Bayesian Probability solving
The above diagram is the Bayesian Network.
I want to find P(M=t|A=t && E=f)
I have followed the follwoing steps.
P(M=t|A=t && E=f) = P(M=t && A=t && E=f)/ P(A=t && E=f)
P(M=t && A=t && E=f)= P(M=t|A=t) *P(A=t|E=f && B=t) *P(E=f) *P(B=t)+ P(M=t|A=t) *P(A=t|E=f && B=f) *P(E=f) *P(B=f)
Now to calculate P(A=t && E=f),
I have followed-
P(A=t && E=f)= P(A=t|E=f)*P(E=f)
Now to calculate P(A=t|E=f),
P(A=t|E=f)=P(A=t|E=f && B=t)*P(B=t)+ P(A=t|E=f && B=f)*P(B=f)
Am I correct? Any kind of help would be appreciated.
Thanks.
probability bayesian-network
asked 2 days ago
jisanjisan
184
add a comment |
The above diagram is the Bayesian Network.
I want to find P(M=t|A=t && E=f)
I have followed the follwoing steps.
P(M=t|A=t && E=f) = P(M=t && A=t && E=f)/ P(A=t && E=f)
P(M=t && A=t && E=f)= P(M=t|A=t) *P(A=t|E=f && B=t) *P(E=f) *P(B=t)+ P(M=t|A=t) *P(A=t|E=f && B=f) *P(E=f) *P(B=f)
Now to calculate P(A=t && E=f),
I have followed-
P(A=t && E=f)= P(A=t|E=f)*P(E=f)
Now to calculate P(A=t|E=f),
P(A=t|E=f)=P(A=t|E=f && B=t)*P(B=t)+ P(A=t|E=f && B=f)*P(B=f)
Am I correct? Any kind of help would be appreciated.
Thanks.
probability bayesian-network
asked 2 days ago
jisanjisan
184
add a comment |
The above diagram is the Bayesian Network.
I want to find P(M=t|A=t && E=f)
I have followed the follwoing steps.
P(M=t|A=t && E=f) = P(M=t && A=t && E=f)/ P(A=t && E=f)
P(M=t && A=t && E=f)= P(M=t|A=t) *P(A=t|E=f && B=t) *P(E=f) *P(B=t)+ P(M=t|A=t) *P(A=t|E=f && B=f) *P(E=f) *P(B=f)
Now to calculate P(A=t && E=f),
I have followed-
P(A=t && E=f)= P(A=t|E=f)*P(E=f)
Now to calculate P(A=t|E=f),
P(A=t|E=f)=P(A=t|E=f && B=t)*P(B=t)+ P(A=t|E=f && B=f)*P(B=f)
Am I correct? Any kind of help would be appreciated.
Thanks.
probability bayesian-network
asked 2 days ago
jisanjisan
184
The above diagram is the Bayesian Network.
I want to find P(M=t|A=t && E=f)
I have followed the follwoing steps.
P(M=t|A=t && E=f) = P(M=t && A=t && E=f)/ P(A=t && E=f)
P(M=t && A=t && E=f)= P(M=t|A=t) *P(A=t|E=f && B=t) *P(E=f) *P(B=t)+ P(M=t|A=t) *P(A=t|E=f && B=f) *P(E=f) *P(B=f)
Now to calculate P(A=t && E=f),
I have followed-
P(A=t && E=f)= P(A=t|E=f)*P(E=f)
Now to calculate P(A=t|E=f),
P(A=t|E=f)=P(A=t|E=f && B=t)*P(B=t)+ P(A=t|E=f && B=f)*P(B=f)
Am I correct? Any kind of help would be appreciated.
Thanks.
probability bayesian-network
probability bayesian-network
asked 2 days ago
jisanjisan
184
asked 2 days ago
jisanjisan
184
edited 2 days ago
jisan
asked 2 days ago
jisanjisan
184
asked 2 days ago
jisanjisan
184
asked 2 days ago
jisanjisan
184
184
add a comment |
add a comment |
1 Answer
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Note that $M$ and $E$ are conditionally independent given A. Thus, the probability you are looking for is already given in your diagram:
$$P(M=t|A=t wedge E=f),,=,,P(M=t|A=t)$$
answered 2 days ago
DavidPMDavidPM
28118
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that $M$ and $E$ are conditionally independent given A. Thus, the probability you are looking for is already given in your diagram:
$$P(M=t|A=t wedge E=f),,=,,P(M=t|A=t)$$
answered 2 days ago
DavidPMDavidPM
28118
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
add a comment |
Note that $M$ and $E$ are conditionally independent given A. Thus, the probability you are looking for is already given in your diagram:
$$P(M=t|A=t wedge E=f),,=,,P(M=t|A=t)$$
answered 2 days ago
DavidPMDavidPM
28118
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
add a comment |
Note that $M$ and $E$ are conditionally independent given A. Thus, the probability you are looking for is already given in your diagram:
$$P(M=t|A=t wedge E=f),,=,,P(M=t|A=t)$$
answered 2 days ago
DavidPMDavidPM
28118
Note that $M$ and $E$ are conditionally independent given A. Thus, the probability you are looking for is already given in your diagram:
$$P(M=t|A=t wedge E=f),,=,,P(M=t|A=t)$$
answered 2 days ago
DavidPMDavidPM
28118
answered 2 days ago
DavidPMDavidPM
28118
answered 2 days ago
DavidPMDavidPM
28118
answered 2 days ago
DavidPMDavidPM
28118
28118
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
add a comment |
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Oh, I missed that. In that case, I would not need this calculations. Straight forward answer 0.70.
– jisan
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
Exactly, you do not need any calculations.
– DavidPM
2 days ago
add a comment |
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