# Math Help - Loaded Die Probability

Hi yall...I need some help with this question. It seems to be kicking my trash some. I know how to calculate normal probability, but with the die being loaded I'm not quite sure what to do. Any help would be terrific. Thanks!

A die is loaded so that each odd number is twice as likely as each even number to appear. Create a probability distribution for such a die. Also find P(G), wehre G is the event that a number greater than 3 occurs on a single roll of the die.

Again..THANKS!

2. Originally Posted by pison
Hi yall...I need some help with this question. It seems to be kicking my trash some. I know how to calculate normal probability, but with the die being loaded I'm not quite sure what to do. Any help would be terrific. Thanks!

A die is loaded so that each odd number is twice as likely as each even number to appear. Create a probability distribution for such a die. Also find P(G), wehre G is the event that a number greater than 3 occurs on a single roll of the die.

Again..THANKS!
so

P(odd)=2P(even)

and

P(odd)+P(even)=1

by sub

3P(even)=1

or P(even)=1/3 and then P(odd)=2/3

Then P(2)+P(4)+P(6)=1/3

ASSUMING that these are equal then

P(2)=P(4)=P(6)=1/9

by a simialr argument

P(1)=P(3)=P(5)=2/9

so then P(G) = P(4)+P(5)+P(6)=1/9+2/9+1/9=4/9

3. thank you for responding so quickly...

could you clarify a little more? i am looking to get a probability distribution out of this. any help?

thanks!

4. So if I understand correctly:

If I say $p(x)$ is the probablity of rolling an $x$, the problem states $p(2)=p(4)=p(6)=\frac{1}{2} p(1)=\frac{1}{2} p(3)=\frac{1}{2} p(5)$. Of course $\sum_{i=1}^6 p(i) = 1$. But this implies that:

$
p(1)=p(3)=p(5) = \frac{2}{9}
,
p(2)=p(4)=p(6) = \frac{1}{9}
$

Then G is the event that a number strictly greater than 3 appears, meaning either a 4 or a 5 or a 6 is rolled. So $P(G) = p(4)+p(5)+p(6) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}$

EDIT: The distribution is exactly those values $p(i)$.