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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.

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# Unique Dice Throwing Probability Question

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