Determine Computability of Joint and Conditional Probabilities Given Few Tables
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Problem
I've encountered the following problem during the introduction lecture of my Machine Learning class. I haven't taken a formal probability course yet, so any help would be appreciated.
Consider the multi-valued, random variables:
C (campus), G (grade), M (major), and Y (year).
None of these variables are independent. Given the probability tables for the following joint, marginal, and conditional probabilities, explain how to compute each probability below given the probabilities from above, or write “impossible.”
$P(Y)$, $P(M)$, $P(G,Y)$, $P(C|Y)$, $P(C,M)$, $P(Y|M)$
Attempt
I'm unsure as to how to combine the given probabilities to derive $P(G | C, M)$ (#2) and what to specify for $P(C=main | Y=freshman)$ given the table for $P(C|Y)$ (#3). Does the order of joint and conditional probabilities matter?
- P(M=compSci, Y=sophomore) = $P(Y|M)P(Y)$
- P(G=B | C=LC, M=business) = ?
- P(C=main | Y=freshman) = ?
- P(G=B, M=bio) = not possible
probability conditional-probability
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
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add a comment |
$begingroup$
Problem
I've encountered the following problem during the introduction lecture of my Machine Learning class. I haven't taken a formal probability course yet, so any help would be appreciated.
Consider the multi-valued, random variables:
C (campus), G (grade), M (major), and Y (year).
None of these variables are independent. Given the probability tables for the following joint, marginal, and conditional probabilities, explain how to compute each probability below given the probabilities from above, or write “impossible.”
$P(Y)$, $P(M)$, $P(G,Y)$, $P(C|Y)$, $P(C,M)$, $P(Y|M)$
Attempt
I'm unsure as to how to combine the given probabilities to derive $P(G | C, M)$ (#2) and what to specify for $P(C=main | Y=freshman)$ given the table for $P(C|Y)$ (#3). Does the order of joint and conditional probabilities matter?
- P(M=compSci, Y=sophomore) = $P(Y|M)P(Y)$
- P(G=B | C=LC, M=business) = ?
- P(C=main | Y=freshman) = ?
- P(G=B, M=bio) = not possible
probability conditional-probability
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
$endgroup$
add a comment |
$begingroup$
Problem
I've encountered the following problem during the introduction lecture of my Machine Learning class. I haven't taken a formal probability course yet, so any help would be appreciated.
Consider the multi-valued, random variables:
C (campus), G (grade), M (major), and Y (year).
None of these variables are independent. Given the probability tables for the following joint, marginal, and conditional probabilities, explain how to compute each probability below given the probabilities from above, or write “impossible.”
$P(Y)$, $P(M)$, $P(G,Y)$, $P(C|Y)$, $P(C,M)$, $P(Y|M)$
Attempt
I'm unsure as to how to combine the given probabilities to derive $P(G | C, M)$ (#2) and what to specify for $P(C=main | Y=freshman)$ given the table for $P(C|Y)$ (#3). Does the order of joint and conditional probabilities matter?
- P(M=compSci, Y=sophomore) = $P(Y|M)P(Y)$
- P(G=B | C=LC, M=business) = ?
- P(C=main | Y=freshman) = ?
- P(G=B, M=bio) = not possible
probability conditional-probability
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
$endgroup$
Problem
I've encountered the following problem during the introduction lecture of my Machine Learning class. I haven't taken a formal probability course yet, so any help would be appreciated.
Consider the multi-valued, random variables:
C (campus), G (grade), M (major), and Y (year).
None of these variables are independent. Given the probability tables for the following joint, marginal, and conditional probabilities, explain how to compute each probability below given the probabilities from above, or write “impossible.”
$P(Y)$, $P(M)$, $P(G,Y)$, $P(C|Y)$, $P(C,M)$, $P(Y|M)$
Attempt
I'm unsure as to how to combine the given probabilities to derive $P(G | C, M)$ (#2) and what to specify for $P(C=main | Y=freshman)$ given the table for $P(C|Y)$ (#3). Does the order of joint and conditional probabilities matter?
- P(M=compSci, Y=sophomore) = $P(Y|M)P(Y)$
- P(G=B | C=LC, M=business) = ?
- P(C=main | Y=freshman) = ?
- P(G=B, M=bio) = not possible
probability conditional-probability
probability conditional-probability
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
asked Jan 19 at 18:25
Anthony KrivonosAnthony Krivonos
21410
21410
add a comment |
add a comment |
1 Answer
1
active
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For (2), the only information about $G$ that you have is $P(G,Y)$, which does not give you any information about the dependence between $G$ and $(C,M)$. So this is impossible.
For (3), you already have the table for $P(C mid Y)$, so $P(C=text{main} mid Y=text{freshman})$ is one entry in this table.
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
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Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
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– Anthony Krivonos
Jan 19 at 18:49
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@AnthonyKrivonos Yes I believe so.
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– angryavian
Jan 19 at 18:58
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
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For (2), the only information about $G$ that you have is $P(G,Y)$, which does not give you any information about the dependence between $G$ and $(C,M)$. So this is impossible.
For (3), you already have the table for $P(C mid Y)$, so $P(C=text{main} mid Y=text{freshman})$ is one entry in this table.
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
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$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
add a comment |
$begingroup$
For (2), the only information about $G$ that you have is $P(G,Y)$, which does not give you any information about the dependence between $G$ and $(C,M)$. So this is impossible.
For (3), you already have the table for $P(C mid Y)$, so $P(C=text{main} mid Y=text{freshman})$ is one entry in this table.
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
add a comment |
$begingroup$
For (2), the only information about $G$ that you have is $P(G,Y)$, which does not give you any information about the dependence between $G$ and $(C,M)$. So this is impossible.
For (3), you already have the table for $P(C mid Y)$, so $P(C=text{main} mid Y=text{freshman})$ is one entry in this table.
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
$endgroup$
For (2), the only information about $G$ that you have is $P(G,Y)$, which does not give you any information about the dependence between $G$ and $(C,M)$. So this is impossible.
For (3), you already have the table for $P(C mid Y)$, so $P(C=text{main} mid Y=text{freshman})$ is one entry in this table.
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
answered Jan 19 at 18:34
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
add a comment |
$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
Ah, makes sense! By the same reasoning for (2), (4) would also be impossible, correct? And (1) can be derived from the givens.
$endgroup$
– Anthony Krivonos
Jan 19 at 18:49
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
$begingroup$
@AnthonyKrivonos Yes I believe so.
$endgroup$
– angryavian
Jan 19 at 18:58
add a comment |
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