Timeline for very simple conditional probability question
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2010 at 20:55 | comment | added | Yemon Choi | Thanks. Well, the original downvote was overly hasty, so it needed reconsidering anyway. | |
| Aug 24, 2010 at 10:59 | comment | added | KalEl | Edited... but feel free to downvote again if you don't agree with the solution! | |
| Aug 24, 2010 at 10:58 | history | edited | KalEl | CC BY-SA 2.5 |
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| Aug 21, 2010 at 17:24 | comment | added | Yemon Choi | @KalEl: I've had a more serious think about the problem and would like (at least for now!) to retract the comment I made above. If you edit your answer then I think I can reverse my downvote (as it stands, too much time has elapsed for me to do so). | |
| Aug 19, 2010 at 12:09 | comment | added | KalEl | @Yemon, I don't see how it leads back to 2/3. @Tom, I skipped the step $P(BG\ and\ G opens)=P(G\ opens|BG)*P(BG)=q*1/2$, which is $1/2*1/2$ with $q=1/2$ ($q$ is probability of G openning the door for a family with BG, as defined in answer). | |
| Aug 18, 2010 at 21:58 | comment | added | Tom LaGatta | The error in that reasoning is that the events $\{BG\}$ and $\{G~opens~door\}$ are not independent, hence the probability of both of them occurring does not factor as the product of each individual probability. | |
| Aug 18, 2010 at 20:49 | comment | added | Yemon Choi | I don't agree with this interpretation of the question. Surely what was intended by the question is the following: given that you have seen a boy open the door, and given that you know there are two children, what is the probability that the other child is a girl? That interpretation seems to lead back to the 2/3 answer. | |
| Aug 18, 2010 at 20:10 | history | edited | KalEl | CC BY-SA 2.5 |
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| Aug 18, 2010 at 19:55 | history | edited | KalEl | CC BY-SA 2.5 |
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| Aug 18, 2010 at 19:51 | vote | accept | Not Bayes | ||
| Aug 18, 2010 at 19:37 | history | answered | KalEl | CC BY-SA 2.5 |