Bayes network probability question
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I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
so far i have gotten
Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48
Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56
not 100% sure or if these are correct or how to proceed when it comes to the next 2
probability bayesian bayes-theorem bayesian-network
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up vote
-1
down vote
favorite
I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
so far i have gotten
Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48
Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56
not 100% sure or if these are correct or how to proceed when it comes to the next 2
probability bayesian bayes-theorem bayesian-network
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
so far i have gotten
Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48
Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56
not 100% sure or if these are correct or how to proceed when it comes to the next 2
probability bayesian bayes-theorem bayesian-network
I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
so far i have gotten
Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48
Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56
not 100% sure or if these are correct or how to proceed when it comes to the next 2
probability bayesian bayes-theorem bayesian-network
probability bayesian bayes-theorem bayesian-network
asked Nov 16 at 22:16
pog
32
32
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1 Answer
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0
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What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$
$$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$
We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.
- Note: double-check results in calculations.
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$
$$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$
We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.
- Note: double-check results in calculations.
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
add a comment |
up vote
0
down vote
accepted
What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$
$$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$
We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.
- Note: double-check results in calculations.
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$
$$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$
We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.
- Note: double-check results in calculations.
What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a),p(a) + p(c|neg a),p(neg a) = 0.48$$
$$p(b) = p(b|a),p(a) + p(b|neg a),p(neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,neg a)=p(b,c|a),p(a)+p(b,c|neg a),p(neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a),p(c|a),p(a) + p(b|neg a),p(c|neg a),p(neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c),p(b,c)=0.192$$
We can calculate $p(neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, neg b)=p(d|neg b,c),p(neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.
- Note: double-check results in calculations.
answered Nov 17 at 0:41
DavidPM
1165
1165
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
add a comment |
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
Hi, I understand this up until "We can calculate p(¬b,c) the same way we did for p(b,c) (giving also 0.24)." when I try to do it I get ''P(¬b|a) P(c|a) P(a) + P (¬b |¬a) P(c | ¬a) P (¬a)'' but some things such as P(¬b|a) are not on the diagram so where are you getting those numbers?
– pog
Nov 18 at 17:42
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
$$p(neg b|a)=1-p(b|a)$$. It's the complement.
– DavidPM
Nov 18 at 19:17
add a comment |
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