A pack of 2n cards, n of which are red, and another n are black. It is divided into two equal parts and a...
$begingroup$
A pack of 2n cards, n of which are red, and another n are black. It is divided into two equal parts and a card is drawn from each part. Find the probability that the cards drawn are of the same colour.
This is what I have done:
(https://i.stack.imgur.com/kEPOJ.jpg)
And this is what the book has done:
(https://i.stack.imgur.com/CTUjR.jpg)
Please say me who is correct.
( I am unable to do formatting in text as I don't have a net connection on PC)
probability
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
$endgroup$
add a comment |
$begingroup$
A pack of 2n cards, n of which are red, and another n are black. It is divided into two equal parts and a card is drawn from each part. Find the probability that the cards drawn are of the same colour.
This is what I have done:
(https://i.stack.imgur.com/kEPOJ.jpg)
And this is what the book has done:
(https://i.stack.imgur.com/CTUjR.jpg)
Please say me who is correct.
( I am unable to do formatting in text as I don't have a net connection on PC)
probability
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
$endgroup$
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47
add a comment |
$begingroup$
A pack of 2n cards, n of which are red, and another n are black. It is divided into two equal parts and a card is drawn from each part. Find the probability that the cards drawn are of the same colour.
This is what I have done:
(https://i.stack.imgur.com/kEPOJ.jpg)
And this is what the book has done:
(https://i.stack.imgur.com/CTUjR.jpg)
Please say me who is correct.
( I am unable to do formatting in text as I don't have a net connection on PC)
probability
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
$endgroup$
A pack of 2n cards, n of which are red, and another n are black. It is divided into two equal parts and a card is drawn from each part. Find the probability that the cards drawn are of the same colour.
This is what I have done:
(https://i.stack.imgur.com/kEPOJ.jpg)
And this is what the book has done:
(https://i.stack.imgur.com/CTUjR.jpg)
Please say me who is correct.
( I am unable to do formatting in text as I don't have a net connection on PC)
probability
probability
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
asked Jan 23 at 5:48
Abhishek GhoshAbhishek Ghosh
658
658
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47
add a comment |
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47
add a comment |
0
active
oldest
votes
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%2f3084131%2fa-pack-of-2n-cards-n-of-which-are-red-and-another-n-are-black-it-is-divided-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3084131%2fa-pack-of-2n-cards-n-of-which-are-red-and-another-n-are-black-it-is-divided-i%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
$begingroup$
Both are the same!
$endgroup$
– Phicar
Jan 23 at 6:38
$begingroup$
Please can you explain how?
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:40
$begingroup$
Well the only different thing is the index in the sum, but for the cases $k=0,k=n$ the argument of the sum is 0. You both got $$sum _{k=0}^n2frac{k(n-k)}{n^2}$$
$endgroup$
– Phicar
Jan 23 at 6:44
$begingroup$
Thanks for helping me👍
$endgroup$
– Abhishek Ghosh
Jan 23 at 6:47