Solving Probability Problem: Red Cube 6 & Sum of Cubes 10

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The probability problem involves rolling two cubes, one red and one white, and determining the likelihood that the red cube shows a 6 while the total sum of both cubes equals 10. The correct calculation indicates that the red cube being a 6 has a probability of 1/6, and the only combination that allows the total to reach 10 is if the white cube shows a 4. Therefore, the probability is calculated as 1/6 for the red cube being a 6 and 1/6 for the white cube being a 4, resulting in a total probability of 1/36. The discussion clarifies that the teacher's interpretation was incorrect because the problem specifies the color of the cube showing 6. The conclusion confirms that the initial calculation of 1/36 is indeed correct.
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The question is:
Two number cubes are rolled--one is red and the other is white. Find the probability that the red cube is a 6 and the sum of the two cubes is 10.

I said that it is 1/6 for being a 6 and since only a 4 can add up to 10 its 1/6*1/6=1/36. But my teacher said it was wrong. Please help!
 
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I must say that I think you are correct and your teacher wrong. Are you sure you two are talking about the same problem? If the problem were just "one cube is 6 and the two sum to 10" then then you could get a 6 on either, and a four on the other: (1/6)(1/6)+ (1/6)(1/6)= 1/18.
 
MACHO-WIMP said:
The question is:
Two number cubes are rolled--one is red and the other is white. Find the probability that the red cube is a 6 and the sum of the two cubes is 10.

I said that it is 1/6 for being a 6 and since only a 4 can add up to 10 its 1/6*1/6=1/36. But my teacher said it was wrong. Please help!

You are correct; your teacher made a mistake. Your teacher would be correct IF you were not told the color of the "6" die---but you *were* told its color.

RGV
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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