Mixing
Is this a trick question in probability?
$begingroup$
Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=frac{1}{3}$
My intuition tells me since $P(A)= frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $frac{1}{3}$ and even more since I looked at this answer.
Is this really just a trick question or is there more to it?
probability probability-theory statistics probability-distributions conditional-probability
edited Jan 24 at 22:40
David G. Stork
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
$endgroup$
add a comment |
$begingroup$
Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=frac{1}{3}$
My intuition tells me since $P(A)= frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $frac{1}{3}$ and even more since I looked at this answer.
Is this really just a trick question or is there more to it?
probability probability-theory statistics probability-distributions conditional-probability
edited Jan 24 at 22:40
David G. Stork
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
$endgroup$
2
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
1
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10
add a comment |
$begingroup$
Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=frac{1}{3}$
My intuition tells me since $P(A)= frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $frac{1}{3}$ and even more since I looked at this answer.
Is this really just a trick question or is there more to it?
probability probability-theory statistics probability-distributions conditional-probability
edited Jan 24 at 22:40
David G. Stork
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
$endgroup$
Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=frac{1}{3}$
My intuition tells me since $P(A)= frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $frac{1}{3}$ and even more since I looked at this answer.
Is this really just a trick question or is there more to it?
probability probability-theory statistics probability-distributions conditional-probability
probability probability-theory statistics probability-distributions conditional-probability
edited Jan 24 at 22:40
David G. Stork
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
edited Jan 24 at 22:40
David G. Stork
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
edited Jan 24 at 22:40
David G. Stork
11.1k41432
edited Jan 24 at 22:40
David G. Stork
11.1k41432
edited Jan 24 at 22:40
David G. Stork
11.1k41432
11.1k41432
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
asked Jan 24 at 22:19
Colin HicksColin Hicks
408210
408210
2
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
1
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10
add a comment |
2
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
1
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10
2
2
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
1
1
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to use Bayes theorem for these kind of exercises :
$$begin{align}P(A mid B) &= frac{P(A cap B)}{P(B)} \&= P(A)& text{since A et B are independent}end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky.
But yes that's it.
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
$endgroup$
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
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%2f3086430%2fis-this-a-trick-question-in-probability%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
$begingroup$
You need to use Bayes theorem for these kind of exercises :
$$begin{align}P(A mid B) &= frac{P(A cap B)}{P(B)} \&= P(A)& text{since A et B are independent}end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky.
But yes that's it.
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
$endgroup$
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
add a comment |
$begingroup$
You need to use Bayes theorem for these kind of exercises :
$$begin{align}P(A mid B) &= frac{P(A cap B)}{P(B)} \&= P(A)& text{since A et B are independent}end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky.
But yes that's it.
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
$endgroup$
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
add a comment |
$begingroup$
You need to use Bayes theorem for these kind of exercises :
$$begin{align}P(A mid B) &= frac{P(A cap B)}{P(B)} \&= P(A)& text{since A et B are independent}end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky.
But yes that's it.
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
$endgroup$
You need to use Bayes theorem for these kind of exercises :
$$begin{align}P(A mid B) &= frac{P(A cap B)}{P(B)} \&= P(A)& text{since A et B are independent}end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky.
But yes that's it.
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
edited Jan 24 at 22:27
Graham Kemp
86.8k43579
86.8k43579
answered Jan 24 at 22:22
ThinkingThinking
1,22716
answered Jan 24 at 22:22
ThinkingThinking
1,22716
answered Jan 24 at 22:22
ThinkingThinking
1,22716
1,22716
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
add a comment |
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
$begingroup$
I was trying to use total probability $P(A)=P(A|B)P(B)+P(A|B')P(B')$ but I couldn't figure out how to solve for P(B) which would give me the answer. I quickly realized however, that this was likely a trick.
$endgroup$
– Colin Hicks
Jan 24 at 22:32
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%2f3086430%2fis-this-a-trick-question-in-probability%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
2
$begingroup$
That’s it! $space$
$endgroup$
– Frpzzd
Jan 24 at 22:20
$begingroup$
Is $B'$ a typo or does it mean something? Possibly the non-occurence of $B$? In that case you answer is still right of course.
$endgroup$
– kmm
Jan 24 at 22:30
1
$begingroup$
@kmm $B', B^c, neg B$ and others are common ways to write the complementary event to $B$, i.e. $Omega setminus B$ where $Omega$ is the sample space.
$endgroup$
– JMoravitz
Jan 24 at 22:31
$begingroup$
No need to tell me. Curious, I've seen $B^c$, $lnot B$, $overline{B}$ but never before $B'$
$endgroup$
– kmm
Jan 27 at 17:10