1. ## I'm missing something

so I've got this question:
A die with faces numbered 1 to 6 is biased so that $P(score=r)=kr$, (r=1,...,6). Find the value of k.

Now my problem is i don't know which numbers it is biased to. supposing that (r=1,...,6) means the numbers from 1 to 6.
Thanks

2. Originally Posted by arze
so I've got this question:
A die with faces numbered 1 to 6 is biased so that $P(score=r)=kr$, (r=1,...,6). Find the value of k.

Now my problem is i don't know which numbers it is biased to. supposing that (r=1,...,6) means the numbers from 1 to 6.
Thanks
You don't need to, you only need to know that:

$P(score=1)+ P(score=2)+ ...+ P(score=6)=1$

CB

3. I'm really sorry i don't get it.

4. Originally Posted by arze
I'm really sorry i don't get it.
When you roll the die it will come up with some face, so the probability of a result is 1, which is the sum of the probabilities of all the possible outcomes.

CB

5. ok. so i know that r can be any number, and P(any number)=1. and what's next?
P(score is r)=kr=1?

6. The sum of the probabilities of all posssible outcomes must be $1$.

So:

$
\sum_{r=1}^6 P(score=r)=\sum_{r=1}^6 k r = 21k=1
$

Hence $k=1/21$.

CB

7. Hello, arze!

You have to understand what was given.

A die with faces numbered 1 to 6 is biased so that $P(\text{score}=r)\:=\:kr,\;\;(r\,=\,1,\,2,\,\hdots ,\,6)$
Find the value of $k.$

We are told that the probability of rolling a number $r$ is a constant $k$ times $r$

So we have these probabilities:

. . . . $\begin{array}{c|c}
r & P(r)\\ \hline
1 & k \\
2 & 2k \\
3 & 3k \\
4 & 4k \\
5 & 5k \\
6 & 6k \end{array}$

The probability that one of the numbers turn up is 100%.

So we have: . $k + 2k + 3k + 4k + 5k + 6k \:=\:1 \quad\Rightarrow\quad 21k \:=\:1$

Therefore: . $k \:=\:\frac{1}{21}$

8. Thank you! i understand now!