A dice is weighted such that the chance to throw a certain eye (1,2,3,4,5 or 6) is proportional with this eye.
If I understand this correctly, that would mean if I throw a 5 ; then the chances to have thrown a 5 are 1,5,10 etc. How would I go about expressing the probability of throwing this 5 (i am confused..).
On
In statistics and probability, proportionality is often denoted as $\propto$. In our case, we have
$\mathbb{P}(k)\propto k \ \iff \ \mathbb{P}(k)=c\cdot k,$
for $k\in \{1,2,\cdots,6\}$ and some fixed constant $c\in \mathbb{R}_{+}$.
A constant $c$ must then satisfy:
$\sum_{i=1}^{6}\mathbb{P}(i)=1 \ \Longrightarrow \ c\cdot(\sum_{i=1}^{6}k)=1$.
It means that $p_{1}:p_{2}:p_{3}:p_{4}:p_{5}:p_{6}=1:2:3:4:5:6$ where $p_{i}$ are the probability of getting number $i$.
You can solve this by using this proportion and the fact that they sum up to $1$.