Probability and expectancy problem
up vote
5
down vote
favorite
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.7k114593
asked Dec 5 at 18:35
user610402
354
add a comment |
up vote
5
down vote
favorite
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.7k114593
asked Dec 5 at 18:35
user610402
354
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
edited Dec 5 at 18:46
greedoid
36.7k114593
asked Dec 5 at 18:35
user610402
354
if we choose a size $25$ subset of the set: ${1,2,....100}$
what is the expectancy of the number of sequential pairs in the subset?
expectancy still confuses me, can anybody help?
probability expected-value
probability expected-value
edited Dec 5 at 18:46
greedoid
36.7k114593
asked Dec 5 at 18:35
user610402
354
edited Dec 5 at 18:46
greedoid
36.7k114593
asked Dec 5 at 18:35
user610402
354
edited Dec 5 at 18:46
greedoid
36.7k114593
edited Dec 5 at 18:46
greedoid
36.7k114593
edited Dec 5 at 18:46
greedoid
36.7k114593
36.7k114593
asked Dec 5 at 18:35
user610402
354
asked Dec 5 at 18:35
user610402
354
asked Dec 5 at 18:35
user610402
354
354
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56
add a comment |
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
40.1k645107
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.7k114593
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027467%2fprobability-and-expectancy-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
40.1k645107
add a comment |
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
40.1k645107
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
40.1k645107
Let's assign a label $1,2 dots 25$ (arbitrary order) to each of the selected numbers. Let $A_{i,j}$ with $1le i<j le 25$ be $1$ if elements with labels $i,j$ have sequential values, $0$ otherwise.
Then $X = sum A_{i,j}$ counts the number of sequential pairs, which is what we want, and
$$E(X)=sum A_{i,j}$$
But $E(X)=sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p frac{99}{binom{100}{2}}$
Where $n_p = binom{25}{2}=300$ is the number of pairs. Then
$$E(X) = frac{300 times 99} { binom{100}{2}}=6$$
answered Dec 5 at 18:52
leonbloy
40.1k645107
answered Dec 5 at 18:52
leonbloy
40.1k645107
answered Dec 5 at 18:52
leonbloy
40.1k645107
answered Dec 5 at 18:52
leonbloy
40.1k645107
40.1k645107
add a comment |
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.7k114593
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.7k114593
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
up vote
7
down vote
up vote
7
down vote
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.7k114593
Let $X_i$ be an indicator random variable that $i$ and $i+1$ ($iin {1,2,...,99}$) are in a choosen subset.
Then $$ P(X_i = 1) = {{98choose 23}over {100choose 25}}= {6over 99}$$ Since $X= X_1+...+X_{99}$ we have $$ E(X) = E(X_1)+...+E(X_{99}) = 99{6over 99} = 6$$
answered Dec 5 at 18:39
greedoid
36.7k114593
edited Dec 5 at 18:59
answered Dec 5 at 18:39
greedoid
36.7k114593
answered Dec 5 at 18:39
greedoid
36.7k114593
answered Dec 5 at 18:39
greedoid
36.7k114593
36.7k114593
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Is that better @copper.hat?
– greedoid
Dec 5 at 18:44
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
Yes thanks. ${}{}$
– copper.hat
Dec 5 at 18:49
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027467%2fprobability-and-expectancy-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Related: math.stackexchange.com/questions/1566418/….
– StubbornAtom
Dec 10 at 18:56