Timeline for Probability question (Birthday problem)
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2017 at 20:43 | history | edited | Henry |
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| Mar 7, 2015 at 21:58 | answer | added | BruceET | timeline score: 0 | |
| May 3, 2012 at 16:17 | vote | accept | Low Scores | ||
| May 3, 2012 at 4:59 | answer | added | André Nicolas | timeline score: 5 | |
| May 3, 2012 at 4:52 | comment | added | yiyi | @LowScores If you are counting the number of ways to get just one match then as Andre Nicolas said, you need to count the number of ways to get one match, but then also make sure there are no other matches by having the rest not be matching. | |
| May 3, 2012 at 4:33 | comment | added | Arturo Magidin | @LowScores: You seem to be trying to both count and compute probabilities at the same time. Can't really do that. It's also false that "there are 23 people and each has a probability of 1/365 of sharing the same birthday". Sharing the same birthday with whom? Again: are you trying to count the number of ways in which exactly two people can share a birthday (so that then you can compute probability as "number of good cases"/"number of total cases", or are you trying to compute probability along the way? It might be simpler if you first count, and then you compute probability. | |
| May 3, 2012 at 4:27 | comment | added | Low Scores | My thinking is that there are 23 people and each have a probability of $\frac{1}{365}$ of sharing the same birthday if I want 2, then I would have to pick 2 people from that group of 23. That is the reason why I multiplied $\frac{1}{365}$ by $\binom{23}{2}$. My calculation for the other 21 people takes into account that every person has a different birthday from the other. Andre seems to think I should consider 22 people and not 21 for the numerator. I've got to think... | |
| May 3, 2012 at 4:17 | comment | added | Arturo Magidin | Why are you multiplying by $1/365^2$? Are you trying to count (as your use of "there are .... ways..." suggests) or probability? And are you taking into account that among the other 21 people two may share the same birthday, though not have it the same day as the two you selected to begin with? I don't understand your throught process... | |
| May 3, 2012 at 4:17 | comment | added | André Nicolas | The idea is good. We need to choose the $2$ people. Then we need to choose a birthday for them. Then we need to choose different birthdays for the others. The numerator will be a little different from yours, we need to choose different birthdays for $22$ "people," our birthday pair and the other people. The denominator is right. | |
| May 3, 2012 at 4:10 | answer | added | yiyi | timeline score: -1 | |
| May 3, 2012 at 4:07 | history | asked | Low Scores | CC BY-SA 3.0 |