Probabilities: 5-sided, 6-sided dice: find prob of same num.

[briany]

I posted this in intermediate algebra but it may be more appropriate here. I am looking for a formula so I can help my daughter with a homework problem. I do not want the answer, but the formula.

Question: If I have two dice, one 5-sided numbered 1-5 and one 6-sided numbered 1-6, what is the probability that if I throw each dice once they will come up the same number?

Thanks for your help.
Brian

 

[Denis]

Re: Probabilities

To have a chance, 6-sider must show from 1 to 5: that's a 5/6 chance.
Then the 5-sider has to match: that's a 1/5 chance.
So probability = (5/6) * (1/5) = 5/30 = 1/6.

Hey Brian, what does a 5-sided dice look like? :wink:

 

[Loren]

Re: Probabilities

I'm not sure there is a formula. I do know of a procedure to get the answer. I'll pose a similar problem to show the procedure.

Suppose we have two dice, one with 4 faces and the other with 6 faces, and assume they are honest. What is the probability of getting one die being twice the other die when rolling each die one time?
We can represent our sample space by listing all the possible outcomes. This can be done by setting up a grid with the possible outcomes of one die across the top and the possible outcomes of the other die down the side.

...1...2...3...4...5...6
1
2
3
4

Now, for each position as a result of the intersection of each outcome, decide where that represents the given condition.
If it does, mark it with an x. If it doesn't mark it with a 0.


...1...2...3...4...5...6
1..0..x...0...0...0...0
2..x..0...0...x...0...0
3..0..0...0...0...0...x
4..0..x...0...0...0...0

Count the number of x's. I get 5.
Count the size of the sample space (all outcomes). I get 4*6=24.
The probability is 5/24.

 

[DrMike]

Re: Probabilities

Hey Brian, what does a 5-sided dice look like? :wink:

It could be a square pyramid or triangular prism with the proportions "just so"... Or a really really thin pentagonal prism.............

 

[Denis]

Re: Probabilities

I'm not sure there is a formula.

Loren, if A,B (A > B) are the number of faces on the 2 dice, then probability of pair = 1 / A ; yes?

A = 11, B = 7 : (7/11) * (1/7) = 1/11
A = 11, B = 3 : (3/11) * (1/3) = 1/11

Works for all cases...