How can I use probability to solve a dice problem with a two-dice table?

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To solve the probability of outcomes when rolling two dice, a table of all possible sums is essential. The probabilities for specific conditions are calculated based on the total outcomes of 36 rolls. The first condition, that the sum is divisible by seven, yields a probability of 1/6, while the second condition regarding even factors is more complex and requires careful consideration of all sums. The third condition, identifying composite sums, results in a probability of 7/12. Understanding the nuances of how to interpret the conditions is crucial for accurate calculations.
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Draw a table for throwing two dice (one red one blue). Find the probability.

1. That the sum is divisible by seven
2. The sum has factors whose sum is even
3. The sum is a composite number.

My solutions:

1. 1/6
2. 7/11
3. 7/12

Work is attached:
Any advice would be great. Thanks in advanced.
 

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npellegrino said:
Draw a table for throwing two dice (one red one blue). Find the probability.

1. That the sum is divisible by seven
2. The sum has factors whose sum is even
3. The sum is a composite number.

My solutions:

1. 1/6
2. 7/11
3. 7/12

Work is attached:
Any advice would be great. Thanks in advanced.

I believe #1 and #3 are correct. Still trying to get my head around #2, but you are probably correct.
 


Yes, 2 was tricky, it's the factors of the sum that are only even. But would I consider all 36 sums or only the 12 (2,3,4,5,6,7,8,9,10,11,12) hmmm
 


You need to consider all the 36 possible rolls, right? If I asked you, "what's the probability that a roll will come up 7?", you wouldn't answer 1/11, you'd answer 1/6 (I hope!). But I could phrase that question in many different ways: "What's the probability that the number on the bottom of one die is equal to the number on top of the other? What's the probability that you will roll the fourth prime number?", etc..

But no way of phrasing the question changes the fact that they want the *probability* that the condition will be fulfilled, which has to be weighted by how frequently the dice fall on the various numbers.
 


Take a look at the sum of 10. You have listed the factors 1,2,5,10. I would call those the divisors of 10. Had I been asked to factor 10 into prime factors I would have written 10 = 2*5. So the first thing to do is to be sure to know what interpretation of the question is intended.

Once you have determined which sums qualify for what you are looking for you can use the pdf of X = the sum of the dice to add P(X = k) for the qualifying values of k.
 
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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