I have a question regarding to an exercise:
A dice is manipulated in such a way, that the probability of rolling a certain score is proportional to the corresponding numerical value (e.g. rolling the score “four” is twice as likely as rolling the score “two” and so on).
Determine the probability of the events:
$1. A$ = odd number $2. B$ = prime number $3. C$ = even number $4. A$, but not $B$
Can help me please somebody help me out? Greets!
First find the probabilities of each outcome:
$$ \text{Let }x=P(6) \\ \text{Then, }P(5)=\frac{5}{6}x,P(4)=\frac{4}{6}x,P(3)=\frac{3}{6}x,\dots P(1)=\frac{1}{6}x $$
(If you are unfamiliar with this notation, then here's how it works. $P$ should be read as 'the probability of'. For example, $P(5)=\frac{5}{6}x$ means that the chance of a $5$ coming up is $\frac{5}{6}$ multiplied by some unknown value $x$.)
The probabilites must add up to $1$. Therefore,
$$ x+\frac{5}{6}x+\frac{4}{6}x+\dots+\frac{1}{6}x=1 \\ \frac{21}{6}x=1 \\ x=\frac{1}{\left(\frac{21}{6}\right)}=\frac{6}{21} $$
And so the probabilities equal to: $$P(6)=\frac{6}{21}, P(5)=\frac{5}{21},P(4)=\frac{4}{21},P(3)=\frac{3}{21},\dots P(1)=\frac{1}{21}$$
From here, it is pretty straightforward. Just add up the probabilites for each of the questions. For a start, I'll do Q2 for you:
$$ P(\text{A prime number being chosen})=\frac{2}{21}+\frac{3}{21}+\frac{5}{21}=\frac{10}{21} $$
If you need any more help, feel free to ask.