probability of choosing the i-th red ball from a basket containing $m$ and $n$ indexed red and black balls...
$begingroup$
A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$
We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).
Does it matter that the black balls are indexed or not?
The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?
probability
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
$endgroup$
add a comment |
$begingroup$
A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$
We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).
Does it matter that the black balls are indexed or not?
The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?
probability
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
$endgroup$
add a comment |
$begingroup$
A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$
We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).
Does it matter that the black balls are indexed or not?
The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?
probability
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
$endgroup$
A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$
We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).
Does it matter that the black balls are indexed or not?
The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?
probability
probability
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
asked Dec 5 '18 at 14:30
John CataldoJohn Cataldo
1,1721316
1,1721316
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.
This because $k$ of the $m+n$ balls are selected.
Btw, this is true for every fixed ball.
No matter whether the black balls are numbered or not.
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
$endgroup$
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
|
show 1 more comment
$begingroup$
The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$
If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
For example,
${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
${R_1, R_3, B_3, B_5}$
count as three separate outcomes if we index the black balls,
but if we don't distinguish the black balls these are all part of the single outcome
${R_1, R_3, 2times B}.$
The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.
This because $k$ of the $m+n$ balls are selected.
Btw, this is true for every fixed ball.
No matter whether the black balls are numbered or not.
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
$endgroup$
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
|
show 1 more comment
$begingroup$
The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.
This because $k$ of the $m+n$ balls are selected.
Btw, this is true for every fixed ball.
No matter whether the black balls are numbered or not.
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
$endgroup$
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
|
show 1 more comment
$begingroup$
The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.
This because $k$ of the $m+n$ balls are selected.
Btw, this is true for every fixed ball.
No matter whether the black balls are numbered or not.
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
$endgroup$
The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.
This because $k$ of the $m+n$ balls are selected.
Btw, this is true for every fixed ball.
No matter whether the black balls are numbered or not.
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
edited Dec 5 '18 at 15:05
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
answered Dec 5 '18 at 14:52
drhabdrhab
99.6k544130
99.6k544130
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
|
show 1 more comment
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
I changed my answer.
$endgroup$
– drhab
Dec 5 '18 at 15:06
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
$endgroup$
– John Cataldo
Dec 5 '18 at 15:09
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Yes. It does not matter whether the black balls wear a label or not.
$endgroup$
– drhab
Dec 5 '18 at 15:10
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
Could you explain a little bit why that's true? Because that was really my question
$endgroup$
– John Cataldo
Dec 5 '18 at 15:11
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
$begingroup$
It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
$endgroup$
– drhab
Dec 5 '18 at 15:29
|
show 1 more comment
$begingroup$
The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$
If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
For example,
${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
${R_1, R_3, B_3, B_5}$
count as three separate outcomes if we index the black balls,
but if we don't distinguish the black balls these are all part of the single outcome
${R_1, R_3, 2times B}.$
The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
$endgroup$
add a comment |
$begingroup$
The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$
If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
For example,
${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
${R_1, R_3, B_3, B_5}$
count as three separate outcomes if we index the black balls,
but if we don't distinguish the black balls these are all part of the single outcome
${R_1, R_3, 2times B}.$
The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
$endgroup$
add a comment |
$begingroup$
The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$
If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
For example,
${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
${R_1, R_3, B_3, B_5}$
count as three separate outcomes if we index the black balls,
but if we don't distinguish the black balls these are all part of the single outcome
${R_1, R_3, 2times B}.$
The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
$endgroup$
The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$
If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
For example,
${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
${R_1, R_3, B_3, B_5}$
count as three separate outcomes if we index the black balls,
but if we don't distinguish the black balls these are all part of the single outcome
${R_1, R_3, 2times B}.$
The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
edited Dec 5 '18 at 14:49
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
answered Dec 5 '18 at 14:43
David KDavid K
53.6k342116
53.6k342116
add a comment |
add a comment |
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