Probability of selecting $j$-th ball from urn is $mathbb{P}(B_jmid A_k)=frac{k}{n}$
$begingroup$
An urn contains $M$ balls, of which $M_1$ are white. A set of $n$ balls is obtained either
by sampling balls with replacement (each ball is replaced in the urn after it is drawn), or
without replacement (a ball once drawn remains out of the urn).
Let $B_j$ be the event that
the ball selected at the $j$-th step is white, and $A_k$ the event that a sample of size $n$ contains
exactly $k$ white balls. Show that $$mathbb{P}(B_jmid A_k)=frac{k}{n}$$both for sampling with replacement and for sampling without replacement.
It seems clear that $mathbb{P}(B_1) =frac{M_1}{M}$ and $mathbb{P}(B_2) = frac{M_1}{M} frac{M-M1}{M-1} + frac{M_1}{M} frac{M_1-1}{M-1} = frac{M_1}{M}$. We can probably suspect that $mathbb{P}(B_j)=frac{M_1}{M}$ as well, if $j leq M_1$.
However, it's not clear for me how to construct the probability $mathbb{P}(B_jmid A_k$) from here.
If I'm not wrong, $$mathbb{P}(A_k) = frac{left( begin{array}{c} M_1\k end{array} right)left( begin{array}{c} M - M_1\ n -k end{array} right)}{left( begin{array}{c} M\n end{array} right)} tag{1}$$due to hypergeometric distribution. From here $$mathbb{P}(B_jmid A_k)=frac{mathbb{P}(B_j cap A_k)}{mathbb{P}(A_k)} $$.
But I'm not sure how to estimate the $mathbb{P}(B_j cap A_k)$. As far as I understand, instead of all possible events as in (1), event ${ B_j cap A_k }$ should cover a part of it where the $B_j$ holds, but not sure how to get to the $frac{k}{n}$ from here.
Any hints or suggestions would be appreciated!
probability combinatorics conditional-probability
$endgroup$
add a comment |
$begingroup$
An urn contains $M$ balls, of which $M_1$ are white. A set of $n$ balls is obtained either
by sampling balls with replacement (each ball is replaced in the urn after it is drawn), or
without replacement (a ball once drawn remains out of the urn).
Let $B_j$ be the event that
the ball selected at the $j$-th step is white, and $A_k$ the event that a sample of size $n$ contains
exactly $k$ white balls. Show that $$mathbb{P}(B_jmid A_k)=frac{k}{n}$$both for sampling with replacement and for sampling without replacement.
It seems clear that $mathbb{P}(B_1) =frac{M_1}{M}$ and $mathbb{P}(B_2) = frac{M_1}{M} frac{M-M1}{M-1} + frac{M_1}{M} frac{M_1-1}{M-1} = frac{M_1}{M}$. We can probably suspect that $mathbb{P}(B_j)=frac{M_1}{M}$ as well, if $j leq M_1$.
However, it's not clear for me how to construct the probability $mathbb{P}(B_jmid A_k$) from here.
If I'm not wrong, $$mathbb{P}(A_k) = frac{left( begin{array}{c} M_1\k end{array} right)left( begin{array}{c} M - M_1\ n -k end{array} right)}{left( begin{array}{c} M\n end{array} right)} tag{1}$$due to hypergeometric distribution. From here $$mathbb{P}(B_jmid A_k)=frac{mathbb{P}(B_j cap A_k)}{mathbb{P}(A_k)} $$.
But I'm not sure how to estimate the $mathbb{P}(B_j cap A_k)$. As far as I understand, instead of all possible events as in (1), event ${ B_j cap A_k }$ should cover a part of it where the $B_j$ holds, but not sure how to get to the $frac{k}{n}$ from here.
Any hints or suggestions would be appreciated!
probability combinatorics conditional-probability
$endgroup$
add a comment |
$begingroup$
An urn contains $M$ balls, of which $M_1$ are white. A set of $n$ balls is obtained either
by sampling balls with replacement (each ball is replaced in the urn after it is drawn), or
without replacement (a ball once drawn remains out of the urn).
Let $B_j$ be the event that
the ball selected at the $j$-th step is white, and $A_k$ the event that a sample of size $n$ contains
exactly $k$ white balls. Show that $$mathbb{P}(B_jmid A_k)=frac{k}{n}$$both for sampling with replacement and for sampling without replacement.
It seems clear that $mathbb{P}(B_1) =frac{M_1}{M}$ and $mathbb{P}(B_2) = frac{M_1}{M} frac{M-M1}{M-1} + frac{M_1}{M} frac{M_1-1}{M-1} = frac{M_1}{M}$. We can probably suspect that $mathbb{P}(B_j)=frac{M_1}{M}$ as well, if $j leq M_1$.
However, it's not clear for me how to construct the probability $mathbb{P}(B_jmid A_k$) from here.
If I'm not wrong, $$mathbb{P}(A_k) = frac{left( begin{array}{c} M_1\k end{array} right)left( begin{array}{c} M - M_1\ n -k end{array} right)}{left( begin{array}{c} M\n end{array} right)} tag{1}$$due to hypergeometric distribution. From here $$mathbb{P}(B_jmid A_k)=frac{mathbb{P}(B_j cap A_k)}{mathbb{P}(A_k)} $$.
But I'm not sure how to estimate the $mathbb{P}(B_j cap A_k)$. As far as I understand, instead of all possible events as in (1), event ${ B_j cap A_k }$ should cover a part of it where the $B_j$ holds, but not sure how to get to the $frac{k}{n}$ from here.
Any hints or suggestions would be appreciated!
probability combinatorics conditional-probability
$endgroup$
An urn contains $M$ balls, of which $M_1$ are white. A set of $n$ balls is obtained either
by sampling balls with replacement (each ball is replaced in the urn after it is drawn), or
without replacement (a ball once drawn remains out of the urn).
Let $B_j$ be the event that
the ball selected at the $j$-th step is white, and $A_k$ the event that a sample of size $n$ contains
exactly $k$ white balls. Show that $$mathbb{P}(B_jmid A_k)=frac{k}{n}$$both for sampling with replacement and for sampling without replacement.
It seems clear that $mathbb{P}(B_1) =frac{M_1}{M}$ and $mathbb{P}(B_2) = frac{M_1}{M} frac{M-M1}{M-1} + frac{M_1}{M} frac{M_1-1}{M-1} = frac{M_1}{M}$. We can probably suspect that $mathbb{P}(B_j)=frac{M_1}{M}$ as well, if $j leq M_1$.
However, it's not clear for me how to construct the probability $mathbb{P}(B_jmid A_k$) from here.
If I'm not wrong, $$mathbb{P}(A_k) = frac{left( begin{array}{c} M_1\k end{array} right)left( begin{array}{c} M - M_1\ n -k end{array} right)}{left( begin{array}{c} M\n end{array} right)} tag{1}$$due to hypergeometric distribution. From here $$mathbb{P}(B_jmid A_k)=frac{mathbb{P}(B_j cap A_k)}{mathbb{P}(A_k)} $$.
But I'm not sure how to estimate the $mathbb{P}(B_j cap A_k)$. As far as I understand, instead of all possible events as in (1), event ${ B_j cap A_k }$ should cover a part of it where the $B_j$ holds, but not sure how to get to the $frac{k}{n}$ from here.
Any hints or suggestions would be appreciated!
probability combinatorics conditional-probability
probability combinatorics conditional-probability
edited Jan 24 at 22:14
Bernard
122k741116
122k741116
asked Jan 24 at 21:56
sjoksjok
353
353
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1 Answer
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$begingroup$
To compute $P(B_jcap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white.
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore,
$$
P(B_jcap A_k)=frac{M_1 binom{M_1-1}{k-1}binom{M-M_1}{n-k}(n-1)!}{frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement.
Alternatively, you could argue that given $A_k$, all $binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=binom{n-1}{k-1}/binom{n}k=k/n$.
$endgroup$
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
add a comment |
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$begingroup$
To compute $P(B_jcap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white.
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore,
$$
P(B_jcap A_k)=frac{M_1 binom{M_1-1}{k-1}binom{M-M_1}{n-k}(n-1)!}{frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement.
Alternatively, you could argue that given $A_k$, all $binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=binom{n-1}{k-1}/binom{n}k=k/n$.
$endgroup$
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
add a comment |
$begingroup$
To compute $P(B_jcap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white.
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore,
$$
P(B_jcap A_k)=frac{M_1 binom{M_1-1}{k-1}binom{M-M_1}{n-k}(n-1)!}{frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement.
Alternatively, you could argue that given $A_k$, all $binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=binom{n-1}{k-1}/binom{n}k=k/n$.
$endgroup$
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
add a comment |
$begingroup$
To compute $P(B_jcap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white.
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore,
$$
P(B_jcap A_k)=frac{M_1 binom{M_1-1}{k-1}binom{M-M_1}{n-k}(n-1)!}{frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement.
Alternatively, you could argue that given $A_k$, all $binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=binom{n-1}{k-1}/binom{n}k=k/n$.
$endgroup$
To compute $P(B_jcap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white.
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore,
$$
P(B_jcap A_k)=frac{M_1 binom{M_1-1}{k-1}binom{M-M_1}{n-k}(n-1)!}{frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement.
Alternatively, you could argue that given $A_k$, all $binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=binom{n-1}{k-1}/binom{n}k=k/n$.
answered Jan 24 at 22:18
Mike EarnestMike Earnest
24.3k22151
24.3k22151
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
add a comment |
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
$begingroup$
Thanks, very clear explanation!
$endgroup$
– sjok
Jan 24 at 22:37
add a comment |
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