# I'm missing something

• Jan 27th 2010, 10:10 PM
arze
I'm missing something
so I've got this question:
A die with faces numbered 1 to 6 is biased so that $\displaystyle P(score=r)=kr$, (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
• Jan 27th 2010, 10:57 PM
CaptainBlack
Quote:

Originally Posted by arze
so I've got this question:
A die with faces numbered 1 to 6 is biased so that $\displaystyle P(score=r)=kr$, (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:

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

CB
• Jan 27th 2010, 11:12 PM
arze
I'm really sorry i don't get it.
• Jan 27th 2010, 11:22 PM
CaptainBlack
Quote:

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
• Jan 27th 2010, 11:29 PM
arze
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?
• Jan 28th 2010, 01:23 AM
CaptainBlack
The sum of the probabilities of all posssible outcomes must be $\displaystyle 1$.

So:

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

Hence $\displaystyle k=1/21$.

CB
• Jan 28th 2010, 05:42 AM
Soroban
Hello, arze!

You have to understand what was given.

Quote:

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

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

So we have these probabilities:

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

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

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

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

• Jan 28th 2010, 05:45 AM
arze
Thank you! i understand now!