Mixing
Probability and Counting
$begingroup$
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
$endgroup$
add a comment |
$begingroup$
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
$endgroup$
add a comment |
$begingroup$
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
$endgroup$
I have just picked up a text on discrete math and its been ages since I have done this so if anyone can show me with steps to correct my fault, that would be so great.
Repair facility has 25 failed keyboards, 6 of which have electrical defects and 19 which have mechanical defects.
(1) How many ways are there to randomly select 5 of these keyboards for a thorough inspection (order does not matter)? 25 choose 5
(2) IN how many ways can a sample of 5 keyboards be selected so that exactly tow have an electrical defect? (25 choose 5)/(6 choose 5)
(3) If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect? (5 choose 4)....?
Am I correct in my thinking logic or way off?
probability discrete-mathematics
probability discrete-mathematics
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
edited Sep 18 '13 at 4:42
hherklj kljkljklj
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
asked Sep 18 '13 at 4:35
hherklj kljkljkljhherklj kljkljklj
5624
5624
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
$endgroup$
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more 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%2f497153%2fprobability-and-counting%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
$begingroup$
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
$endgroup$
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
$begingroup$
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
$endgroup$
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
$begingroup$
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
$endgroup$
For part 2, you need exactly two out of five so there are ${6 choose 2}$ ways to do that leaving ${19choose 3}$ for the remaining mechanically failed devices. So in total there are ${6choose 2}{19choose 3}$ total ways. Now for part three, think of what it means to say "at least" and use the same process. So you need a sample with exactly 4 plus a sample with 5....
For cases where we are sampling without replacement, a hyper geometric distribution is used. Say you want the probability of $K$ out $N$ successes with a sample size $n$. Then the mass function looks like
$$P(X<k)=sum_{i=0}^k{frac{{Kchoose i}{N-K choose n-i}}{N choose n}}$$
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
edited Sep 18 '13 at 11:26
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
answered Sep 18 '13 at 4:42
Eleven-ElevenEleven-Eleven
5,49072659
5,49072659
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more comment
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
$begingroup$
So the for 3, by "at least", would it be (5 choose 4)*(19 choose 1)? Or am I thinking of "at least" from the wrong perspective?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:49
2
2
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
Wrong perspective... You have 19 mechanically failed keyboards and 6 electrically failed keyboards. So it should be ${6choose 1}{19choose 4}$, but at least means you can have 4 or 5 mechanically defective keyboards in your sample.... Can you finish it then?
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 4:55
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
So then if I am leaning in the right direction, would it be ((6 choose 1)*(19 choose 4))/(5 choose 4) ?
$endgroup$
– hherklj kljkljklj
Sep 18 '13 at 4:58
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
No...you have a new criterion to meet with the second case of 5 mechanically failed keyboards so that is ${19choose 5}$. Now you must sum the two cases. This is counting , not probability so there should be no division in your method. If you were looking for the probabilities of these cases you divide by ${25choose 5}$. This is sampling without replacement and calls for the hyper geometric distribution.
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:04
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
$begingroup$
Okay so I saw part 3 calls for probability so you should have $frac{{19choose 4}{6choose 1}+{19choose 5}{6choose 0}}{{25choose 5}}$
$endgroup$
– Eleven-Eleven
Sep 18 '13 at 5:08
|
show 1 more 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%2f497153%2fprobability-and-counting%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
