Suppose a channel where 0 or 1 is sent, what is probability 1 was sent given 1 was received?
$begingroup$
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 '18 at 4:48
platty
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
$endgroup$
add a comment |
$begingroup$
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 '18 at 4:48
platty
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
$endgroup$
add a comment |
$begingroup$
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
edited Dec 3 '18 at 4:48
platty
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
$endgroup$
I have the following probability.
probability a $0$ is sent is $0.4$
probability a $1$ is sent is $0.6$
probability that due to noise, a $0$ is changed to a $1$ during transmission is $0.2$
probability that due to noise, a $1$ is changed to a $0$ during transmission is $0.1$
Here is what I think should be the solution:
Let $A$ be $1$ was received, $B$ be $1$ was sent
$P(B) = 0.6$
$P(A|B) = 0.6 * 0.9 = 0.54$
$P(A) = 0.6 * 0.9 + 0.4 * 0.2 = 0.62$
Thus using $$P(B|A) = frac{P(A|B)P(B)}{P(A)}$$
$P(B|A) = 52.26%$
Is my solution correct?
probability proof-verification conditional-probability
probability proof-verification conditional-probability
edited Dec 3 '18 at 4:48
platty
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
edited Dec 3 '18 at 4:48
platty
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
edited Dec 3 '18 at 4:48
platty
3,370320
edited Dec 3 '18 at 4:48
platty
3,370320
edited Dec 3 '18 at 4:48
platty
3,370320
3,370320
asked Dec 3 '18 at 4:44
CupCup
276
asked Dec 3 '18 at 4:44
CupCup
276
asked Dec 3 '18 at 4:44
CupCup
276
276
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
$endgroup$
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%2f3023642%2fsuppose-a-channel-where-0-or-1-is-sent-what-is-probability-1-was-sent-given-1-w%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
$begingroup$
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
$endgroup$
Not quite. $P(A | B) = 0.9$; what you wrote above is actually $P(A cap B)$. Everything else looks fine, so you should end up with $P(B mid A) = frac{0.9(0.6)}{0.62} approx 0.87$.
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
answered Dec 3 '18 at 4:47
plattyplatty
3,370320
3,370320
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%2f3023642%2fsuppose-a-channel-where-0-or-1-is-sent-what-is-probability-1-was-sent-given-1-w%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
