Can you compute $P(A)$ if you know $P(A|B), P(A|C), P(B)$ and $P(C)$?
$begingroup$
The agent is described as either loyal or not-loyal. The probability that the agent is loyal is
given by ‘p.’ The negotiations can reach two outcomes for the country: favorable and
unfavorable. The outcome of the negotiations may give an indication of whether the
agent is loyal or not.
Suppose that the negotiations have an unfavorable outcome for the country. Suppose that
$p($loyal$)=.8$, $p($unfavorable$|$loyal$)=.5$, and $p($unfavorable$|$not loyal$)=.7$. Use Bayes
Theorem to calculate the probability that the agent is not loyal.
My question is: how can I use Bayes Theorem in this case when I don't know $p($unfavourable$)$ or $p($favourable$)$? Is it even possible to compute $p($not loyal$│$unfavourable$)$ from the data given?
probability probability-theory
edited Dec 9 '18 at 4:38
dantopa
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
$endgroup$
add a comment |
$begingroup$
The agent is described as either loyal or not-loyal. The probability that the agent is loyal is
given by ‘p.’ The negotiations can reach two outcomes for the country: favorable and
unfavorable. The outcome of the negotiations may give an indication of whether the
agent is loyal or not.
Suppose that the negotiations have an unfavorable outcome for the country. Suppose that
$p($loyal$)=.8$, $p($unfavorable$|$loyal$)=.5$, and $p($unfavorable$|$not loyal$)=.7$. Use Bayes
Theorem to calculate the probability that the agent is not loyal.
My question is: how can I use Bayes Theorem in this case when I don't know $p($unfavourable$)$ or $p($favourable$)$? Is it even possible to compute $p($not loyal$│$unfavourable$)$ from the data given?
probability probability-theory
edited Dec 9 '18 at 4:38
dantopa
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
$endgroup$
$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13
add a comment |
$begingroup$
The agent is described as either loyal or not-loyal. The probability that the agent is loyal is
given by ‘p.’ The negotiations can reach two outcomes for the country: favorable and
unfavorable. The outcome of the negotiations may give an indication of whether the
agent is loyal or not.
Suppose that the negotiations have an unfavorable outcome for the country. Suppose that
$p($loyal$)=.8$, $p($unfavorable$|$loyal$)=.5$, and $p($unfavorable$|$not loyal$)=.7$. Use Bayes
Theorem to calculate the probability that the agent is not loyal.
My question is: how can I use Bayes Theorem in this case when I don't know $p($unfavourable$)$ or $p($favourable$)$? Is it even possible to compute $p($not loyal$│$unfavourable$)$ from the data given?
probability probability-theory
edited Dec 9 '18 at 4:38
dantopa
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
$endgroup$
The agent is described as either loyal or not-loyal. The probability that the agent is loyal is
given by ‘p.’ The negotiations can reach two outcomes for the country: favorable and
unfavorable. The outcome of the negotiations may give an indication of whether the
agent is loyal or not.
Suppose that the negotiations have an unfavorable outcome for the country. Suppose that
$p($loyal$)=.8$, $p($unfavorable$|$loyal$)=.5$, and $p($unfavorable$|$not loyal$)=.7$. Use Bayes
Theorem to calculate the probability that the agent is not loyal.
My question is: how can I use Bayes Theorem in this case when I don't know $p($unfavourable$)$ or $p($favourable$)$? Is it even possible to compute $p($not loyal$│$unfavourable$)$ from the data given?
probability probability-theory
probability probability-theory
edited Dec 9 '18 at 4:38
dantopa
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
edited Dec 9 '18 at 4:38
dantopa
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
edited Dec 9 '18 at 4:38
dantopa
6,46942243
edited Dec 9 '18 at 4:38
dantopa
6,46942243
edited Dec 9 '18 at 4:38
dantopa
6,46942243
6,46942243
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
asked Dec 9 '18 at 3:46
Alex SimionAlex Simion
204
204
$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13
add a comment |
$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13
$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13
$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The law of total probability says:
$$P(text{unfavorable})=P(text{unfavorable}|text{loyal})P(text{loyal})+P(text{unfavorable}|text{not loyal})P(text{not loyal})=.5cdot .8+.7cdot .2$$
Does this help?
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
$endgroup$
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
The law of total probability says:
$$P(text{unfavorable})=P(text{unfavorable}|text{loyal})P(text{loyal})+P(text{unfavorable}|text{not loyal})P(text{not loyal})=.5cdot .8+.7cdot .2$$
Does this help?
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
$endgroup$
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
add a comment |
$begingroup$
The law of total probability says:
$$P(text{unfavorable})=P(text{unfavorable}|text{loyal})P(text{loyal})+P(text{unfavorable}|text{not loyal})P(text{not loyal})=.5cdot .8+.7cdot .2$$
Does this help?
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
$endgroup$
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
add a comment |
$begingroup$
The law of total probability says:
$$P(text{unfavorable})=P(text{unfavorable}|text{loyal})P(text{loyal})+P(text{unfavorable}|text{not loyal})P(text{not loyal})=.5cdot .8+.7cdot .2$$
Does this help?
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
$endgroup$
The law of total probability says:
$$P(text{unfavorable})=P(text{unfavorable}|text{loyal})P(text{loyal})+P(text{unfavorable}|text{not loyal})P(text{not loyal})=.5cdot .8+.7cdot .2$$
Does this help?
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
answered Dec 9 '18 at 4:24
Zachary SelkZachary Selk
593311
593311
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
add a comment |
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
$begingroup$
Yes, thanks. Forgot about P(A)=P(A|B)P(B) + P(A|C)P(C)
$endgroup$
– Alex Simion
Dec 9 '18 at 4:58
add a comment |
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$begingroup$
I'm not asked to compute P(not loyal), but P(not loyal│unfavourable). Note that it is said that we shall start from the supposition that the negotiations have an unfavourable outcome. Theoretically, the solution should be: p(not loyal│unfavourable)=(p(unfavourable│not loyal)*p(not loyal))/(p(unfavourable)) Unfortunately, I can't get around p(unfavourable). Am I missing something?
$endgroup$
– Alex Simion
Dec 9 '18 at 4:13