Timeline for How to calculate the probability of getting 2 heads wen the coins is biased?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2019 at 10:28 | comment | added | Daria | Thank you for guiding me through. | |
| Jun 25, 2019 at 10:27 | comment | added | gunes | Yes, that is right. | |
| Jun 25, 2019 at 10:25 | vote | accept | Daria | ||
| Jun 25, 2019 at 10:25 | |||||
| Jun 25, 2019 at 10:23 | comment | added | Daria | As for the denominator we have (0.7)^2(0.5) + (0.4)^2(0.5), right? | |
| Jun 25, 2019 at 10:18 | comment | added | gunes | Yes, the numerator in the Bayes formulation is $(0.7)^2(0.5)$ | |
| Jun 25, 2019 at 10:17 | comment | added | Daria | Then we get as for numerator: 0.7^2* 0.5, don't we? | |
| Jun 25, 2019 at 10:15 | comment | added | gunes | You need the prior when substituting $P(C_1)$ and $P(C_2)$ | |
| Jun 25, 2019 at 10:15 | comment | added | Daria | Then I can't understand why do we need prior here. | |
| Jun 25, 2019 at 10:14 | comment | added | gunes | $P(E|C_1)$ means the probability of two heads (i.e. HH) when know we use coin 1. | |
| Jun 25, 2019 at 10:12 | comment | added | Daria | How do you get P (E given C1) to be equal to 0.7^2? | |
| Jun 25, 2019 at 10:09 | history | answered | gunes | CC BY-SA 4.0 |