Unique Dice Throwing Probability Question

[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.

 

[gopher_p]

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 1, 2, 3 and then sum those probabilities. .

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

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

 

[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(D_1-5)

Probability of rolling a 6
4x = P(D₆)

BUT => P(D_1-5) + P(D₆) =1

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

So we end up with

P(D_1-5) = 1/5
P(D₆) = 4/5

So my new answers are
part a) (4/5)
part b) (1/5)

For part c) I've come up with

P(D_1-2) = 1/5*2
P(D_3-5) = 1/5*3
P(D₆) = 4/5

P(at least a 3) = P(D_3-5) + P(D₆) - P(D_1-2)
P(D_>=3) = 3/5 + 4/5 - 2/5 = 1

I am on the right track with this?

 

[statdad]

"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.

 

[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?