Mixing
Finite Math (Probability/Venn Diagram)
$begingroup$
13) A survey revealed that $25%$ of people are entertained by reading books, $39%$ are entertained by watching TV, and $36%$ are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.
Is this problem stated correctly? How can $36%$ of the people be entertained by both books and TV, when only $25%$ of the people are entertained by reading books?
EDIT
Here are two other questions from the exam, that the instructor said followed the same logic as the question above.
14) Of the coffee makers sold in an appliance store, $5.0%$ have either a faulty switch or a defective cord, $1.6%$ have a faulty switch, and $0.2%$ have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
15) A survey of senior citizens at a doctor's office shows that $42%$ take blood pressure-lowering medication, $45%$ take cholesterol-lowering medication, and $13%$ take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.
probability
asked Apr 5 '14 at 2:17
JesseJesse
521111
$endgroup$
add a comment |
$begingroup$
13) A survey revealed that $25%$ of people are entertained by reading books, $39%$ are entertained by watching TV, and $36%$ are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.
Is this problem stated correctly? How can $36%$ of the people be entertained by both books and TV, when only $25%$ of the people are entertained by reading books?
EDIT
Here are two other questions from the exam, that the instructor said followed the same logic as the question above.
14) Of the coffee makers sold in an appliance store, $5.0%$ have either a faulty switch or a defective cord, $1.6%$ have a faulty switch, and $0.2%$ have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
15) A survey of senior citizens at a doctor's office shows that $42%$ take blood pressure-lowering medication, $45%$ take cholesterol-lowering medication, and $13%$ take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.
probability
asked Apr 5 '14 at 2:17
JesseJesse
521111
$endgroup$
add a comment |
$begingroup$
13) A survey revealed that $25%$ of people are entertained by reading books, $39%$ are entertained by watching TV, and $36%$ are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.
Is this problem stated correctly? How can $36%$ of the people be entertained by both books and TV, when only $25%$ of the people are entertained by reading books?
EDIT
Here are two other questions from the exam, that the instructor said followed the same logic as the question above.
14) Of the coffee makers sold in an appliance store, $5.0%$ have either a faulty switch or a defective cord, $1.6%$ have a faulty switch, and $0.2%$ have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
15) A survey of senior citizens at a doctor's office shows that $42%$ take blood pressure-lowering medication, $45%$ take cholesterol-lowering medication, and $13%$ take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.
probability
asked Apr 5 '14 at 2:17
JesseJesse
521111
$endgroup$
13) A survey revealed that $25%$ of people are entertained by reading books, $39%$ are entertained by watching TV, and $36%$ are entertained by both books and TV. What is the probability that a person will be entertained by either books or TV? Express the answer as a percentage.
Is this problem stated correctly? How can $36%$ of the people be entertained by both books and TV, when only $25%$ of the people are entertained by reading books?
EDIT
Here are two other questions from the exam, that the instructor said followed the same logic as the question above.
14) Of the coffee makers sold in an appliance store, $5.0%$ have either a faulty switch or a defective cord, $1.6%$ have a faulty switch, and $0.2%$ have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
15) A survey of senior citizens at a doctor's office shows that $42%$ take blood pressure-lowering medication, $45%$ take cholesterol-lowering medication, and $13%$ take both medications. What is the probability that senior citizen takes either blood pressure-lowering or cholesterol-lowering medication? Express the answer as a percentage.
probability
probability
asked Apr 5 '14 at 2:17
JesseJesse
521111
asked Apr 5 '14 at 2:17
JesseJesse
521111
edited Apr 5 '14 at 20:45
Jesse
asked Apr 5 '14 at 2:17
JesseJesse
521111
asked Apr 5 '14 at 2:17
JesseJesse
521111
asked Apr 5 '14 at 2:17
JesseJesse
521111
521111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem is stated correctly. Usually, there'd be a problem: $25%$ like TV, $75%$ like books, and $50%$ are entertained by both. In this case, since $25+39+36=100$, I am led to believe that the question should, when stated more rigorously, read:
A survey revealed that $25%$ of people are entertained only by reading books, $39%$ are entertained by only by watching TV, and $36%$ are entertained only by both books and TV. What is the probability that a person will be entertained by either books or TV?
Other than that, the question is fine.
Edit: In my example, i.e., $25%$ like TV, $75%$ like books, and $50%$ are entertained by both,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.75-0.50=0.5=50%$$
Similarly, for your problem,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.39-0.36=0.28=28%$$
$endgroup$
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
add a comment |
$begingroup$
You use the inclusion-exclusion principle. $Pr(A cup B) = Pr(A) + Pr(B) - Pr(A cap B)$. Your exam questions are worded correctly and don't have the same problem as the original question you posed did. It always must be true that $Pr(A cap B) le min(Pr(A), Pr(B))$, since $A cap B subseteq A$ and $A cap B subseteq B$.
For question
14) Let $F$ denote the event of a faulty switch, and $D$ denote the event of a defective cord. You are told, $P(F cup D) = 0.05, P(F) = 0.016$, and $P(Fcap D) = 0.002$. From the inclusion-exclusion principle you know
$P(F cup D) = P(F) + P(D) - P(F cap D) Rightarrow P(D) = P(F cup D) - P(F) + P(F cap D) =$ $0.05 - 0.016 + 0.002 = 0.036$.
Question 15) Follow's a similar procedure (apply inclusion-exclusion principle) to get $.42 + .45 - 0.13 = 0.74$.
answered Apr 5 '14 at 3:58
GridGrid
411512
$endgroup$
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
|
show 6 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem is stated correctly. Usually, there'd be a problem: $25%$ like TV, $75%$ like books, and $50%$ are entertained by both. In this case, since $25+39+36=100$, I am led to believe that the question should, when stated more rigorously, read:
A survey revealed that $25%$ of people are entertained only by reading books, $39%$ are entertained by only by watching TV, and $36%$ are entertained only by both books and TV. What is the probability that a person will be entertained by either books or TV?
Other than that, the question is fine.
Edit: In my example, i.e., $25%$ like TV, $75%$ like books, and $50%$ are entertained by both,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.75-0.50=0.5=50%$$
Similarly, for your problem,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.39-0.36=0.28=28%$$
$endgroup$
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
add a comment |
$begingroup$
The problem is stated correctly. Usually, there'd be a problem: $25%$ like TV, $75%$ like books, and $50%$ are entertained by both. In this case, since $25+39+36=100$, I am led to believe that the question should, when stated more rigorously, read:
A survey revealed that $25%$ of people are entertained only by reading books, $39%$ are entertained by only by watching TV, and $36%$ are entertained only by both books and TV. What is the probability that a person will be entertained by either books or TV?
Other than that, the question is fine.
Edit: In my example, i.e., $25%$ like TV, $75%$ like books, and $50%$ are entertained by both,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.75-0.50=0.5=50%$$
Similarly, for your problem,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.39-0.36=0.28=28%$$
$endgroup$
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
add a comment |
$begingroup$
The problem is stated correctly. Usually, there'd be a problem: $25%$ like TV, $75%$ like books, and $50%$ are entertained by both. In this case, since $25+39+36=100$, I am led to believe that the question should, when stated more rigorously, read:
A survey revealed that $25%$ of people are entertained only by reading books, $39%$ are entertained by only by watching TV, and $36%$ are entertained only by both books and TV. What is the probability that a person will be entertained by either books or TV?
Other than that, the question is fine.
Edit: In my example, i.e., $25%$ like TV, $75%$ like books, and $50%$ are entertained by both,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.75-0.50=0.5=50%$$
Similarly, for your problem,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.39-0.36=0.28=28%$$
$endgroup$
The problem is stated correctly. Usually, there'd be a problem: $25%$ like TV, $75%$ like books, and $50%$ are entertained by both. In this case, since $25+39+36=100$, I am led to believe that the question should, when stated more rigorously, read:
A survey revealed that $25%$ of people are entertained only by reading books, $39%$ are entertained by only by watching TV, and $36%$ are entertained only by both books and TV. What is the probability that a person will be entertained by either books or TV?
Other than that, the question is fine.
Edit: In my example, i.e., $25%$ like TV, $75%$ like books, and $50%$ are entertained by both,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.75-0.50=0.5=50%$$
Similarly, for your problem,
$$mathrm{Pr(TVcup Books)}=mathrm{Pr(TV)}+mathrm{Pr(Books)}-mathrm{Pr(TVcap Books)}\
implies mathrm{Pr(TVcup Books)}=0.25+0.39-0.36=0.28=28%$$
edited Apr 5 '14 at 4:16
answered Apr 5 '14 at 2:30
user122283
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
add a comment |
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
I feel the question would only make sense, if it was stated "rigorously" as you put, otherwise it seems ambiguous. Anyways what answer do you get? And in your given example, do you mind walking through how to solve it?
$endgroup$
– Jesse
Apr 5 '14 at 3:32
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
@Jesse See my edited answer.
$endgroup$
– user122283
Apr 5 '14 at 4:16
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
$begingroup$
When $P(A cap B) > P(A)$ does that imply that $P(A)$ is the probability for $A$ to strictly occur, otherwise the probability of obtaining $A$, be it solely $A$ or $A$ with $B$ is greater than the given $P(A)$? Which sounds odd... help me out.
$endgroup$
– Jesse
Apr 5 '14 at 20:12
add a comment |
$begingroup$
You use the inclusion-exclusion principle. $Pr(A cup B) = Pr(A) + Pr(B) - Pr(A cap B)$. Your exam questions are worded correctly and don't have the same problem as the original question you posed did. It always must be true that $Pr(A cap B) le min(Pr(A), Pr(B))$, since $A cap B subseteq A$ and $A cap B subseteq B$.
For question
14) Let $F$ denote the event of a faulty switch, and $D$ denote the event of a defective cord. You are told, $P(F cup D) = 0.05, P(F) = 0.016$, and $P(Fcap D) = 0.002$. From the inclusion-exclusion principle you know
$P(F cup D) = P(F) + P(D) - P(F cap D) Rightarrow P(D) = P(F cup D) - P(F) + P(F cap D) =$ $0.05 - 0.016 + 0.002 = 0.036$.
Question 15) Follow's a similar procedure (apply inclusion-exclusion principle) to get $.42 + .45 - 0.13 = 0.74$.
answered Apr 5 '14 at 3:58
GridGrid
411512
$endgroup$
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
|
show 6 more comments
$begingroup$
You use the inclusion-exclusion principle. $Pr(A cup B) = Pr(A) + Pr(B) - Pr(A cap B)$. Your exam questions are worded correctly and don't have the same problem as the original question you posed did. It always must be true that $Pr(A cap B) le min(Pr(A), Pr(B))$, since $A cap B subseteq A$ and $A cap B subseteq B$.
For question
14) Let $F$ denote the event of a faulty switch, and $D$ denote the event of a defective cord. You are told, $P(F cup D) = 0.05, P(F) = 0.016$, and $P(Fcap D) = 0.002$. From the inclusion-exclusion principle you know
$P(F cup D) = P(F) + P(D) - P(F cap D) Rightarrow P(D) = P(F cup D) - P(F) + P(F cap D) =$ $0.05 - 0.016 + 0.002 = 0.036$.
Question 15) Follow's a similar procedure (apply inclusion-exclusion principle) to get $.42 + .45 - 0.13 = 0.74$.
answered Apr 5 '14 at 3:58
GridGrid
411512
$endgroup$
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
|
show 6 more comments
$begingroup$
You use the inclusion-exclusion principle. $Pr(A cup B) = Pr(A) + Pr(B) - Pr(A cap B)$. Your exam questions are worded correctly and don't have the same problem as the original question you posed did. It always must be true that $Pr(A cap B) le min(Pr(A), Pr(B))$, since $A cap B subseteq A$ and $A cap B subseteq B$.
For question
14) Let $F$ denote the event of a faulty switch, and $D$ denote the event of a defective cord. You are told, $P(F cup D) = 0.05, P(F) = 0.016$, and $P(Fcap D) = 0.002$. From the inclusion-exclusion principle you know
$P(F cup D) = P(F) + P(D) - P(F cap D) Rightarrow P(D) = P(F cup D) - P(F) + P(F cap D) =$ $0.05 - 0.016 + 0.002 = 0.036$.
Question 15) Follow's a similar procedure (apply inclusion-exclusion principle) to get $.42 + .45 - 0.13 = 0.74$.
answered Apr 5 '14 at 3:58
GridGrid
411512
$endgroup$
You use the inclusion-exclusion principle. $Pr(A cup B) = Pr(A) + Pr(B) - Pr(A cap B)$. Your exam questions are worded correctly and don't have the same problem as the original question you posed did. It always must be true that $Pr(A cap B) le min(Pr(A), Pr(B))$, since $A cap B subseteq A$ and $A cap B subseteq B$.
For question
14) Let $F$ denote the event of a faulty switch, and $D$ denote the event of a defective cord. You are told, $P(F cup D) = 0.05, P(F) = 0.016$, and $P(Fcap D) = 0.002$. From the inclusion-exclusion principle you know
$P(F cup D) = P(F) + P(D) - P(F cap D) Rightarrow P(D) = P(F cup D) - P(F) + P(F cap D) =$ $0.05 - 0.016 + 0.002 = 0.036$.
Question 15) Follow's a similar procedure (apply inclusion-exclusion principle) to get $.42 + .45 - 0.13 = 0.74$.
answered Apr 5 '14 at 3:58
GridGrid
411512
edited Apr 5 '14 at 20:57
answered Apr 5 '14 at 3:58
GridGrid
411512
answered Apr 5 '14 at 3:58
GridGrid
411512
answered Apr 5 '14 at 3:58
GridGrid
411512
411512
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
|
show 6 more comments
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
How would you show this on a Venn Diagram?
$endgroup$
– Jesse
Apr 5 '14 at 18:26
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
So is the question asking for the probability of the symmetric difference or the union ?
$endgroup$
– Jesse
Apr 5 '14 at 19:35
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
The question is asking about the union, you want to include the middle portion. While symmetric difference in this case would be $Pr(A) + Pr(B) - 2*Pr(A cap B)$. Looking back my previous Venn Diagram explanation it seems it's slightly incorrect, ignore it. Sanath's description of the question is better, the original wording isn't the best. In any case, the question is just testing the 'inclusion-exclusion principle'.
$endgroup$
– Grid
Apr 5 '14 at 20:10
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
This was an exam question, so if the wording is not good I would like to challenge it, but if the question is correct by the context/logic, I need to know how the logic leads to an unambiguous answer. Namely, what does $P(A cap B) > P(A)$ imply?
$endgroup$
– Jesse
Apr 5 '14 at 20:21
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
$begingroup$
^ I lied haha, sorry I don't know what I was thinking you're right it's obviously not possible. See updated answer. Do challenge your instructor.
$endgroup$
– Grid
Apr 5 '14 at 20:27
|
show 6 more comments
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