Probability of Circuit Flowing












0












$begingroup$



An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.



Find the probability that relays A and D are conductive if the circuit is flowing.




My solution:



Let $A, B, C, D$, be the events that the corresponding relay is conductive.



Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.



$$
P(Flow) = P((A cap C) cup (B cap D))
$$



then



$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$



therefore



$$
P(Flow) = .9639
$$



I know everything up to here is right. The problem comes with the conditional probability aspect:



$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$



My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:



$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$



However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
    $endgroup$
    – amd
    Dec 6 '18 at 23:51


















0












$begingroup$



An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.



Find the probability that relays A and D are conductive if the circuit is flowing.




My solution:



Let $A, B, C, D$, be the events that the corresponding relay is conductive.



Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.



$$
P(Flow) = P((A cap C) cup (B cap D))
$$



then



$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$



therefore



$$
P(Flow) = .9639
$$



I know everything up to here is right. The problem comes with the conditional probability aspect:



$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$



My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:



$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$



However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
    $endgroup$
    – amd
    Dec 6 '18 at 23:51
















0












0








0





$begingroup$



An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.



Find the probability that relays A and D are conductive if the circuit is flowing.




My solution:



Let $A, B, C, D$, be the events that the corresponding relay is conductive.



Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.



$$
P(Flow) = P((A cap C) cup (B cap D))
$$



then



$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$



therefore



$$
P(Flow) = .9639
$$



I know everything up to here is right. The problem comes with the conditional probability aspect:



$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$



My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:



$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$



However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?










share|cite|improve this question









$endgroup$





An electrical circuit can go from the starting point to the ending point through one of two paths: through relays A and C, or through relays B and D. The relays are independent and have a probability of $0.9$ of being conductive. It is considered to be flowing if either of these paths are conductive.



Find the probability that relays A and D are conductive if the circuit is flowing.




My solution:



Let $A, B, C, D$, be the events that the corresponding relay is conductive.



Since the relays are independent, the probability of any two relays being both conductive is $.9^2 = .81$.



$$
P(Flow) = P((A cap C) cup (B cap D))
$$



then



$$
P(overline{Flow}) = P(overline{(A cap C) cup (B cap D)}) = P(overline{(A cap C)} cap overline{(B cap D)}) = (1 - .81)^2 = 0.0361
$$



therefore



$$
P(Flow) = .9639
$$



I know everything up to here is right. The problem comes with the conditional probability aspect:



$$
P(A cap D | Flow) = frac{P((A cap D) cap Flow)}{P(Flow)} = frac{P(Flow|Acap D) P(A cap D)}{P(Flow)}
$$



My intuition here tells me that, in order for the circuit to be flowing given that A and D are both working, then either B or C must be working. As such:



$$
P(A cap D | Flow) = frac{P(B cup C) P(A cap D)}{P(Flow)} = frac{.99 cdot .81}{.9639} = .8319
$$



However this is the wrong answer (correct answer is $.916$). Where did my logic go wrong?







probability conditional-probability bayes-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 23:30









Bryden CBryden C

30418




30418












  • $begingroup$
    Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
    $endgroup$
    – amd
    Dec 6 '18 at 23:51




















  • $begingroup$
    Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
    $endgroup$
    – amd
    Dec 6 '18 at 23:51


















$begingroup$
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
$endgroup$
– amd
Dec 6 '18 at 23:51






$begingroup$
Your value looks correct to me. This problem is small enough that you can verify it with a direct case analysis.
$endgroup$
– amd
Dec 6 '18 at 23:51












0






active

oldest

votes











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029212%2fprobability-of-circuit-flowing%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029212%2fprobability-of-circuit-flowing%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Berounka

Sphinx de Gizeh

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...