Timeline for Card picking dependant probability without replacement - P(6,6,Red)
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2019 at 9:08 | comment | added | ArsenBerk | I thank you too for your correction and you are welcome :) | |
| Jun 10, 2019 at 8:45 | comment | added | Lewis Morris | Thank you so much, you've shed some light on something I was having trouble understanding before. Amazing !! :) | |
| Jun 10, 2019 at 8:08 | history | edited | ArsenBerk | CC BY-SA 4.0 |
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| Jun 10, 2019 at 8:08 | comment | added | ArsenBerk | Oh, that is mistake on my part, I am really sorry for that. Yes you are absolutely right, in my answer too, it should be $\frac{26}{50}$ and $\frac{25}{50}$ in case 1 and case 2 respectively. Seems like you checked my answer here :) | |
| Jun 10, 2019 at 4:50 | comment | added | Lewis Morris | I've had to change it around a bit because the probability of picking a red card is higher as there could be between 24-26 red cards not 11-13 (two suits, not one). I believe that for case 3 it would be 2/52 * 1/51 * 24/50 ≈ 0.04% (slightly lower than case 1 as there are less red cards to choose from) . I hope this means the overall probability of P(6,6,Red) ≈ 0.2262% | |
| Jun 10, 2019 at 4:43 | comment | added | Lewis Morris | Thank you so much for your explanation. Its really helped me and I think I have the answer. | |
| Jun 9, 2019 at 19:08 | history | answered | ArsenBerk | CC BY-SA 4.0 |