Conditional probability (urns and balls)












5












$begingroup$


From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







share|cite|improve this question











$endgroup$












  • $begingroup$
    In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    $endgroup$
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • $begingroup$
    Yes, the calculation looks fine to me.
    $endgroup$
    – DavidPM
    Nov 27 '18 at 19:38










  • $begingroup$
    Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    $endgroup$
    – Makhayama
    Nov 27 '18 at 19:40
















5












$begingroup$


From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







share|cite|improve this question











$endgroup$












  • $begingroup$
    In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    $endgroup$
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • $begingroup$
    Yes, the calculation looks fine to me.
    $endgroup$
    – DavidPM
    Nov 27 '18 at 19:38










  • $begingroup$
    Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    $endgroup$
    – Makhayama
    Nov 27 '18 at 19:40














5












5








5


1



$begingroup$


From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins







share|cite|improve this question











$endgroup$




From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears.
What is the probability that:
1) second white ball appears on the second step
2) second white ball appears on the third step
3) second white ball appears on the k-th step



My solution is the following.
The probability of the second white ball is
$$Pr(X=2)=frac{6}{18}frac{5}{17}$$



The probability of the third white ball is
$$Pr(X=3)=frac{6}{18}frac{12}{17}frac{5}{16}+ frac{12}{18}frac{6}{17}frac{5}{16} = frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$



Therefore, the probability
$$Pr(X=k)=frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$



Could somebody, please, check my solution, especially the third part. Thank you.






probability balls-in-bins




probability balls-in-bins






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 11:07







Makhayama

















asked Nov 27 '18 at 19:26









MakhayamaMakhayama

262




262












  • $begingroup$
    In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    $endgroup$
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • $begingroup$
    Yes, the calculation looks fine to me.
    $endgroup$
    – DavidPM
    Nov 27 '18 at 19:38










  • $begingroup$
    Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    $endgroup$
    – Makhayama
    Nov 27 '18 at 19:40


















  • $begingroup$
    In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
    $endgroup$
    – N. F. Taussig
    Nov 27 '18 at 19:34












  • $begingroup$
    Yes, the calculation looks fine to me.
    $endgroup$
    – DavidPM
    Nov 27 '18 at 19:38










  • $begingroup$
    Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
    $endgroup$
    – Makhayama
    Nov 27 '18 at 19:40
















$begingroup$
In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 19:34






$begingroup$
In the $Pr(x = 3)$ case, you meant to say the probability that the second white ball is selected on the third draw.
$endgroup$
– N. F. Taussig
Nov 27 '18 at 19:34














$begingroup$
Yes, the calculation looks fine to me.
$endgroup$
– DavidPM
Nov 27 '18 at 19:38




$begingroup$
Yes, the calculation looks fine to me.
$endgroup$
– DavidPM
Nov 27 '18 at 19:38












$begingroup$
Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
$endgroup$
– Makhayama
Nov 27 '18 at 19:40




$begingroup$
Yes, exactly. The first white ball can be selected either on the first or second draw and the second white ball is selected on the third draw. Thus, Pr(x=3). X is the random variable showing the number of the step on which the second white ball is selected.
$endgroup$
– Makhayama
Nov 27 '18 at 19:40










1 Answer
1






active

oldest

votes


















1












$begingroup$

I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    $endgroup$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • $begingroup$
    @GrahamKemp An elegant alternative approach.
    $endgroup$
    – callculus
    Nov 28 '18 at 2:39











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016199%2fconditional-probability-urns-and-balls%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









1












$begingroup$

I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    $endgroup$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • $begingroup$
    @GrahamKemp An elegant alternative approach.
    $endgroup$
    – callculus
    Nov 28 '18 at 2:39
















1












$begingroup$

I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    $endgroup$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • $begingroup$
    @GrahamKemp An elegant alternative approach.
    $endgroup$
    – callculus
    Nov 28 '18 at 2:39














1












1








1





$begingroup$

I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.






share|cite|improve this answer









$endgroup$



I used the hypergeometric distribution for the $(k-1)$-th draw.



The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is



$$frac{binom{6}{1}cdot binom{12}{k-2}}{binom{18}{k-1}}=frac{6cdot 12!cdot (k-1)!cdot (19-k)!}{18!cdot (k-2)!cdot (14-k)!}=frac{6cdot 12!cdot (k-1)cdot (19-k)!}{18!cdot (14-k)!}$$



At the k-th draw we have to pick a white ball. The probability is $frac{5}{19-k}$. In total I get



$$P(X=k)=frac{5cdot 6cdot 12!cdot (k-1)cdot (18-k)!}{18!cdot (14-k)!}$$



This is the same result you got.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 21:02









callculuscallculus

17.9k31427




17.9k31427








  • 1




    $begingroup$
    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    $endgroup$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • $begingroup$
    @GrahamKemp An elegant alternative approach.
    $endgroup$
    – callculus
    Nov 28 '18 at 2:39














  • 1




    $begingroup$
    Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
    $endgroup$
    – Graham Kemp
    Nov 28 '18 at 1:56












  • $begingroup$
    @GrahamKemp An elegant alternative approach.
    $endgroup$
    – callculus
    Nov 28 '18 at 2:39








1




1




$begingroup$
Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
$endgroup$
– Graham Kemp
Nov 28 '18 at 1:56






$begingroup$
Alternatively you seek the probability that the six white balls are distibuted one into the first $k−1$ places, one into the next place, and four into the remaining $18−k$places. Same result: $$mathsf P(X=k)=dfrac{dbinom{k−1}1dbinom11dbinom{18−k}4}{dbinom{18}6}=dfrac{6!~(k−1)~12!~(18−k)!}{4!~18!~(14−k)!}$$
$endgroup$
– Graham Kemp
Nov 28 '18 at 1:56














$begingroup$
@GrahamKemp An elegant alternative approach.
$endgroup$
– callculus
Nov 28 '18 at 2:39




$begingroup$
@GrahamKemp An elegant alternative approach.
$endgroup$
– callculus
Nov 28 '18 at 2:39


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016199%2fconditional-probability-urns-and-balls%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more