Mixing
Probability distribution in lie detectors
$begingroup$
If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph, lied on the application
for an applicant who passed the polygraph, was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?
probability
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
$endgroup$
add a comment |
$begingroup$
If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph, lied on the application
for an applicant who passed the polygraph, was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?
probability
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
$endgroup$
$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54
add a comment |
$begingroup$
If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph, lied on the application
for an applicant who passed the polygraph, was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?
probability
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
$endgroup$
If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph, lied on the application
for an applicant who passed the polygraph, was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?
probability
probability
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
asked Oct 4 '13 at 22:23
Lolly NielsenLolly Nielsen
161
161
$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54
add a comment |
$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54
$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Draw a tree diagram as follows. From the "root," draw two downward going branches. Label the end of one $L$, for "lied on the application," and the end of the other $T$.
Write $0.01$ along the branch that goes to $L$, and $0.99$ along the branch that goes to $T$.
From the node labelled $L$, draw two downward-going branches, leading to two nodes. Label the end of one branch $P$, for "passed the test," and the other $F$. Along the branch that leads from $L$ to $P$, write $0.1$. Along the branch that leads to $F$, write $0.9$.
From the node labelled $T$, draw two downward-going branches, leading to two new nodes you will label $P$ and $F$. Along the branch that leads from $T$ to $P$, write $0.9$. Along the other branch, write $0.1$.
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Draw a tree diagram as follows. From the "root," draw two downward going branches. Label the end of one $L$, for "lied on the application," and the end of the other $T$.
Write $0.01$ along the branch that goes to $L$, and $0.99$ along the branch that goes to $T$.
From the node labelled $L$, draw two downward-going branches, leading to two nodes. Label the end of one branch $P$, for "passed the test," and the other $F$. Along the branch that leads from $L$ to $P$, write $0.1$. Along the branch that leads to $F$, write $0.9$.
From the node labelled $T$, draw two downward-going branches, leading to two new nodes you will label $P$ and $F$. Along the branch that leads from $T$ to $P$, write $0.9$. Along the other branch, write $0.1$.
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
add a comment |
$begingroup$
Draw a tree diagram as follows. From the "root," draw two downward going branches. Label the end of one $L$, for "lied on the application," and the end of the other $T$.
Write $0.01$ along the branch that goes to $L$, and $0.99$ along the branch that goes to $T$.
From the node labelled $L$, draw two downward-going branches, leading to two nodes. Label the end of one branch $P$, for "passed the test," and the other $F$. Along the branch that leads from $L$ to $P$, write $0.1$. Along the branch that leads to $F$, write $0.9$.
From the node labelled $T$, draw two downward-going branches, leading to two new nodes you will label $P$ and $F$. Along the branch that leads from $T$ to $P$, write $0.9$. Along the other branch, write $0.1$.
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
add a comment |
$begingroup$
Draw a tree diagram as follows. From the "root," draw two downward going branches. Label the end of one $L$, for "lied on the application," and the end of the other $T$.
Write $0.01$ along the branch that goes to $L$, and $0.99$ along the branch that goes to $T$.
From the node labelled $L$, draw two downward-going branches, leading to two nodes. Label the end of one branch $P$, for "passed the test," and the other $F$. Along the branch that leads from $L$ to $P$, write $0.1$. Along the branch that leads to $F$, write $0.9$.
From the node labelled $T$, draw two downward-going branches, leading to two new nodes you will label $P$ and $F$. Along the branch that leads from $T$ to $P$, write $0.9$. Along the other branch, write $0.1$.
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
$endgroup$
Draw a tree diagram as follows. From the "root," draw two downward going branches. Label the end of one $L$, for "lied on the application," and the end of the other $T$.
Write $0.01$ along the branch that goes to $L$, and $0.99$ along the branch that goes to $T$.
From the node labelled $L$, draw two downward-going branches, leading to two nodes. Label the end of one branch $P$, for "passed the test," and the other $F$. Along the branch that leads from $L$ to $P$, write $0.1$. Along the branch that leads to $F$, write $0.9$.
From the node labelled $T$, draw two downward-going branches, leading to two new nodes you will label $P$ and $F$. Along the branch that leads from $T$ to $P$, write $0.9$. Along the other branch, write $0.1$.
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
answered Oct 4 '13 at 23:07
André NicolasAndré Nicolas
454k36432819
454k36432819
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
add a comment |
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
Thanks, Andre. I see what you are saying and it makes very good sense. I'll work from here and I think I can answer all the questions from this. Thanks.
$endgroup$
– Lolly Nielsen
Oct 4 '13 at 23:39
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
$begingroup$
You are welcome. If you run into difficulties, please leave a message, asking an explicitly stated question.
$endgroup$
– André Nicolas
Oct 4 '13 at 23:41
add a comment |
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$begingroup$
I think what you really want to ask is what are the probabilities, unconditioned on the applicant.
$endgroup$
– David Simmons
Oct 4 '13 at 22:47
$begingroup$
Most of these things you need to define; for example, when you say that the lie detector is 90% accurate, does that mean that it has a .9 probability of giving a correct answer regardless of whether the examined person lied or not?
$endgroup$
– Jonathan Y.
Oct 4 '13 at 22:54