What is the probability that exactly one event will occur if…
$begingroup$
Two events $A$ and $B$ have probabilities respectively of $p(A)$ and $p(B)$.
What is the probability that exactly one event will occur if
- $A$ and $B$ are mutually exclusive
- $A$ and $B$ are not mutually exclusive, but dependent
- $A$ and $B$ are not mutually exclusive, but independent
Attempt:
- $P(A)+P(B)$
- $P(A)P(B^c)+P(B)P(A^c)$
- $P(Acap B^c)+P(A^ccap B)$
probability-theory
asked Jul 11 '17 at 4:40
akse232akse232
204
$endgroup$
add a comment |
$begingroup$
Two events $A$ and $B$ have probabilities respectively of $p(A)$ and $p(B)$.
What is the probability that exactly one event will occur if
- $A$ and $B$ are mutually exclusive
- $A$ and $B$ are not mutually exclusive, but dependent
- $A$ and $B$ are not mutually exclusive, but independent
Attempt:
- $P(A)+P(B)$
- $P(A)P(B^c)+P(B)P(A^c)$
- $P(Acap B^c)+P(A^ccap B)$
probability-theory
asked Jul 11 '17 at 4:40
akse232akse232
204
$endgroup$
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51
add a comment |
$begingroup$
Two events $A$ and $B$ have probabilities respectively of $p(A)$ and $p(B)$.
What is the probability that exactly one event will occur if
- $A$ and $B$ are mutually exclusive
- $A$ and $B$ are not mutually exclusive, but dependent
- $A$ and $B$ are not mutually exclusive, but independent
Attempt:
- $P(A)+P(B)$
- $P(A)P(B^c)+P(B)P(A^c)$
- $P(Acap B^c)+P(A^ccap B)$
probability-theory
asked Jul 11 '17 at 4:40
akse232akse232
204
$endgroup$
Two events $A$ and $B$ have probabilities respectively of $p(A)$ and $p(B)$.
What is the probability that exactly one event will occur if
- $A$ and $B$ are mutually exclusive
- $A$ and $B$ are not mutually exclusive, but dependent
- $A$ and $B$ are not mutually exclusive, but independent
Attempt:
- $P(A)+P(B)$
- $P(A)P(B^c)+P(B)P(A^c)$
- $P(Acap B^c)+P(A^ccap B)$
probability-theory
probability-theory
asked Jul 11 '17 at 4:40
akse232akse232
204
asked Jul 11 '17 at 4:40
akse232akse232
204
asked Jul 11 '17 at 4:40
akse232akse232
204
asked Jul 11 '17 at 4:40
akse232akse232
204
asked Jul 11 '17 at 4:40
akse232akse232
204
204
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51
add a comment |
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$ P(Acap B^c) + P(Bcap A^c)
= P(A) + P(B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) = P(A) + P(B) - 2P(A)P(B)$
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
$endgroup$
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%2f2354435%2fwhat-is-the-probability-that-exactly-one-event-will-occur-if%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
$begingroup$
$ P(Acap B^c) + P(Bcap A^c)
= P(A) + P(B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) = P(A) + P(B) - 2P(A)P(B)$
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
$endgroup$
add a comment |
$begingroup$
$ P(Acap B^c) + P(Bcap A^c)
= P(A) + P(B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) = P(A) + P(B) - 2P(A)P(B)$
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
$endgroup$
add a comment |
$begingroup$
$ P(Acap B^c) + P(Bcap A^c)
= P(A) + P(B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) = P(A) + P(B) - 2P(A)P(B)$
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
$endgroup$
$ P(Acap B^c) + P(Bcap A^c)
= P(A) + P(B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) $$P(Acap B^c) + P(Bcap A^c)
= P(Acup B) - P(Acap B) = P(A) + P(B) - 2P(A)P(B)$
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
answered Jul 11 '17 at 23:10
user3658307user3658307
5,0083947
5,0083947
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%2f2354435%2fwhat-is-the-probability-that-exactly-one-event-will-occur-if%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
$begingroup$
You mixed 2 and 3, and the independent case could be simplified a bit
$endgroup$
– Hagen von Eitzen
Jul 11 '17 at 4:46
$begingroup$
$P(A cap B^c) + P(A^c cap B)$ is true for all three, but in case (2) it can't be simplified.
$endgroup$
– Robert Israel
Jul 11 '17 at 4:51