(New page: This is the approach I took to solving this problem: ---- P(R2 | R1) = P(R2_1 n R1_1)P(1-side)/P(R2_1) + P(R2_2 n R1_2)P(2-side)/P(R2_2) + P(R2_3 n R1_3)P(3-side)/P(R2_3) After plugging...) |
(No difference)
|
Latest revision as of 17:03, 16 September 2008
This is the approach I took to solving this problem:
P(R2 | R1) = P(R2_1 n R1_1)P(1-side)/P(R2_1) + P(R2_2 n R1_2)P(2-side)/P(R2_2) + P(R2_3 n R1_3)P(3-side)/P(R2_3)
After plugging in the probabilities you are left with:
P(R2 | R1) = ((1/6)(1/6)(1/3)/(1/6)) + ((1/3)(1/3)(1/3)/(1/3)) + ((1/2)(1/2)(1/3)/(1/2))
= (1/18) + (1/9) + (1/6) = 1/3
I am not entirely sure if that is correct, so it would be helpful if someone checks the logic behind my approach and corrects me if I am wrong. Thanks a lot!
Jared McNealis
