If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraWhat is the probability of an event happening in some interval given probability of it in x interval?what is the probability that they getting on the bus?probability question shooting starThree machines are in use certain number of minutes each hour. What is the probability that at least one will be in use at a given moment of the day?What will the probability of scoring prime no. of goals in a game?Probability of takeover announcement in the next hourProbability of an event occurring within a smaller time interval if one knows the probability of occurrence over a larger time intervalSimple probability calculationsWhat is the probability that some random event won't happen in the next 10 minutes given it happened exactly twice in the last 120 minutes?Find the total probability of at least one take-off distaster happening over one yearSolving simple Probability with variational inference
What's parked in Mil Moscow helicopter plant?
Co-worker works way more than he should
Password Generator in batch
What if Force was not Mass times Acceleration?
"Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English?
Are all CP/M-80 implementations binary compatible?
Is accepting an invalid credit card number a security issue?
std::is_constructible on incomplete types
What is this word supposed to be?
Map material from china not allowed to leave the country
What was Apollo 13's "Little Jolt" after MECO?
Can you stand up from being prone using Skirmisher outside of your turn?
What *exactly* is electrical current, voltage, and resistance?
What is the ongoing value of the Kanban board to the developers as opposed to management
PIC mathematical operations weird problem
Do I need to protect SFP ports and optics from dust/contaminants? If so, how?
Is this homebrew racial feat, Stonehide, balanced?
Does Mathematica have an implementation of the Poisson Binomial Distribution?
Suing a Police Officer Instead of the Police Department
I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?
Additive group of local rings
Is Bran literally the world's memory?
Function to calculate red-edgeNDVI in Google Earth Engine
"Rubric" as meaning "signature" or "personal mark" -- is this accepted usage?
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraWhat is the probability of an event happening in some interval given probability of it in x interval?what is the probability that they getting on the bus?probability question shooting starThree machines are in use certain number of minutes each hour. What is the probability that at least one will be in use at a given moment of the day?What will the probability of scoring prime no. of goals in a game?Probability of takeover announcement in the next hourProbability of an event occurring within a smaller time interval if one knows the probability of occurrence over a larger time intervalSimple probability calculationsWhat is the probability that some random event won't happen in the next 10 minutes given it happened exactly twice in the last 120 minutes?Find the total probability of at least one take-off distaster happening over one yearSolving simple Probability with variational inference
$begingroup$
Poorly worded title but I don't know what the nature of this probability question is called.
I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:
The chance of the dog not barking in a given hour is 1-84% = 16%
If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.
Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.
Question1: Is this correct?
Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:
p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.
Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.
91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.
What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?
probability
$endgroup$
add a comment |
$begingroup$
Poorly worded title but I don't know what the nature of this probability question is called.
I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:
The chance of the dog not barking in a given hour is 1-84% = 16%
If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.
Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.
Question1: Is this correct?
Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:
p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.
Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.
91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.
What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?
probability
$endgroup$
1
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41
add a comment |
$begingroup$
Poorly worded title but I don't know what the nature of this probability question is called.
I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:
The chance of the dog not barking in a given hour is 1-84% = 16%
If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.
Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.
Question1: Is this correct?
Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:
p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.
Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.
91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.
What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?
probability
$endgroup$
Poorly worded title but I don't know what the nature of this probability question is called.
I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?
Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:
The chance of the dog not barking in a given hour is 1-84% = 16%
If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.
Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.
Question1: Is this correct?
Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:
p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.
Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.
91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.
What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?
probability
probability
edited Apr 20 at 3:40
YuiTo Cheng
2,82141139
2,82141139
asked Apr 19 at 21:32
Doug FirDoug Fir
46818
46818
1
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41
add a comment |
1
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41
1
1
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.
However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.
$endgroup$
add a comment |
$begingroup$
The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.
We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.
If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.
New contributor
$endgroup$
add a comment |
$begingroup$
The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.
Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.
$endgroup$
add a comment |
Your Answer
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%2f3194064%2fif-the-probability-of-a-dog-barking-one-or-more-times-in-a-given-hour-is-84-th%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.
However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.
$endgroup$
add a comment |
$begingroup$
The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.
However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.
$endgroup$
add a comment |
$begingroup$
The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.
However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.
$endgroup$
The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.
However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.
answered Apr 19 at 21:40
kccukccu
11.8k11231
11.8k11231
add a comment |
add a comment |
$begingroup$
The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.
We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.
If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.
New contributor
$endgroup$
add a comment |
$begingroup$
The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.
We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.
If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.
New contributor
$endgroup$
add a comment |
$begingroup$
The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.
We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.
If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.
New contributor
$endgroup$
The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.
We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.
If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.
New contributor
New contributor
answered Apr 19 at 21:44
ItsJustLogsBroItsJustLogsBro
461
461
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.
Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.
$endgroup$
add a comment |
$begingroup$
The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.
Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.
$endgroup$
add a comment |
$begingroup$
The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.
Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.
$endgroup$
The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.
Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.
answered Apr 19 at 21:47
herb steinbergherb steinberg
3,2632311
3,2632311
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.
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%2f3194064%2fif-the-probability-of-a-dog-barking-one-or-more-times-in-a-given-hour-is-84-th%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
$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
Apr 19 at 21:41