The probability of Bus A arriving before Bus BExpected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables
Can the US President recognize Israel’s sovereignty over the Golan Heights for the USA or does that need an act of Congress?
Pre-mixing cryogenic fuels and using only one fuel tank
What does chmod -u do?
What should you do if you miss a job interview (deliberately)?
Extract more than nine arguments that occur periodically in a sentence to use in macros in order to typset
I'm the sea and the sun
Temporarily disable WLAN internet access for children, but allow it for adults
What happens if you are holding an Iron Flask with a demon inside and walk into an Antimagic Field?
Moving brute-force search to FPGA
Is there a RAID 0 Equivalent for RAM?
Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?
Can disgust be a key component of horror?
Does malloc reserve more space while allocating memory?
Is this toilet slogan correct usage of the English language?
Mimic lecturing on blackboard, facing audience
Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?
The probability of Bus A arriving before Bus B
What is going on with 'gets(stdin)' on the site coderbyte?
Why can Carol Danvers change her suit colours in the first place?
How could a planet have erratic days?
Strong empirical falsification of quantum mechanics based on vacuum energy density
Why does the Sun have different day lengths, but not the gas giants?
What's the difference between releasing hormones and tropic hormones?
When were female captains banned from Starfleet?
The probability of Bus A arriving before Bus B
Expected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
add a comment |
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago
add a comment |
$begingroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
New contributor
$endgroup$
Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?
My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?
probability
probability
New contributor
New contributor
edited 1 hour ago
IrinaS
New contributor
asked 1 hour ago
IrinaSIrinaS
62
62
New contributor
New contributor
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago
add a comment |
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
Here is the graph:
$hspace2cm$
Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
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
);
);
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3158927%2fthe-probability-of-bus-a-arriving-before-bus-b%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
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
add a comment |
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
add a comment |
$begingroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
$endgroup$
Let $A_e$ ($A$ early) be the event that $A$ arrives before $3$pm.
Let $B_ell$ be the event that $B$ arrives after $4$pm.
Let $C$ be the union : $C=A_e cup B_ell$.
Let $X$ be the event of interest ( $A$ arrives before $B$).
What we know (don't we?) that is $P(X | C)=1$ and $P(X | overlineC)=0.5$
Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$
Can you go on from here ?
answered 1 hour ago
leonbloyleonbloy
41.8k647108
41.8k647108
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
add a comment |
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
$begingroup$
Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
$endgroup$
– IrinaS
4 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
Here is the graph:
$hspace2cm$
Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
Here is the graph:
$hspace2cm$
Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$
$endgroup$
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
add a comment |
$begingroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
Here is the graph:
$hspace2cm$
Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$
$endgroup$
Guide:
1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.
2) The total area of the rectangle and square is $5$, so pdf is $1/5$.
3) Find total area of the square and rectangle above the line, which is $4.5$.
5) Finally, the required probability is $4.5cdot 1/5=9/10$.
Here is the graph:
$hspace2cm$
Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$
edited 5 mins ago
answered 1 hour ago
farruhotafarruhota
21.4k2841
21.4k2841
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
add a comment |
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
$begingroup$
do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
$endgroup$
– IrinaS
15 mins ago
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
add a comment |
$begingroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
$endgroup$
First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.
You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.
So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.
answered 1 hour ago
Robert ShoreRobert Shore
3,410323
3,410323
add a comment |
add a comment |
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3158927%2fthe-probability-of-bus-a-arriving-before-bus-b%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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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
$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
1 hour ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
44 mins ago