Urn problem involving removement of half of the balls
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An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?
It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.
probability probability-theory
asked Jan 10 at 20:01
user1337user1337
48210
$endgroup$
add a comment |
$begingroup$
An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?
It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.
probability probability-theory
asked Jan 10 at 20:01
user1337user1337
48210
$endgroup$
1
$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
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To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
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– amd
Jan 10 at 20:04
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Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10
$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
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– rtybase
Jan 10 at 20:29
$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42
add a comment |
$begingroup$
An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?
It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.
probability probability-theory
asked Jan 10 at 20:01
user1337user1337
48210
$endgroup$
An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?
It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.
probability probability-theory
probability probability-theory
asked Jan 10 at 20:01
user1337user1337
48210
asked Jan 10 at 20:01
user1337user1337
48210
edited Jan 10 at 20:12
user1337
asked Jan 10 at 20:01
user1337user1337
48210
asked Jan 10 at 20:01
user1337user1337
48210
asked Jan 10 at 20:01
user1337user1337
48210
48210
1
$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04
$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10
$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29
$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42
add a comment |
1
$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04
$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10
$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29
$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42
1
1
$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04
$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04
$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10
$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10
$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29
$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29
$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42
$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note the events like
$B={text{tangerine ball is extracted with the 2nd extraction}}$- $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$
And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.
And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
add a comment |
$begingroup$
The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.
Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.
answered Jan 10 at 21:18
lulululu
43.6k25081
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
Note the events like
$B={text{tangerine ball is extracted with the 2nd extraction}}$- $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$
And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.
And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
add a comment |
$begingroup$
Note the events like
$B={text{tangerine ball is extracted with the 2nd extraction}}$- $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$
And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.
And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
$endgroup$
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
add a comment |
$begingroup$
Note the events like
$B={text{tangerine ball is extracted with the 2nd extraction}}$- $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$
And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.
And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
$endgroup$
Note the events like
$B={text{tangerine ball is extracted with the 2nd extraction}}$- $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$
And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.
And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
edited Jan 10 at 21:36
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
answered Jan 10 at 20:46
rtybasertybase
11.5k31534
11.5k31534
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
add a comment |
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
$begingroup$
The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
$endgroup$
– user1337
Jan 10 at 21:00
1
1
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
$begingroup$
@user1337 updated ...
$endgroup$
– rtybase
Jan 10 at 21:28
add a comment |
$begingroup$
The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.
Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.
answered Jan 10 at 21:18
lulululu
43.6k25081
$endgroup$
add a comment |
$begingroup$
The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.
Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.
answered Jan 10 at 21:18
lulululu
43.6k25081
$endgroup$
add a comment |
$begingroup$
The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.
Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.
answered Jan 10 at 21:18
lulululu
43.6k25081
$endgroup$
The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.
Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.
answered Jan 10 at 21:18
lulululu
43.6k25081
edited Jan 10 at 23:53
answered Jan 10 at 21:18
lulululu
43.6k25081
answered Jan 10 at 21:18
lulululu
43.6k25081
answered Jan 10 at 21:18
lulululu
43.6k25081
43.6k25081
add a comment |
add a comment |
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$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02
$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
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– amd
Jan 10 at 20:04
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Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
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– user1337
Jan 10 at 20:10
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Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
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– rtybase
Jan 10 at 20:29
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Thanks very much. Could you put it as an answer?
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– user1337
Jan 10 at 20:42