Timeline for Probability of selection containing certain options
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2020 at 11:30 | comment | added | Gertjan Brouwer | I know I need to multiply by 3, because of it being ordered. I however still don't see the difference between the two examples. I'll work through the book to get to your way of calculating probabilities and see if I understand at that point. I'll get back to you on this. Thanks a lot | |
| Sep 26, 2020 at 11:27 | vote | accept | Gertjan Brouwer | ||
| Sep 25, 2020 at 23:36 | comment | added | BruceET |
Two Hearts and a Diamond: Unordered, choose(13,2)*13/choose(52,3) returns $0.04588235.$ Ordered, (13*12*13)/(52*51*50) returns $0.01529412,$ which needs to be multiplied by $3$ to arrange the cards HHD once you've got them. // If you write ${13 \choose 2}$ as $13!/(11!\cdot 2!),$ etc. maybe you can track the difference between the two methods.
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| Sep 25, 2020 at 23:25 | comment | added | BruceET | As soon as you write $\frac{5}{24} \times \frac{7}{23} \times \frac{6}{22} = 0.0173$ you are looking at ordered samples. That's why you need the modification of the answer book. For the the card example you're treating all Hearts as the same, so there is no worrying about order. // By your approach you'd need to consider order for the probability of two Hearts and a Diamond. | |
| Sep 25, 2020 at 23:06 | comment | added | Gertjan Brouwer | Thank you for your answer. I have never seen probabilities being calculated this way. It will probably take me a couple of days to get to that section. Do you mind adding a part where you explain using my way of calculating probabilities? | |
| Sep 25, 2020 at 19:48 | history | edited | BruceET | CC BY-SA 4.0 |
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| Sep 25, 2020 at 19:37 | history | edited | BruceET | CC BY-SA 4.0 |
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| Sep 25, 2020 at 19:32 | history | answered | BruceET | CC BY-SA 4.0 |