Mixing
find the probability that two experiments matches in at least one number
$begingroup$
Suppose you have an urn with $N$ different balls numbered from $1$ to $N$. One experiment consist in sustracting $k$ balls without reposition.
Find the probability that if you perform 2 experiments there will be at least one number appearing in both experiments, for example: in $1,3,5,7,9$ and $1,2,10,11,12$ both experiments contain the number $1$.
I'm not sure if the answer has a simple form because I asked this to myself. I was playing a game and I needed the answer for the case $N=74$ and $k=5$ ( I don't know the answer for this case yet either).
The total number of possible experiments is $binom{N}{k}$, therefore if I perform $2$ experiments the total cases is given by $binom{N}{k}^2$. I don't know how to compute the favorable cases.
My idea is to start with an specific experiment (let's say $1,2,3,4,5$) and compute the number of experiments that contain any of this numbers.
probability combinatorics
asked Jan 23 at 16:19
CarvalCarval
204
$endgroup$
add a comment |
$begingroup$
Suppose you have an urn with $N$ different balls numbered from $1$ to $N$. One experiment consist in sustracting $k$ balls without reposition.
Find the probability that if you perform 2 experiments there will be at least one number appearing in both experiments, for example: in $1,3,5,7,9$ and $1,2,10,11,12$ both experiments contain the number $1$.
I'm not sure if the answer has a simple form because I asked this to myself. I was playing a game and I needed the answer for the case $N=74$ and $k=5$ ( I don't know the answer for this case yet either).
The total number of possible experiments is $binom{N}{k}$, therefore if I perform $2$ experiments the total cases is given by $binom{N}{k}^2$. I don't know how to compute the favorable cases.
My idea is to start with an specific experiment (let's say $1,2,3,4,5$) and compute the number of experiments that contain any of this numbers.
probability combinatorics
asked Jan 23 at 16:19
CarvalCarval
204
$endgroup$
add a comment |
$begingroup$
Suppose you have an urn with $N$ different balls numbered from $1$ to $N$. One experiment consist in sustracting $k$ balls without reposition.
Find the probability that if you perform 2 experiments there will be at least one number appearing in both experiments, for example: in $1,3,5,7,9$ and $1,2,10,11,12$ both experiments contain the number $1$.
I'm not sure if the answer has a simple form because I asked this to myself. I was playing a game and I needed the answer for the case $N=74$ and $k=5$ ( I don't know the answer for this case yet either).
The total number of possible experiments is $binom{N}{k}$, therefore if I perform $2$ experiments the total cases is given by $binom{N}{k}^2$. I don't know how to compute the favorable cases.
My idea is to start with an specific experiment (let's say $1,2,3,4,5$) and compute the number of experiments that contain any of this numbers.
probability combinatorics
asked Jan 23 at 16:19
CarvalCarval
204
$endgroup$
Suppose you have an urn with $N$ different balls numbered from $1$ to $N$. One experiment consist in sustracting $k$ balls without reposition.
Find the probability that if you perform 2 experiments there will be at least one number appearing in both experiments, for example: in $1,3,5,7,9$ and $1,2,10,11,12$ both experiments contain the number $1$.
I'm not sure if the answer has a simple form because I asked this to myself. I was playing a game and I needed the answer for the case $N=74$ and $k=5$ ( I don't know the answer for this case yet either).
The total number of possible experiments is $binom{N}{k}$, therefore if I perform $2$ experiments the total cases is given by $binom{N}{k}^2$. I don't know how to compute the favorable cases.
My idea is to start with an specific experiment (let's say $1,2,3,4,5$) and compute the number of experiments that contain any of this numbers.
probability combinatorics
probability combinatorics
asked Jan 23 at 16:19
CarvalCarval
204
asked Jan 23 at 16:19
CarvalCarval
204
asked Jan 23 at 16:19
CarvalCarval
204
asked Jan 23 at 16:19
CarvalCarval
204
asked Jan 23 at 16:19
CarvalCarval
204
204
add a comment |
add a comment |
1 Answer
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$begingroup$
You can just imagine that you draw the balls $1$ to $k$ in the first draw. Otherwise you can renumber the balls. Now ask the chance you don't get a number from $1$ to $k$ in the second draw and subtract from $1$. You get
$$1-frac {N-k choose k}{N choose k}$$
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can just imagine that you draw the balls $1$ to $k$ in the first draw. Otherwise you can renumber the balls. Now ask the chance you don't get a number from $1$ to $k$ in the second draw and subtract from $1$. You get
$$1-frac {N-k choose k}{N choose k}$$
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
You can just imagine that you draw the balls $1$ to $k$ in the first draw. Otherwise you can renumber the balls. Now ask the chance you don't get a number from $1$ to $k$ in the second draw and subtract from $1$. You get
$$1-frac {N-k choose k}{N choose k}$$
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
You can just imagine that you draw the balls $1$ to $k$ in the first draw. Otherwise you can renumber the balls. Now ask the chance you don't get a number from $1$ to $k$ in the second draw and subtract from $1$. You get
$$1-frac {N-k choose k}{N choose k}$$
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
$endgroup$
You can just imagine that you draw the balls $1$ to $k$ in the first draw. Otherwise you can renumber the balls. Now ask the chance you don't get a number from $1$ to $k$ in the second draw and subtract from $1$. You get
$$1-frac {N-k choose k}{N choose k}$$
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
answered Jan 23 at 16:22
Ross MillikanRoss Millikan
299k24200374
299k24200374
add a comment |
add a comment |
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