Mixing
Probability of Tom being late to work
$begingroup$
So I have this problem here.
The probability that Tom is late for work on a rainy day is 0.4 and the probability that he is late for work on a non-rainy day is 0.1.
Can I state that P(Tom being late for work)=P(Late|Rain)+P(Late|No rain)=0.4+0.1=0.5 ?
probability probability-theory conditional-probability
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
$endgroup$
add a comment |
$begingroup$
So I have this problem here.
The probability that Tom is late for work on a rainy day is 0.4 and the probability that he is late for work on a non-rainy day is 0.1.
Can I state that P(Tom being late for work)=P(Late|Rain)+P(Late|No rain)=0.4+0.1=0.5 ?
probability probability-theory conditional-probability
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
$endgroup$
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28
add a comment |
$begingroup$
So I have this problem here.
The probability that Tom is late for work on a rainy day is 0.4 and the probability that he is late for work on a non-rainy day is 0.1.
Can I state that P(Tom being late for work)=P(Late|Rain)+P(Late|No rain)=0.4+0.1=0.5 ?
probability probability-theory conditional-probability
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
$endgroup$
So I have this problem here.
The probability that Tom is late for work on a rainy day is 0.4 and the probability that he is late for work on a non-rainy day is 0.1.
Can I state that P(Tom being late for work)=P(Late|Rain)+P(Late|No rain)=0.4+0.1=0.5 ?
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
asked Jan 30 at 18:20
The Poor JewThe Poor Jew
657
657
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28
add a comment |
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming that "the probability that tom is late for work on a rainy day" is in reference to $Pr(text{Late}mid text{Rain})$ then no. You need to know the probability of it being a rainy day. The correct statement would be:
$Pr(text{Late}) = Pr(text{Rain})Pr(text{Late}|text{Rain}) + Pr(text{No rain})Pr(text{Late}|text{No rain})$
This is the law of total probability.
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
another person has already answered this correctly, but I just wanted to help you think through this. Imagine that everyday is rainy: there would be a 40 percent chance that he is late on a daily basis. Now imagine that it is never rainy: there is a 10 percent chance that he is late. So, having a %50 percent chance of being late is impossible, the answer must be between these two values (0.1-0.4).
answered Jan 30 at 18:30
user639944user639944
111
$endgroup$
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
});
}
});
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%2f3093909%2fprobability-of-tom-being-late-to-work%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
$begingroup$
Assuming that "the probability that tom is late for work on a rainy day" is in reference to $Pr(text{Late}mid text{Rain})$ then no. You need to know the probability of it being a rainy day. The correct statement would be:
$Pr(text{Late}) = Pr(text{Rain})Pr(text{Late}|text{Rain}) + Pr(text{No rain})Pr(text{Late}|text{No rain})$
This is the law of total probability.
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Assuming that "the probability that tom is late for work on a rainy day" is in reference to $Pr(text{Late}mid text{Rain})$ then no. You need to know the probability of it being a rainy day. The correct statement would be:
$Pr(text{Late}) = Pr(text{Rain})Pr(text{Late}|text{Rain}) + Pr(text{No rain})Pr(text{Late}|text{No rain})$
This is the law of total probability.
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
$endgroup$
add a comment |
$begingroup$
Assuming that "the probability that tom is late for work on a rainy day" is in reference to $Pr(text{Late}mid text{Rain})$ then no. You need to know the probability of it being a rainy day. The correct statement would be:
$Pr(text{Late}) = Pr(text{Rain})Pr(text{Late}|text{Rain}) + Pr(text{No rain})Pr(text{Late}|text{No rain})$
This is the law of total probability.
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
$endgroup$
Assuming that "the probability that tom is late for work on a rainy day" is in reference to $Pr(text{Late}mid text{Rain})$ then no. You need to know the probability of it being a rainy day. The correct statement would be:
$Pr(text{Late}) = Pr(text{Rain})Pr(text{Late}|text{Rain}) + Pr(text{No rain})Pr(text{Late}|text{No rain})$
This is the law of total probability.
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
answered Jan 30 at 18:24
JMoravitzJMoravitz
48.7k43988
48.7k43988
add a comment |
add a comment |
$begingroup$
another person has already answered this correctly, but I just wanted to help you think through this. Imagine that everyday is rainy: there would be a 40 percent chance that he is late on a daily basis. Now imagine that it is never rainy: there is a 10 percent chance that he is late. So, having a %50 percent chance of being late is impossible, the answer must be between these two values (0.1-0.4).
answered Jan 30 at 18:30
user639944user639944
111
$endgroup$
add a comment |
$begingroup$
another person has already answered this correctly, but I just wanted to help you think through this. Imagine that everyday is rainy: there would be a 40 percent chance that he is late on a daily basis. Now imagine that it is never rainy: there is a 10 percent chance that he is late. So, having a %50 percent chance of being late is impossible, the answer must be between these two values (0.1-0.4).
answered Jan 30 at 18:30
user639944user639944
111
$endgroup$
add a comment |
$begingroup$
another person has already answered this correctly, but I just wanted to help you think through this. Imagine that everyday is rainy: there would be a 40 percent chance that he is late on a daily basis. Now imagine that it is never rainy: there is a 10 percent chance that he is late. So, having a %50 percent chance of being late is impossible, the answer must be between these two values (0.1-0.4).
answered Jan 30 at 18:30
user639944user639944
111
$endgroup$
another person has already answered this correctly, but I just wanted to help you think through this. Imagine that everyday is rainy: there would be a 40 percent chance that he is late on a daily basis. Now imagine that it is never rainy: there is a 10 percent chance that he is late. So, having a %50 percent chance of being late is impossible, the answer must be between these two values (0.1-0.4).
answered Jan 30 at 18:30
user639944user639944
111
answered Jan 30 at 18:30
user639944user639944
111
answered Jan 30 at 18:30
user639944user639944
111
answered Jan 30 at 18:30
user639944user639944
111
111
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%2f3093909%2fprobability-of-tom-being-late-to-work%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
No, we need to know the respective probability of raining and not raining to calculate this
$endgroup$
– Peter Foreman
Jan 30 at 18:24
$begingroup$
It can't be. Say the chance he is late on a rainy day is $0.9$ and the chance he is late on a non-rainy day is also $0.9$. Would the chance he is late be $1.8?$
$endgroup$
– Ross Millikan
Jan 30 at 18:28