Mixing
Probability theorem question
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The bank has been robbed by one thief. Police has set n-suspects in a line (from 1 to n). The suspects say's true with chance p and they lie with chance (1-p). They asked the n'th suspect if (n-1) suspect has robbed the bank, then they asked (n-1) suspect if (n-2) suspect has robbed the bank and so on until the first suspect. First suspect said that he didn't robbed the bank. What is the probability that the first suspect didn't robbed the bank, depending to what the n-th suspect said. (Conditional probability, the n-th suspect can said that (n-1) suspect has robbed the bank or not).
discrete-mathematics
asked Jan 8 at 22:54
JohnJohn
32
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add a comment |
$begingroup$
The bank has been robbed by one thief. Police has set n-suspects in a line (from 1 to n). The suspects say's true with chance p and they lie with chance (1-p). They asked the n'th suspect if (n-1) suspect has robbed the bank, then they asked (n-1) suspect if (n-2) suspect has robbed the bank and so on until the first suspect. First suspect said that he didn't robbed the bank. What is the probability that the first suspect didn't robbed the bank, depending to what the n-th suspect said. (Conditional probability, the n-th suspect can said that (n-1) suspect has robbed the bank or not).
discrete-mathematics
asked Jan 8 at 22:54
JohnJohn
32
$endgroup$
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Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43
add a comment |
$begingroup$
The bank has been robbed by one thief. Police has set n-suspects in a line (from 1 to n). The suspects say's true with chance p and they lie with chance (1-p). They asked the n'th suspect if (n-1) suspect has robbed the bank, then they asked (n-1) suspect if (n-2) suspect has robbed the bank and so on until the first suspect. First suspect said that he didn't robbed the bank. What is the probability that the first suspect didn't robbed the bank, depending to what the n-th suspect said. (Conditional probability, the n-th suspect can said that (n-1) suspect has robbed the bank or not).
discrete-mathematics
asked Jan 8 at 22:54
JohnJohn
32
$endgroup$
The bank has been robbed by one thief. Police has set n-suspects in a line (from 1 to n). The suspects say's true with chance p and they lie with chance (1-p). They asked the n'th suspect if (n-1) suspect has robbed the bank, then they asked (n-1) suspect if (n-2) suspect has robbed the bank and so on until the first suspect. First suspect said that he didn't robbed the bank. What is the probability that the first suspect didn't robbed the bank, depending to what the n-th suspect said. (Conditional probability, the n-th suspect can said that (n-1) suspect has robbed the bank or not).
discrete-mathematics
discrete-mathematics
asked Jan 8 at 22:54
JohnJohn
32
asked Jan 8 at 22:54
JohnJohn
32
asked Jan 8 at 22:54
JohnJohn
32
asked Jan 8 at 22:54
JohnJohn
32
asked Jan 8 at 22:54
JohnJohn
32
32
$begingroup$
Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43
add a comment |
$begingroup$
Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43
$begingroup$
Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using the definition of conditional probability,
$$P(text{$1$ didn't rob} mid text{$n$ said $n-1$ robbed}) = frac{P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))}{P(text{$n$ said $n-1$ robbed})}.$$
The problem ought to have given you a prior probability for each suspect being the robber. With no further knowledge, one might assume each suspect is equally likely to be the robber.
$$begin{align}
&P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))
\
&= sum_{i=2}^n P((text{$n$ said $n-1$ robbed})cap (text{$i$ robbed}))
\
&= (n-1)frac{1}{n}(1-p) + frac{1}{n}cdot p.
end{align}$$
The denominator can be computed similarly.
The other conditional probability $P(text{$1$ didn't rob} mid text{$n$ said $n-1$ did not rob})$ can be computed in a similar fashion.
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
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add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
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$begingroup$
Using the definition of conditional probability,
$$P(text{$1$ didn't rob} mid text{$n$ said $n-1$ robbed}) = frac{P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))}{P(text{$n$ said $n-1$ robbed})}.$$
The problem ought to have given you a prior probability for each suspect being the robber. With no further knowledge, one might assume each suspect is equally likely to be the robber.
$$begin{align}
&P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))
\
&= sum_{i=2}^n P((text{$n$ said $n-1$ robbed})cap (text{$i$ robbed}))
\
&= (n-1)frac{1}{n}(1-p) + frac{1}{n}cdot p.
end{align}$$
The denominator can be computed similarly.
The other conditional probability $P(text{$1$ didn't rob} mid text{$n$ said $n-1$ did not rob})$ can be computed in a similar fashion.
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
Using the definition of conditional probability,
$$P(text{$1$ didn't rob} mid text{$n$ said $n-1$ robbed}) = frac{P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))}{P(text{$n$ said $n-1$ robbed})}.$$
The problem ought to have given you a prior probability for each suspect being the robber. With no further knowledge, one might assume each suspect is equally likely to be the robber.
$$begin{align}
&P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))
\
&= sum_{i=2}^n P((text{$n$ said $n-1$ robbed})cap (text{$i$ robbed}))
\
&= (n-1)frac{1}{n}(1-p) + frac{1}{n}cdot p.
end{align}$$
The denominator can be computed similarly.
The other conditional probability $P(text{$1$ didn't rob} mid text{$n$ said $n-1$ did not rob})$ can be computed in a similar fashion.
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
Using the definition of conditional probability,
$$P(text{$1$ didn't rob} mid text{$n$ said $n-1$ robbed}) = frac{P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))}{P(text{$n$ said $n-1$ robbed})}.$$
The problem ought to have given you a prior probability for each suspect being the robber. With no further knowledge, one might assume each suspect is equally likely to be the robber.
$$begin{align}
&P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))
\
&= sum_{i=2}^n P((text{$n$ said $n-1$ robbed})cap (text{$i$ robbed}))
\
&= (n-1)frac{1}{n}(1-p) + frac{1}{n}cdot p.
end{align}$$
The denominator can be computed similarly.
The other conditional probability $P(text{$1$ didn't rob} mid text{$n$ said $n-1$ did not rob})$ can be computed in a similar fashion.
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
$endgroup$
Using the definition of conditional probability,
$$P(text{$1$ didn't rob} mid text{$n$ said $n-1$ robbed}) = frac{P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))}{P(text{$n$ said $n-1$ robbed})}.$$
The problem ought to have given you a prior probability for each suspect being the robber. With no further knowledge, one might assume each suspect is equally likely to be the robber.
$$begin{align}
&P((text{$n$ said $n-1$ robbed}) cap (text{$1$ didn't rob}))
\
&= sum_{i=2}^n P((text{$n$ said $n-1$ robbed})cap (text{$i$ robbed}))
\
&= (n-1)frac{1}{n}(1-p) + frac{1}{n}cdot p.
end{align}$$
The denominator can be computed similarly.
The other conditional probability $P(text{$1$ didn't rob} mid text{$n$ said $n-1$ did not rob})$ can be computed in a similar fashion.
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
answered Jan 8 at 23:28
angryavianangryavian
40.6k23380
40.6k23380
add a comment |
add a comment |
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$begingroup$
Is this homework or a real world problem?
$endgroup$
– copper.hat
Jan 8 at 22:56
$begingroup$
It's my own idea.
$endgroup$
– John
Jan 9 at 13:43