drawing balls from urn without out repetition and with repetition
$begingroup$
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
$endgroup$
add a comment |
$begingroup$
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
$endgroup$
add a comment |
$begingroup$
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
$endgroup$
So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?
Now I just want to make sure that without replacement that it would be $$frac{{3choose1}times{7choose5}times{2choose1}times{7choose1}}{{21choose5}}$$
Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?
discrete-mathematics
discrete-mathematics
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
asked Nov 29 '18 at 2:41
GeraltGeralt
8917
8917
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
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%2f3018082%2fdrawing-balls-from-urn-without-out-repetition-and-with-repetition%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
$begingroup$
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
add a comment |
$begingroup$
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
$endgroup$
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
add a comment |
$begingroup$
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
$endgroup$
Assuming that you mean $binom{7}{4}$ instead of $binom{7}{5}$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $frac{1}{3}$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $3 times 2 times 5$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $left(frac{1}{3}right)^5$, to get $frac{10}{81}$.
You would not want to use stars and bars here, because not all combinations of balls are equally likely (rather than getting all 5 red, it's more likely to have 2 red, 2 blue, and 1 white). So counting the total number of distinct groups of 5 balls isn't going to help you obtain the probability.
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
answered Nov 29 '18 at 6:27
plattyplatty
3,370320
3,370320
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
add a comment |
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
Could you explain further why you multiply by 5.
$endgroup$
– Geralt
Nov 29 '18 at 12:36
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
$begingroup$
This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely.
$endgroup$
– platty
Nov 29 '18 at 15:11
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%2f3018082%2fdrawing-balls-from-urn-without-out-repetition-and-with-repetition%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
