Help... Elizabeth has two identical number cubes. Both cubes have faces numbered 1 thourgh 6. if Elizabeth rolls each cube once, what is the probability that the sum of the two numbers on the top faces will be 10. Choices are 1/36 1/12 1/10 1/9
@ospreytriple
Hope you don't mind
Not at all. How many different combinations are there that add up to 10?
5+5=10 9+1=10 8+2=10 7+3=10 6+4=10
Is that right?
But each cube is numbered 1 through 6, so you can't use any numbers greater than 6
Oh yeah
5+5=10 6+4=10
Umm.. Is there more? I'm not sure...
Right. But here's a tricky bit about probability and counting. You also have to consider the order that the numbers come up. So the first cube could say 4 and the second cube could say 6. BUT the first cube could say 6 and the second cube could say 4. They gave to be counted as two different outcomes.
So how about 5+5 6+4 4+6 Understand?
Yes
Ok. Now how many possible outcomes are there. Remember that each cube has six possible outcomes. How many possible outcomes are there. Remember the fundamental counting principle
2?
I'm not sure...
Ahh.. Wait so 1
Because 4+6 and 6+4 are still the same right
Here's a hint: 1 & 1 1 & 2 1 & 3 up to 6 & 4 6 & 5 6 & 6
I still don't get it
OK remember earlier that, when you had two events, you multiplied the possibilities together?
Yeah
Well, there are six possibilities on the first cube and six possibilities on the second cube. How many possibilities in total?
36
so is it 1/36?
Correct. And three of those possibilities give a sum of 10. What's the probability?
5+5?
There are 3 successful outcomes out of a total of 36 outcomes. What's the probability?
@Madi_Lee
3/36 x/36?
Just 3/36. Now you have to reduce it to lowest terms
which is 1/36
Nope. Divide both numerator and denominator by 3 and what do you get?
\[\frac{ 3 }{ 36 } = \frac{ 3\times1 }{ 3\times12 }=?\]
Huh?
Oh nevermind 1 minute
1/12
Exactly. Well done.
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