Drawing/sampling with replacement (Probability/Combinatorics)
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We have an urn with $w$ white marbles and $b$ black marbles and we draw $n$ marbles with replacement. We want to know the probability of drawing $k$ white marbles ($k<n$) and apparently we use the following formula:
$$frac{binom{n}{k}w^kb^{n - k}}{(w + b)^n}$$
What is the logic/intuition behind it?
probability combinatorics
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
$endgroup$
add a comment |
$begingroup$
We have an urn with $w$ white marbles and $b$ black marbles and we draw $n$ marbles with replacement. We want to know the probability of drawing $k$ white marbles ($k<n$) and apparently we use the following formula:
$$frac{binom{n}{k}w^kb^{n - k}}{(w + b)^n}$$
What is the logic/intuition behind it?
probability combinatorics
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
$endgroup$
1
$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47
add a comment |
$begingroup$
We have an urn with $w$ white marbles and $b$ black marbles and we draw $n$ marbles with replacement. We want to know the probability of drawing $k$ white marbles ($k<n$) and apparently we use the following formula:
$$frac{binom{n}{k}w^kb^{n - k}}{(w + b)^n}$$
What is the logic/intuition behind it?
probability combinatorics
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
$endgroup$
We have an urn with $w$ white marbles and $b$ black marbles and we draw $n$ marbles with replacement. We want to know the probability of drawing $k$ white marbles ($k<n$) and apparently we use the following formula:
$$frac{binom{n}{k}w^kb^{n - k}}{(w + b)^n}$$
What is the logic/intuition behind it?
probability combinatorics
probability combinatorics
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
edited Dec 4 '18 at 11:11
N. F. Taussig
44.4k93357
44.4k93357
asked Dec 3 '18 at 17:43
E.KE.K
6
asked Dec 3 '18 at 17:43
E.KE.K
6
asked Dec 3 '18 at 17:43
E.KE.K
6
6
1
$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47
add a comment |
1
$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47
1
1
$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47
$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47
add a comment |
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$begingroup$
The formula is essentially that of the binomial distribution. en.wikipedia.org/wiki/Binomial_distribution
$endgroup$
– herb steinberg
Dec 3 '18 at 17:47