1. ## statistics

hey

i'm slightly confused with how i go about doing this question

A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.

There is another question exactly the same except the score being obtained is inversely proportional.

Thanks. any help would be great

2. Originally Posted by lra11
hey

i'm slightly confused with how i go about doing this question

A cubical dice is biased so that the probability of any particular score between 1 and 6 (inclusive) being obtained is proportional to that score. Find the probability of scoring a 1.

There is another question exactly the same except the score being obtained is inversely proportional.

Thanks. any help would be great
For some constant $k$, you have the following probabilities:

$p(1)=k;~~~
p(2)=2k;~~~
p(3)=3k;~~~
p(4)=4k;~~~
p(5)=5k;~~~
p(6)=6k$

The sum of the probabilities is $1$, so $k+2k+3k+4k+5k+6k=1 \implies 21k=1 \implies k=\frac{1}{21}$. Therefore $p(1)=\frac{1}{21}$.

As for the second, you get:

$p(1)=\frac{k}{1};~~~
p(2)=\frac{k}{2};~~~
p(3)=\frac{k}{3};~~~
p(4)=\frac{k}{4};~~~
p(5)=\frac{k}{5};~~~
p(6)=\frac{k}{6}$

and then $\frac{k}{1}+\frac{k}{2}+\frac{k}{3}+\frac{k}{4}+\f rac{k}{5}+\frac{k}{6}=1 \implies \frac{147k}{60}=21 \implies k=\frac{20}{49}$. Therefore, $p(1)=\frac{20}{49}$

3. ## thanks

thanks, i really understand now

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# A dice is biased so that the probability of any score is proprotional to that score, find the probbability of one

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