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[jh311]

Here's a probability question I came across. I get how to do most of it all the way up to the last step. please help.

Four dice are rolled. What's the probability of getting exactly two "ones" and two "sixes" in any order?

so first i would start by saying each even it is independent, and start by doing (1/6)^4.

next figure out the ways that you can roll 2 "ones" and 2 "sixes"
1616, 1166, 6161, 6611, 1661, 6116 = 6ways

so we have: 6(1/6)^4.

Here's the question: The answer says that 6(1/6)^4 = (1/6)^3. What's the simple math step that i'm missing in order to rewrite it to the simplified format? I'm really confused????

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[RCT PC CRN]

6*1/6*(1/6*1/6*1/6) = (1/6)^3 - Is that what you were looking for?

 

[jh311]

Well I guess that's the only way to do it that makes sense. I'm pretty sure that's right.

(1/6)^4 = 1/1296 and when you multiply it by 6 you get 6/1296= .00462

when it's written as (1/6)^3 = 1/216 = .00462

I just didn't realize that you could re-arrange it that way from 6 x (1/6 x 1/6 x 1/6 x 1/6), or 6(1/6)^4...... to 6 x 1/6 (1/6 x 1/6 x 1/6) or (1/6)^3.

THANKS!!

 

[Streetwolf]

6 = (1/6)^-1

For the math way you were probably looking to see...

 

[dentalWorks]