G'day, I've recently been given a probability question to solve and I'm not 100% on my approach towards it. It goes as follows; A dice is found to be weighted so that the chance of throwing a 6 is four times the chance of throwing a 1. Chances of throwing numbers other than 6 are equally likely. a) What is the probability of throwing a 6? b) What is the probability of throwing a 3? c) What is the probability of throwing at least a 3? For my approach I'm basically assuming the chance of throwing anything except a six is 1/6, which means the chance of throwing a 6 if 4/6. So that answers part a) and b), although I feel that the answer is to simple. And then to answer part c) I just multiply the chances of throwing a 3, 4, 5, 6 independently, and then sum those probabilities. . Any comments on this approach or help with the question would be greatly appreciated.