Mixing
Bayes' theorem : lie-detector machine
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Assuming there is lie-detector which can find out 95% of all lies correctly and of all true statements classifies 98% as true. Now we know that only one person would lie among 300. If the detector says a person is lying, what is the probability that this person is lying?
Assuming that X = {person is lying}, D = {detector finds a lie}, then we need to know p(X|D) given that p(X) = 1/300. Now I'm stuck with the meaning of 95% and 98%, is p(D|X) = 0.95*0.98?
probability bayes-theorem
edited Jan 18 at 21:41
Gil Beckers
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
$endgroup$
add a comment |
$begingroup$
Assuming there is lie-detector which can find out 95% of all lies correctly and of all true statements classifies 98% as true. Now we know that only one person would lie among 300. If the detector says a person is lying, what is the probability that this person is lying?
Assuming that X = {person is lying}, D = {detector finds a lie}, then we need to know p(X|D) given that p(X) = 1/300. Now I'm stuck with the meaning of 95% and 98%, is p(D|X) = 0.95*0.98?
probability bayes-theorem
edited Jan 18 at 21:41
Gil Beckers
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
$endgroup$
$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
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@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10
add a comment |
$begingroup$
Assuming there is lie-detector which can find out 95% of all lies correctly and of all true statements classifies 98% as true. Now we know that only one person would lie among 300. If the detector says a person is lying, what is the probability that this person is lying?
Assuming that X = {person is lying}, D = {detector finds a lie}, then we need to know p(X|D) given that p(X) = 1/300. Now I'm stuck with the meaning of 95% and 98%, is p(D|X) = 0.95*0.98?
probability bayes-theorem
edited Jan 18 at 21:41
Gil Beckers
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
$endgroup$
Assuming there is lie-detector which can find out 95% of all lies correctly and of all true statements classifies 98% as true. Now we know that only one person would lie among 300. If the detector says a person is lying, what is the probability that this person is lying?
Assuming that X = {person is lying}, D = {detector finds a lie}, then we need to know p(X|D) given that p(X) = 1/300. Now I'm stuck with the meaning of 95% and 98%, is p(D|X) = 0.95*0.98?
probability bayes-theorem
probability bayes-theorem
edited Jan 18 at 21:41
Gil Beckers
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
edited Jan 18 at 21:41
Gil Beckers
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
edited Jan 18 at 21:41
Gil Beckers
156
edited Jan 18 at 21:41
Gil Beckers
156
edited Jan 18 at 21:41
Gil Beckers
156
156
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
asked Jan 18 at 18:20
Jiayan YangJiayan Yang
1
1
$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
$begingroup$
@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10
add a comment |
$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
$begingroup$
@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10
$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
$begingroup$
@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively. let $bar X$ and $bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$Pr[D mid X] = 0.95, quad Pr[bar D mid bar X] = 0.98, quad Pr[X] = frac{1}{300}.$$ We want to compute $Pr[X mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied. Bayes' rule gives
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D]},$$ the numerator of which is already known. The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$Pr[D] = Pr[D mid X]Pr[X] + Pr[D mid bar X]Pr[bar X].$$
The first term is the same as the numerator. The second term is computed by noting $$Pr[bar X] = 1 - Pr[X], \ Pr[D mid bar X] = 1 - Pr[bar D mid bar X].$$ Therefore, in terms of the given probabilities, we have
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D mid X]Pr[X] + (1 - Pr[bar D mid bar X])(1 - Pr[X])}.$$
answered Jan 18 at 23:11
heropupheropup
64k762102
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add a comment |
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$begingroup$
Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively. let $bar X$ and $bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$Pr[D mid X] = 0.95, quad Pr[bar D mid bar X] = 0.98, quad Pr[X] = frac{1}{300}.$$ We want to compute $Pr[X mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied. Bayes' rule gives
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D]},$$ the numerator of which is already known. The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$Pr[D] = Pr[D mid X]Pr[X] + Pr[D mid bar X]Pr[bar X].$$
The first term is the same as the numerator. The second term is computed by noting $$Pr[bar X] = 1 - Pr[X], \ Pr[D mid bar X] = 1 - Pr[bar D mid bar X].$$ Therefore, in terms of the given probabilities, we have
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D mid X]Pr[X] + (1 - Pr[bar D mid bar X])(1 - Pr[X])}.$$
answered Jan 18 at 23:11
heropupheropup
64k762102
$endgroup$
add a comment |
$begingroup$
Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively. let $bar X$ and $bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$Pr[D mid X] = 0.95, quad Pr[bar D mid bar X] = 0.98, quad Pr[X] = frac{1}{300}.$$ We want to compute $Pr[X mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied. Bayes' rule gives
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D]},$$ the numerator of which is already known. The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$Pr[D] = Pr[D mid X]Pr[X] + Pr[D mid bar X]Pr[bar X].$$
The first term is the same as the numerator. The second term is computed by noting $$Pr[bar X] = 1 - Pr[X], \ Pr[D mid bar X] = 1 - Pr[bar D mid bar X].$$ Therefore, in terms of the given probabilities, we have
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D mid X]Pr[X] + (1 - Pr[bar D mid bar X])(1 - Pr[X])}.$$
answered Jan 18 at 23:11
heropupheropup
64k762102
$endgroup$
add a comment |
$begingroup$
Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively. let $bar X$ and $bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$Pr[D mid X] = 0.95, quad Pr[bar D mid bar X] = 0.98, quad Pr[X] = frac{1}{300}.$$ We want to compute $Pr[X mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied. Bayes' rule gives
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D]},$$ the numerator of which is already known. The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$Pr[D] = Pr[D mid X]Pr[X] + Pr[D mid bar X]Pr[bar X].$$
The first term is the same as the numerator. The second term is computed by noting $$Pr[bar X] = 1 - Pr[X], \ Pr[D mid bar X] = 1 - Pr[bar D mid bar X].$$ Therefore, in terms of the given probabilities, we have
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D mid X]Pr[X] + (1 - Pr[bar D mid bar X])(1 - Pr[X])}.$$
answered Jan 18 at 23:11
heropupheropup
64k762102
$endgroup$
Using your notation for events, $X$ and $D$ represent the events that a person is lying, and a lie is detected, respectively. let $bar X$ and $bar D$ represent the complementary events; namely, a person tells the truth, and a lie is not detected, respectively.
We are given $$Pr[D mid X] = 0.95, quad Pr[bar D mid bar X] = 0.98, quad Pr[X] = frac{1}{300}.$$ We want to compute $Pr[X mid D]$, the conditional probability that, given the detector indicates a lie, that the person actually lied. Bayes' rule gives
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D]},$$ the numerator of which is already known. The denominator, the unconditional probability of a lie being detected, is found by an application of the law of total probability:
$$Pr[D] = Pr[D mid X]Pr[X] + Pr[D mid bar X]Pr[bar X].$$
The first term is the same as the numerator. The second term is computed by noting $$Pr[bar X] = 1 - Pr[X], \ Pr[D mid bar X] = 1 - Pr[bar D mid bar X].$$ Therefore, in terms of the given probabilities, we have
$$Pr[X mid D] = frac{Pr[D mid X]Pr[X]}{Pr[D mid X]Pr[X] + (1 - Pr[bar D mid bar X])(1 - Pr[X])}.$$
answered Jan 18 at 23:11
heropupheropup
64k762102
answered Jan 18 at 23:11
heropupheropup
64k762102
answered Jan 18 at 23:11
heropupheropup
64k762102
answered Jan 18 at 23:11
heropupheropup
64k762102
64k762102
add a comment |
add a comment |
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$begingroup$
$P(D|X)=0.95$ and $P(neg D|neg X)=0.98$
$endgroup$
– John Douma
Jan 18 at 18:25
$begingroup$
@John Douma Can you help check if my thinking is right or not ? $$ p(X|D) = p(X,D)/p(D) = frac{p(D|X)p(X)}{p(D|X)p(X)+p(D|bar{X})p(bar{X}) = (0.95*1/300)/(0.95*1/300+ 0.02*299/300)= ..$$
$endgroup$
– Jiayan Yang
Jan 18 at 18:55
$begingroup$
@JiayanYang Yes, that is exactly correct
$endgroup$
– Bram28
Jan 18 at 19:11
$begingroup$
Yes, it looks good. Thanks @Bram28.
$endgroup$
– John Douma
Jan 18 at 21:10