How to find probability of occurrence of $E$ given $A$ said $E$ occured?
$begingroup$
$A$ speaks the truth with probability $p$. An event $E$ occurs with probability $alpha$. I have to find the probability of occurrence of $E$ given that $A$ said $E$ occured.
My approach:
$P[E|A_E] = frac{P[A_E|E]P[E]}{P[A_E|E]P[E] + P[A_E|E^c]P[E^c]}$
Now I have to find $P[A_E|E]$ and $P[A_E|E^c]$ from the following information
$P[A_E|E]P[E] + P[{A_E}^c|E^c]P[E^c] = p$,
$P[A_E|E^c] + P[{A_E}^c|E^c] = 1$,
$P[A_E|E] + P[{A_E}^c|E] = 1$
But I could not find any way. Please help.
probability bayes-theorem
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
$endgroup$
add a comment |
$begingroup$
$A$ speaks the truth with probability $p$. An event $E$ occurs with probability $alpha$. I have to find the probability of occurrence of $E$ given that $A$ said $E$ occured.
My approach:
$P[E|A_E] = frac{P[A_E|E]P[E]}{P[A_E|E]P[E] + P[A_E|E^c]P[E^c]}$
Now I have to find $P[A_E|E]$ and $P[A_E|E^c]$ from the following information
$P[A_E|E]P[E] + P[{A_E}^c|E^c]P[E^c] = p$,
$P[A_E|E^c] + P[{A_E}^c|E^c] = 1$,
$P[A_E|E] + P[{A_E}^c|E] = 1$
But I could not find any way. Please help.
probability bayes-theorem
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
$endgroup$
add a comment |
$begingroup$
$A$ speaks the truth with probability $p$. An event $E$ occurs with probability $alpha$. I have to find the probability of occurrence of $E$ given that $A$ said $E$ occured.
My approach:
$P[E|A_E] = frac{P[A_E|E]P[E]}{P[A_E|E]P[E] + P[A_E|E^c]P[E^c]}$
Now I have to find $P[A_E|E]$ and $P[A_E|E^c]$ from the following information
$P[A_E|E]P[E] + P[{A_E}^c|E^c]P[E^c] = p$,
$P[A_E|E^c] + P[{A_E}^c|E^c] = 1$,
$P[A_E|E] + P[{A_E}^c|E] = 1$
But I could not find any way. Please help.
probability bayes-theorem
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
$endgroup$
$A$ speaks the truth with probability $p$. An event $E$ occurs with probability $alpha$. I have to find the probability of occurrence of $E$ given that $A$ said $E$ occured.
My approach:
$P[E|A_E] = frac{P[A_E|E]P[E]}{P[A_E|E]P[E] + P[A_E|E^c]P[E^c]}$
Now I have to find $P[A_E|E]$ and $P[A_E|E^c]$ from the following information
$P[A_E|E]P[E] + P[{A_E}^c|E^c]P[E^c] = p$,
$P[A_E|E^c] + P[{A_E}^c|E^c] = 1$,
$P[A_E|E] + P[{A_E}^c|E] = 1$
But I could not find any way. Please help.
probability bayes-theorem
probability bayes-theorem
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
edited Dec 25 '18 at 15:13
Ethan Bolker
42.8k549113
42.8k549113
asked Dec 25 '18 at 15:10
PritamPritam
266
asked Dec 25 '18 at 15:10
PritamPritam
266
asked Dec 25 '18 at 15:10
PritamPritam
266
266
add a comment |
add a comment |
1 Answer
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$begingroup$
If A speaks truth with probability p, if he says E occurred it did occur with probability $alpha$. If A does not speak the truth with probability (1-p), if he says E occurred then it did not occur with probability $1-alpha$.
Now $$P(E/text{A says E occurred}) = frac{alpha p }{alpha p + (1-alpha)(1-p)}$$
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
If A speaks truth with probability p, if he says E occurred it did occur with probability $alpha$. If A does not speak the truth with probability (1-p), if he says E occurred then it did not occur with probability $1-alpha$.
Now $$P(E/text{A says E occurred}) = frac{alpha p }{alpha p + (1-alpha)(1-p)}$$
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
$begingroup$
If A speaks truth with probability p, if he says E occurred it did occur with probability $alpha$. If A does not speak the truth with probability (1-p), if he says E occurred then it did not occur with probability $1-alpha$.
Now $$P(E/text{A says E occurred}) = frac{alpha p }{alpha p + (1-alpha)(1-p)}$$
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
add a comment |
$begingroup$
If A speaks truth with probability p, if he says E occurred it did occur with probability $alpha$. If A does not speak the truth with probability (1-p), if he says E occurred then it did not occur with probability $1-alpha$.
Now $$P(E/text{A says E occurred}) = frac{alpha p }{alpha p + (1-alpha)(1-p)}$$
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
$endgroup$
If A speaks truth with probability p, if he says E occurred it did occur with probability $alpha$. If A does not speak the truth with probability (1-p), if he says E occurred then it did not occur with probability $1-alpha$.
Now $$P(E/text{A says E occurred}) = frac{alpha p }{alpha p + (1-alpha)(1-p)}$$
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
answered Dec 25 '18 at 15:16
Satish RamanathanSatish Ramanathan
9,92531323
9,92531323
add a comment |
add a comment |
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