The probabilities of outcomes of throwing a weighted die 
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.
 A: Let the probability of throwing a $1$ be $p$. By what we were told, the probability of $2$ is also $p$, and the same is true for $3,4,5$. The probability of $6$ is $4p$. So $5p+4p=1$, and $p=\frac{1}{9}$. 
Then for example the probability of throwing at most $2$ is $\frac{2}{9}$, so the probability of at least $3$ is $\frac{7}{9}$. 
A: The probability or throwing a $6$ is $\frac{4}{9}$.  The chance of throwing anything except a $6$ is not $\frac{1}{6}$, as you said, but $\frac{1}{9}$.  Check this is true.  $b$ follows immediately from the above statements. See if you can figure out part $c$ with this information. 
A: If the chance of throwing each number other than a six is $1/6$, and the chance of throwing a six is $4/6$ then the chance of throwing any number at all is $5\times 1/6+4/6 = 9/6$.  Oppsie.
So that can't be right. It has to be $1$, so multiply everything by $2/3$. 
Then the chance of throwing each number other than a six must be $1/9$ and the chance of throwing a six must be $4/9$, and he chance of throwing any number at all is $1$; as it should be.
Now use this and the method you suggest to find the probability of throwing at least a three.
