What is the conditional probability distribution for who wrote this particular exam?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
A small course has 3 students, $B$, $C$ and $D$. Based on the previous
testscores of these students we know that $B$ usually scores $80$%
correct on the exam questions, $C$ scores $60$% and $D$ scores $40$%.
The anonymous test that is now corrected has $4$ correct answers out
of $8$. What is the conditional probability distribution for who took
this particular exam?
This just doesn't want to work. I know Baye's theorem is of use here and possibly the Law of total probability.
Let $B=1, C=2$ and $D=3.$ Now also let $A_i, i=1,2,3$ be the event that person $A_i$ took the test and $X= $number of correctly answered question on the test. We are thus looking for
$$P(A_i|X=4)=fracP(X=4P(X=4).$$
This did not make anything simpler. The only thing I now in the RHS is that $P(A_i)=1/3.$
probability probability-theory
asked May 29 at 18:11
Parseval
2,6501615
add a comment |Â
up vote
2
down vote
favorite
A small course has 3 students, $B$, $C$ and $D$. Based on the previous
testscores of these students we know that $B$ usually scores $80$%
correct on the exam questions, $C$ scores $60$% and $D$ scores $40$%.
The anonymous test that is now corrected has $4$ correct answers out
of $8$. What is the conditional probability distribution for who took
this particular exam?
This just doesn't want to work. I know Baye's theorem is of use here and possibly the Law of total probability.
Let $B=1, C=2$ and $D=3.$ Now also let $A_i, i=1,2,3$ be the event that person $A_i$ took the test and $X= $number of correctly answered question on the test. We are thus looking for
$$P(A_i|X=4)=fracP(X=4P(X=4).$$
This did not make anything simpler. The only thing I now in the RHS is that $P(A_i)=1/3.$
probability probability-theory
asked May 29 at 18:11
Parseval
2,6501615
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A small course has 3 students, $B$, $C$ and $D$. Based on the previous
testscores of these students we know that $B$ usually scores $80$%
correct on the exam questions, $C$ scores $60$% and $D$ scores $40$%.
The anonymous test that is now corrected has $4$ correct answers out
of $8$. What is the conditional probability distribution for who took
this particular exam?
This just doesn't want to work. I know Baye's theorem is of use here and possibly the Law of total probability.
Let $B=1, C=2$ and $D=3.$ Now also let $A_i, i=1,2,3$ be the event that person $A_i$ took the test and $X= $number of correctly answered question on the test. We are thus looking for
$$P(A_i|X=4)=fracP(X=4P(X=4).$$
This did not make anything simpler. The only thing I now in the RHS is that $P(A_i)=1/3.$
probability probability-theory
asked May 29 at 18:11
Parseval
2,6501615
A small course has 3 students, $B$, $C$ and $D$. Based on the previous
testscores of these students we know that $B$ usually scores $80$%
correct on the exam questions, $C$ scores $60$% and $D$ scores $40$%.
The anonymous test that is now corrected has $4$ correct answers out
of $8$. What is the conditional probability distribution for who took
this particular exam?
This just doesn't want to work. I know Baye's theorem is of use here and possibly the Law of total probability.
Let $B=1, C=2$ and $D=3.$ Now also let $A_i, i=1,2,3$ be the event that person $A_i$ took the test and $X= $number of correctly answered question on the test. We are thus looking for
$$P(A_i|X=4)=fracP(X=4P(X=4).$$
This did not make anything simpler. The only thing I now in the RHS is that $P(A_i)=1/3.$
probability probability-theory
asked May 29 at 18:11
Parseval
2,6501615
edited Aug 17 at 19:50
asked May 29 at 18:11
Parseval
2,6501615
asked May 29 at 18:11
Parseval
2,6501615
asked May 29 at 18:11
Parseval
2,6501615
2,6501615
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44
add a comment |Â
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$P(X=4|A) = 8choose 4 0.8^4(1-0.8)^4\
P(X=4|B) = 8choose 4 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = frac P(X=4P(X=4$
answered May 29 at 18:22
Doug M
39.4k31749
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
 |Â
show 1 more comment
up vote
3
down vote
Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam.
answered May 29 at 18:25
Acccumulation
5,2162515
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$P(X=4|A) = 8choose 4 0.8^4(1-0.8)^4\
P(X=4|B) = 8choose 4 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = frac P(X=4P(X=4$
answered May 29 at 18:22
Doug M
39.4k31749
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
 |Â
show 1 more comment
up vote
2
down vote
accepted
$P(X=4|A) = 8choose 4 0.8^4(1-0.8)^4\
P(X=4|B) = 8choose 4 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = frac P(X=4P(X=4$
answered May 29 at 18:22
Doug M
39.4k31749
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$P(X=4|A) = 8choose 4 0.8^4(1-0.8)^4\
P(X=4|B) = 8choose 4 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = frac P(X=4P(X=4$
answered May 29 at 18:22
Doug M
39.4k31749
$P(X=4|A) = 8choose 4 0.8^4(1-0.8)^4\
P(X=4|B) = 8choose 4 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = frac P(X=4P(X=4$
answered May 29 at 18:22
Doug M
39.4k31749
edited May 29 at 18:31
answered May 29 at 18:22
Doug M
39.4k31749
answered May 29 at 18:22
Doug M
39.4k31749
answered May 29 at 18:22
Doug M
39.4k31749
39.4k31749
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
 |Â
show 1 more comment
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
I believe you mean P(X = 4|A) in the first line, rather than P(X = 4|B).
– Acccumulation
May 29 at 18:29
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Probably, and I think I will change it. But the first sentence says that the students in the class are $B,C$ and $D$
– Doug M
May 29 at 18:30
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
Whoops, I missed that the students' names start at B. That's a bit confusing.
– Acccumulation
May 29 at 18:34
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
The whole question is poorly edited. I probably should have just voted to close, but I really don't like closing people's questions.
– Doug M
May 29 at 18:35
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
@DougM - This answer is correct. But If you are using Baye's, where does $P(B)$ go? And why could you split up the denominator like that? Can you elaborate on the steps inbetween?
– Parseval
May 29 at 20:39
 |Â
show 1 more comment
up vote
3
down vote
Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam.
answered May 29 at 18:25
Acccumulation
5,2162515
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
add a comment |Â
up vote
3
down vote
Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam.
answered May 29 at 18:25
Acccumulation
5,2162515
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam.
answered May 29 at 18:25
Acccumulation
5,2162515
Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam.
answered May 29 at 18:25
Acccumulation
5,2162515
edited May 29 at 18:33
answered May 29 at 18:25
Acccumulation
5,2162515
answered May 29 at 18:25
Acccumulation
5,2162515
answered May 29 at 18:25
Acccumulation
5,2162515
5,2162515
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
add a comment |Â
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
Thanks, In the exam writers solution it seems he assumes all priors are 1/3. Thanks for the linguistic input also, edited accordingly!
– Parseval
May 29 at 20:36
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2800910%2fwhat-is-the-conditional-probability-distribution-for-who-wrote-this-particular-e%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Perhaps the title should be revised?
– Michael Burr
May 29 at 18:13
@MichaelBurr. Done! Sorry about that.
– Parseval
May 29 at 19:44