Find the probability that the distance between $a$ and $b$ is greater than $3$.
$begingroup$
Two real numbers $a$ and $b$ are selected at random along the x-axis such that
$−2 leq b leq 0$ and $0 leq a leq 3$.
Can someone explain how to approach this problem?
probability
edited Jan 15 at 9:30
Shubham Johri
5,082717
asked Jan 15 at 8:38
makmak
1
$endgroup$
add a comment |
$begingroup$
Two real numbers $a$ and $b$ are selected at random along the x-axis such that
$−2 leq b leq 0$ and $0 leq a leq 3$.
Can someone explain how to approach this problem?
probability
edited Jan 15 at 9:30
Shubham Johri
5,082717
asked Jan 15 at 8:38
makmak
1
$endgroup$
add a comment |
$begingroup$
Two real numbers $a$ and $b$ are selected at random along the x-axis such that
$−2 leq b leq 0$ and $0 leq a leq 3$.
Can someone explain how to approach this problem?
probability
edited Jan 15 at 9:30
Shubham Johri
5,082717
asked Jan 15 at 8:38
makmak
1
$endgroup$
Two real numbers $a$ and $b$ are selected at random along the x-axis such that
$−2 leq b leq 0$ and $0 leq a leq 3$.
Can someone explain how to approach this problem?
probability
probability
edited Jan 15 at 9:30
Shubham Johri
5,082717
asked Jan 15 at 8:38
makmak
1
edited Jan 15 at 9:30
Shubham Johri
5,082717
asked Jan 15 at 8:38
makmak
1
edited Jan 15 at 9:30
Shubham Johri
5,082717
edited Jan 15 at 9:30
Shubham Johri
5,082717
edited Jan 15 at 9:30
Shubham Johri
5,082717
5,082717
asked Jan 15 at 8:38
makmak
1
asked Jan 15 at 8:38
makmak
1
asked Jan 15 at 8:38
makmak
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would have attempted it this way: let us first find the probability when there are finitely many choices for $b$. Say $bin Asubset[-2,0]$ and that $|A|=n<infty$.
The required probability is $1-P(a-ble3)$.
The probability of chosing $b=xin A$ is $1/n$.
The probability that $ale b+3$ is$$P(a-ble3|b=x)=frac{l[0,x+3]}{l[0,3]}=frac{x+3}3$$where $l[c,d]=d-c$ is the length of the interval $[c,d]$. Then you have,$$P(a-ble3)=sum_1^n P(a-ble3|b=x_i)times P(b=x_i)\=sum_1^nfrac{x_i+3}3timesfrac1n=frac{3n+sum_1^nx_i}{3n}=1+frac{sum_1^nx_i}{3n}$$The answer we require is just $$P(a-bge3)=1-Big[1+frac{sum_1^nx_i}{3n}Big]=-frac{sum_1^nx_i}{3n}$$So the solution to your original problem is the value of the limit$$lim_{ntoinfty}-frac{sum_1^nx_i}{3n}$$To solve the limit, write $x_i$ as $-2+j/n$ with $j$ ranging from $0to2n$. The limit then transforms to$$lim_{ntoinfty}-frac{sum_1^{2n}-2+j/n}{3n}=int_0^2frac{2-x}3dx=4/3-2/3=2/3$$
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
$endgroup$
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%2f3074200%2ffind-the-probability-that-the-distance-between-a-and-b-is-greater-than-3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would have attempted it this way: let us first find the probability when there are finitely many choices for $b$. Say $bin Asubset[-2,0]$ and that $|A|=n<infty$.
The required probability is $1-P(a-ble3)$.
The probability of chosing $b=xin A$ is $1/n$.
The probability that $ale b+3$ is$$P(a-ble3|b=x)=frac{l[0,x+3]}{l[0,3]}=frac{x+3}3$$where $l[c,d]=d-c$ is the length of the interval $[c,d]$. Then you have,$$P(a-ble3)=sum_1^n P(a-ble3|b=x_i)times P(b=x_i)\=sum_1^nfrac{x_i+3}3timesfrac1n=frac{3n+sum_1^nx_i}{3n}=1+frac{sum_1^nx_i}{3n}$$The answer we require is just $$P(a-bge3)=1-Big[1+frac{sum_1^nx_i}{3n}Big]=-frac{sum_1^nx_i}{3n}$$So the solution to your original problem is the value of the limit$$lim_{ntoinfty}-frac{sum_1^nx_i}{3n}$$To solve the limit, write $x_i$ as $-2+j/n$ with $j$ ranging from $0to2n$. The limit then transforms to$$lim_{ntoinfty}-frac{sum_1^{2n}-2+j/n}{3n}=int_0^2frac{2-x}3dx=4/3-2/3=2/3$$
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
I would have attempted it this way: let us first find the probability when there are finitely many choices for $b$. Say $bin Asubset[-2,0]$ and that $|A|=n<infty$.
The required probability is $1-P(a-ble3)$.
The probability of chosing $b=xin A$ is $1/n$.
The probability that $ale b+3$ is$$P(a-ble3|b=x)=frac{l[0,x+3]}{l[0,3]}=frac{x+3}3$$where $l[c,d]=d-c$ is the length of the interval $[c,d]$. Then you have,$$P(a-ble3)=sum_1^n P(a-ble3|b=x_i)times P(b=x_i)\=sum_1^nfrac{x_i+3}3timesfrac1n=frac{3n+sum_1^nx_i}{3n}=1+frac{sum_1^nx_i}{3n}$$The answer we require is just $$P(a-bge3)=1-Big[1+frac{sum_1^nx_i}{3n}Big]=-frac{sum_1^nx_i}{3n}$$So the solution to your original problem is the value of the limit$$lim_{ntoinfty}-frac{sum_1^nx_i}{3n}$$To solve the limit, write $x_i$ as $-2+j/n$ with $j$ ranging from $0to2n$. The limit then transforms to$$lim_{ntoinfty}-frac{sum_1^{2n}-2+j/n}{3n}=int_0^2frac{2-x}3dx=4/3-2/3=2/3$$
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
I would have attempted it this way: let us first find the probability when there are finitely many choices for $b$. Say $bin Asubset[-2,0]$ and that $|A|=n<infty$.
The required probability is $1-P(a-ble3)$.
The probability of chosing $b=xin A$ is $1/n$.
The probability that $ale b+3$ is$$P(a-ble3|b=x)=frac{l[0,x+3]}{l[0,3]}=frac{x+3}3$$where $l[c,d]=d-c$ is the length of the interval $[c,d]$. Then you have,$$P(a-ble3)=sum_1^n P(a-ble3|b=x_i)times P(b=x_i)\=sum_1^nfrac{x_i+3}3timesfrac1n=frac{3n+sum_1^nx_i}{3n}=1+frac{sum_1^nx_i}{3n}$$The answer we require is just $$P(a-bge3)=1-Big[1+frac{sum_1^nx_i}{3n}Big]=-frac{sum_1^nx_i}{3n}$$So the solution to your original problem is the value of the limit$$lim_{ntoinfty}-frac{sum_1^nx_i}{3n}$$To solve the limit, write $x_i$ as $-2+j/n$ with $j$ ranging from $0to2n$. The limit then transforms to$$lim_{ntoinfty}-frac{sum_1^{2n}-2+j/n}{3n}=int_0^2frac{2-x}3dx=4/3-2/3=2/3$$
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
$endgroup$
I would have attempted it this way: let us first find the probability when there are finitely many choices for $b$. Say $bin Asubset[-2,0]$ and that $|A|=n<infty$.
The required probability is $1-P(a-ble3)$.
The probability of chosing $b=xin A$ is $1/n$.
The probability that $ale b+3$ is$$P(a-ble3|b=x)=frac{l[0,x+3]}{l[0,3]}=frac{x+3}3$$where $l[c,d]=d-c$ is the length of the interval $[c,d]$. Then you have,$$P(a-ble3)=sum_1^n P(a-ble3|b=x_i)times P(b=x_i)\=sum_1^nfrac{x_i+3}3timesfrac1n=frac{3n+sum_1^nx_i}{3n}=1+frac{sum_1^nx_i}{3n}$$The answer we require is just $$P(a-bge3)=1-Big[1+frac{sum_1^nx_i}{3n}Big]=-frac{sum_1^nx_i}{3n}$$So the solution to your original problem is the value of the limit$$lim_{ntoinfty}-frac{sum_1^nx_i}{3n}$$To solve the limit, write $x_i$ as $-2+j/n$ with $j$ ranging from $0to2n$. The limit then transforms to$$lim_{ntoinfty}-frac{sum_1^{2n}-2+j/n}{3n}=int_0^2frac{2-x}3dx=4/3-2/3=2/3$$
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
edited Jan 15 at 10:43
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
answered Jan 15 at 9:29
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
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.
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%2f3074200%2ffind-the-probability-that-the-distance-between-a-and-b-is-greater-than-3%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
