Why do we use $nchoose k$ for a binomial distribution instead of $n+k-1choose k$?

up vote
0
down vote

favorite

I am trying to get my head around this. In my understanding a binomial distribution uses replacement and $nchoose k$ precisely states that there's no repetition and that's not the case with a coin toss for instance.

Thanks in advance.

edited Aug 23 at 2:53

asked Aug 23 at 2:19

Julio Arriaga

add a commentÂ |Â

up vote
0
down vote

favorite

Thanks in advance.

edited Aug 23 at 2:53

asked Aug 23 at 2:19

Julio Arriaga

add a commentÂ |Â

up vote
0
down vote

favorite

Thanks in advance.

edited Aug 23 at 2:53

asked Aug 23 at 2:19

Julio Arriaga

Thanks in advance.

edited Aug 23 at 2:53

asked Aug 23 at 2:19

Julio Arriaga

edited Aug 23 at 2:53

asked Aug 23 at 2:19

Julio Arriaga

asked Aug 23 at 2:19

Julio Arriaga

asked Aug 23 at 2:19

Julio Arriaga

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

In a binomial distribution, the choice of $k$ is the choice of which trials succeed. There can be no repetition because the same trial cannot succeed multiple times.

answered Aug 23 at 2:22

Y. Forman

11.2k423

add a commentÂ |Â

up vote
1
down vote

The probability of getting exactly $k$ successes in $n$ trials is given by the probability mass function:

$Pr(X=k)=n choose kp^k(1-p)^n-k$

i.e. of out $n$ choose $exactly$ $k$ to succeed with probability $p$.

answered Aug 23 at 2:27

Andrew Allen

1197

add a commentÂ |Â

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2891626%2fwhy-do-we-use-n-choose-k-for-a-binomial-distribution-instead-of-nk-1-choo%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

In a binomial distribution, the choice of $k$ is the choice of which trials succeed. There can be no repetition because the same trial cannot succeed multiple times.

answered Aug 23 at 2:22

Y. Forman

11.2k423

add a commentÂ |Â

up vote
1
down vote

accepted

In a binomial distribution, the choice of $k$ is the choice of which trials succeed. There can be no repetition because the same trial cannot succeed multiple times.

answered Aug 23 at 2:22

Y. Forman

11.2k423

add a commentÂ |Â

up vote
1
down vote

accepted

In a binomial distribution, the choice of $k$ is the choice of which trials succeed. There can be no repetition because the same trial cannot succeed multiple times.

answered Aug 23 at 2:22

Y. Forman

11.2k423

In a binomial distribution, the choice of $k$ is the choice of which trials succeed. There can be no repetition because the same trial cannot succeed multiple times.

answered Aug 23 at 2:22

Y. Forman

11.2k423

answered Aug 23 at 2:22

Y. Forman

11.2k423

answered Aug 23 at 2:22

Y. Forman

11.2k423

answered Aug 23 at 2:22

Y. Forman

11.2k423

add a commentÂ |Â

up vote
1
down vote

The probability of getting exactly $k$ successes in $n$ trials is given by the probability mass function:

$Pr(X=k)=n choose kp^k(1-p)^n-k$

i.e. of out $n$ choose $exactly$ $k$ to succeed with probability $p$.

answered Aug 23 at 2:27

Andrew Allen

1197

add a commentÂ |Â

up vote
1
down vote

The probability of getting exactly $k$ successes in $n$ trials is given by the probability mass function:

$Pr(X=k)=n choose kp^k(1-p)^n-k$

i.e. of out $n$ choose $exactly$ $k$ to succeed with probability $p$.

answered Aug 23 at 2:27

Andrew Allen

1197

add a commentÂ |Â

up vote
1
down vote

The probability of getting exactly $k$ successes in $n$ trials is given by the probability mass function:

$Pr(X=k)=n choose kp^k(1-p)^n-k$

i.e. of out $n$ choose $exactly$ $k$ to succeed with probability $p$.

answered Aug 23 at 2:27

Andrew Allen

1197

The probability of getting exactly $k$ successes in $n$ trials is given by the probability mass function:

$Pr(X=k)=n choose kp^k(1-p)^n-k$

i.e. of out $n$ choose $exactly$ $k$ to succeed with probability $p$.

answered Aug 23 at 2:27

Andrew Allen

1197

answered Aug 23 at 2:27

Andrew Allen

1197

answered Aug 23 at 2:27

Andrew Allen

1197

answered Aug 23 at 2:27

Andrew Allen

1197

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

搜尋此網誌

Vtyjkyuk