comes from A and is defective) + P(textbattery comes from B and battery is defective).$$
Bayes' rule tells us
$$P(textbattery comes from A and battery is defective)=P(textbattery is defective | textbattery comes from A)P(textbattery comes from A) = 0.6*0.06=.036.$$
and
$$P(textbattery comes from B and battery is defective)=P(textbattery is defective | textbattery comes from B)P(textbattery comes from B)=0.4*0.04=0.016.$$
Therefore
$$P(textselecting a defective battery)=0.036+0.016=0.054.$$
This gives
$$P(textselecting a non-defective battery)=1-0.054=0.946.$$
answered Mar 16 at 4:17
dlnBdlnB
93712
$endgroup$
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
|
show 1 more comment
$begingroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(textselecting a defective battery)=P(text(battery comes from A and battery is defective) or (battery comes from B and battery is defective)).$$
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
$$P(textselecting a defective battery)=P(textBattery comes from A and is defective) + P(textbattery comes from B and battery is defective).$$
Bayes' rule tells us
$$P(textbattery comes from A and battery is defective)=P(textbattery is defective | textbattery comes from A)P(textbattery comes from A) = 0.6*0.06=.036.$$
and
$$P(textbattery comes from B and battery is defective)=P(textbattery is defective | textbattery comes from B)P(textbattery comes from B)=0.4*0.04=0.016.$$
Therefore
$$P(textselecting a defective battery)=0.036+0.016=0.054.$$
This gives
$$P(textselecting a non-defective battery)=1-0.054=0.946.$$
answered Mar 16 at 4:17
dlnBdlnB
93712
$endgroup$
Your solution is incorrect. This problem requires a simple application of Bayes' rule.
$$P(textselecting a defective battery)=P(text(battery comes from A and battery is defective) or (battery comes from B and battery is defective)).$$
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
$$P(textselecting a defective battery)=P(textBattery comes from A and is defective) + P(textbattery comes from B and battery is defective).$$
Bayes' rule tells us
$$P(textbattery comes from A and battery is defective)=P(textbattery is defective | textbattery comes from A)P(textbattery comes from A) = 0.6*0.06=.036.$$
and
$$P(textbattery comes from B and battery is defective)=P(textbattery is defective | textbattery comes from B)P(textbattery comes from B)=0.4*0.04=0.016.$$
Therefore
$$P(textselecting a defective battery)=0.036+0.016=0.054.$$
This gives
$$P(textselecting a non-defective battery)=1-0.054=0.946.$$
answered Mar 16 at 4:17
dlnBdlnB
93712
edited Mar 16 at 4:50
answered Mar 16 at 4:17
dlnBdlnB
93712
answered Mar 16 at 4:17
dlnBdlnB
93712
answered Mar 16 at 4:17
dlnBdlnB
93712
93712
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
|
show 1 more comment
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Thanks got it..
$endgroup$
– prajyal sengupta
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Please accept the answer if you are satisfied. I am happy to clarify if you have any questions.
$endgroup$
– dlnB
Mar 16 at 4:34
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
Its very clear . I was almost there but forgot about the mutually exclusive logic
$endgroup$
– prajyal sengupta
Mar 16 at 4:36
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:49
$begingroup$
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
Mar 16 at 4:52
$begingroup$
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
Mar 16 at 4:52
|
show 1 more comment
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