Recent content by raptor1770

OK based on your suggestion I calculated the probability of rolling a 6 based on Ron getting the loaded die and the unloaded die. I think it's right please let me know. If Ron got the loaded die: (1/6 * 1) + (1/3 * 0) = 1/6 If Ron got the unloaded die: (1/6 * 1/2) + (1/3 * 1/2) = 1/4...

For question 4 I think the solution is something like this. Let L = Loaded U = Unloaded p = probability The probability of rolling 6 given 6 is = \frac{p(6|6) p(6)}{p(6|L) p(L) + p(6|U) p(U)} This is derived using Baye's theorem. I think the numerator is incorrect...

Well I'm really at a loss as to where to start but I'll give you what I got so far. The probability of rolling a six on a normal die is 1/6 on a fixed die it is 1/3 The probability that Fran picked the loaded die is 1/3; the regular die 2/3. So the total probability of rolling a 6 is...

This is part of the study guide for an exam and I'm not sure how to start. Homework Statement Consider three identical-looking dice. Two of the dice are ordinary fair dice (six equally-likely faces, numbered 1,2,...,6), but the third die is "loaded" (the face that ordinarily has a 1 has...