Probability of exactly one defective unit
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(textDefect=1) = P(textDefect)times P(textNot defect)times P(textNot defect) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28224439
asked Dec 9 at 10:45
CruZ
436
add a comment |
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(textDefect=1) = P(textDefect)times P(textNot defect)times P(textNot defect) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28224439
asked Dec 9 at 10:45
CruZ
436
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(textDefect=1) = P(textDefect)times P(textNot defect)times P(textNot defect) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
edited Dec 9 at 13:33
amWhy
191k28224439
asked Dec 9 at 10:45
CruZ
436
Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.
What is the probability that exactly one unit is defective?
My answer would be
$P(textDefect=1) = P(textDefect)times P(textNot defect)times P(textNot defect) = 5/100 times 95/99 times 94/98$
However, I am not sure whether or not this is correct or not. Can someone verify?
probability
probability
edited Dec 9 at 13:33
amWhy
191k28224439
asked Dec 9 at 10:45
CruZ
436
edited Dec 9 at 13:33
amWhy
191k28224439
asked Dec 9 at 10:45
CruZ
436
edited Dec 9 at 13:33
amWhy
191k28224439
edited Dec 9 at 13:33
amWhy
191k28224439
edited Dec 9 at 13:33
amWhy
191k28224439
191k28224439
asked Dec 9 at 10:45
CruZ
436
asked Dec 9 at 10:45
CruZ
436
asked Dec 9 at 10:45
CruZ
436
436
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
1
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Your answer should be $$fracbinom51binom952binom1003$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom51$
- Choose two non-defective ones: $binom952$
- Chose any three: $binom1003$
$$P(mbox"exactly 1 defective") = fracbinom51cdot binom952binom1003$$
answered Dec 9 at 10:55
trancelocation
8,9051521
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%2f3032258%2fprobability-of-exactly-one-defective-unit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Your answer should be $$fracbinom51binom952binom1003$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
accepted
Your answer should be $$fracbinom51binom952binom1003$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Your answer should be $$fracbinom51binom952binom1003$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
Your answer should be $$fracbinom51binom952binom1003$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
answered Dec 9 at 10:56
BelowAverageIntelligence
5221213
5221213
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
– CruZ
Dec 9 at 10:59
1
1
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
sorry, the 500 was a typo...
– BelowAverageIntelligence
Dec 9 at 11:00
1
1
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
@CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
– Quintec
Dec 9 at 14:59
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
Ah thank you for clearing that up, my mistake! Cheers!
– CruZ
Dec 9 at 18:07
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom51$
- Choose two non-defective ones: $binom952$
- Chose any three: $binom1003$
$$P(mbox"exactly 1 defective") = fracbinom51cdot binom952binom1003$$
answered Dec 9 at 10:55
trancelocation
8,9051521
add a comment |
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom51$
- Choose two non-defective ones: $binom952$
- Chose any three: $binom1003$
$$P(mbox"exactly 1 defective") = fracbinom51cdot binom952binom1003$$
answered Dec 9 at 10:55
trancelocation
8,9051521
add a comment |
up vote
7
down vote
up vote
7
down vote
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom51$
- Choose two non-defective ones: $binom952$
- Chose any three: $binom1003$
$$P(mbox"exactly 1 defective") = fracbinom51cdot binom952binom1003$$
answered Dec 9 at 10:55
trancelocation
8,9051521
Here is a suggestion how to proceed as ordering does not play a role
- Choose one defective item: $binom51$
- Choose two non-defective ones: $binom952$
- Chose any three: $binom1003$
$$P(mbox"exactly 1 defective") = fracbinom51cdot binom952binom1003$$
answered Dec 9 at 10:55
trancelocation
8,9051521
answered Dec 9 at 10:55
trancelocation
8,9051521
answered Dec 9 at 10:55
trancelocation
8,9051521
answered Dec 9 at 10:55
trancelocation
8,9051521
8,9051521
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3032258%2fprobability-of-exactly-one-defective-unit%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
1
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
Dec 9 at 10:49
1
Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = DDNUDNDUNDD, where D is af defective unit and N is a non-defective unit?
– CruZ
Dec 9 at 10:53
2
Yes. You will get the same answer mentioned below.
– Thomas Shelby
Dec 9 at 11:03