# Probability Distribution of 6-Sided Die

• Apr 2nd 2009, 09:53 PM
toop
Probability Distribution of 6-Sided Die
A 6-sided die is weighted so that 1,2, and 3 show up equally likely, 4, 5, 6 show up equally likely, and 4 shows up twice as often as 1.

a. Find the probability distribution for when this die is rolled.

b. What is the probability that, when this die is rolled, that an even number shows up? What is the probability that a prime number (2, 3, 5) shows up?

a. 1, 2, 3, 5, and 6 have the probability of 1/6 ... and 4 = 1/3
b. Probability an even number is rolled = 1/2 ... Odd number = 2/3

I'm taking a complete shot in the dark at this so any help would be appreciated. Is there any specific formula I should be using?
• Apr 2nd 2009, 10:37 PM
Moo
Hello,
Quote:

Originally Posted by toop
A 6-sided die is weighted so that 1,2, and 3 show up equally likely, 4, 5, 6 show up equally likely, and 4 shows up twice as often as 1.

a. Find the probability distribution for when this die is rolled.

b. What is the probability that, when this die is rolled, that an even number shows up? What is the probability that a prime number (2, 3, 5) shows up?

a. 1, 2, 3, 5, and 6 have the probability of 1/6 ... and 4 = 1/3

Good attempt, but that's... not the solution...

You should remember that the sum of probabilities must be 1. And $\displaystyle 5 \cdot \frac 16+\frac 13=\frac 76 \neq 1$

In order to solve that, let a be the probability for 1.
We know that 1,2,3 show up equally, that is to say with the probability a.
We know that 4 shows up twice as often as 1, that is to say with the probability 2a.
We know that 4,5,6 show up equally, that is to say with the probability 2a.

Now the sum of probabilities is :
$\displaystyle a+a+a+2a+2a+2a=9a$
And we know that this sum must be 1. So what is the value of a ? :)

Quote:

b. Probability an even number is rolled = 1/2 ... Odd number = 2/3

I'm taking a complete shot in the dark at this so any help would be appreciated. Is there any specific formula I should be using?
Now, with these new values, you can try this 2nd question successfully (Nod)
I see that you had the correct reasoning for finding 1/2 and 2/3.
Again, since "getting an even number" and "getting an odd number" are complementary, the sum of their probability should be 1.

Does this help ?
• Apr 2nd 2009, 10:58 PM
toop
Ah, thank you very much. I had a feeling it needed to equal 1, but couldn't figure out how.

Anyways, what I have now is...

1, 2, 3 = 1/9
4, 5, 6 = 2/9

Even # = 6/9 = 2/3
Odd # = 4/9
• Apr 2nd 2009, 11:03 PM
Moo
Quote:

Originally Posted by toop
Ah, thank you very much. I had a feeling it needed to equal 1, but couldn't figure out how.

Anyways, what I have now is...

1, 2, 3 = 1/9
4, 5, 6 = 2/9

Even # = 6/9 = 2/3
Odd # = 4/9

4/9+6/9 is not 1!

even # : {2,4,6}. You want the probability of getting 2,4 or 6. Since these are disjoint events, the probability of getting one or another is the sum of the probabilities. 1/9+2/9+2/9=...

And then you're asked for prime numbers, {2,3,5}. Same reasoning here :
$\displaystyle \mathbb{P}(X=2 \text{ or } X=3 \text{ or } X=5)=\mathbb{P}(X=2)+\mathbb{P}(X=3)+\mathbb{P}(X= 5)$
• Apr 2nd 2009, 11:06 PM
toop
Oops!

I meant:

evens = 5/9
odds = 4/9
• Apr 2nd 2009, 11:07 PM
Moo
Quote:

Originally Posted by toop
Oops!

I meant:

evens = 5/9
odds = 4/9

that's better ! (Clapping)
• Apr 2nd 2009, 11:12 PM
toop
Haha. Thanks a lot.

I even double checked my first outcome and tried to make sure I didn't make any little mistakes...I guess better to mess up now than on the exam.