# Unique Dice Throwing Probability Question

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1. Oct 8, 2014

### JoshP-hillips

G'day,

I've recently been given a probability question to solve and I'm not 100% on my approach towards it.
It goes as follows;

A dice is found to be weighted so that the chance of throwing a 6 is four times the chance of throwing a 1. Chances of throwing numbers other than 6 are equally likely.

a) What is the probability of throwing a 6?
b) What is the probability of throwing a 3?
c) What is the probability of throwing at least a 3?

For my approach I'm basically assuming the chance of throwing anything except a six is 1/6, which means the chance of throwing a 6 if 4/6.
So that answers part a) and b), although I feel that the answer is to simple.
And then to answer part c) I just multiply the chances of throwing a 3, 4, 5, 6 independently, and then sum those probabilities. .

Any comments on this approach or help with the question would be greatly appreciated.

Last edited: Oct 8, 2014
2. Oct 8, 2014

### gopher_p

What should the sum of the probabilities of all of the possible outcomes be? Is that true in your model?

3. Oct 9, 2014

### JoshP-hillips

Well initially I thought the probability would be (1/6)*5 + (4/6) => 9/6

The 1/6 being the chance of rolling a 1-5
The 4/6 being the chance of rolling a 6 (It being 4 times more likely)

But then I came up with a more logical answer;

Probability of rolling a 1-5
x = P(D1-5)

Probability of rolling a 6
4x = P(D6)

BUT => P(D1-5) + P(D6) =1

Therefore;
x + 4x = 1
5x = 1
x = 1/5

So we end up with

P(D1-5) = 1/5
P(D6) = 4/5

part a) (4/5)
part b) (1/5)

For part c) I've come up with

P(D1-2) = 1/5*2
P(D3-5) = 1/5*3
P(D6) = 4/5

P(at least a 3) = P(D3-5) + P(D6) - P(D1-2)
P(D>=3) = 3/5 + 4/5 - 2/5 = 1

I am on the right track with this?

Last edited: Oct 9, 2014
4. Oct 9, 2014

"A dice is found to be weighted so that the chance of throwing a 6 is four times the chance of throwing a 1. Chances of throwing numbers other than 6 are equally likely."

Your question isn't clearly written. There are two ways to interpret these parts.
a) If p1 = a, then p6 = 4a. If the chances of throwing numbers other than 6 are equally likely is taken directly, this means that also p2 = p3 = p4 = p5 = a as well. If this is true, then it is very easy to find a.

b) If p1 = a, then p6 = 4a. If the second statement applies only the the numbers 2, 3, 4, 5, then all you have from that is that p2 = p3 = p4 = p5 = b, and you don't have enough information to find both a and b.

I will guess you mean the first situation, but you need to be clear about it.

5. Oct 9, 2014

### gopher_p

What should $P(D_6)$ be in terms of $P(D_1)$ ?

What should $P(D_2)$ be in terms of $P(D_1)$ ?

What should $P(D_1)+P(D_2)+P(D_3)+P(D_4)+P(D_5)+P(D_6)$ be?