Two six-sided dice each have the numbers 1 through 6 on their faces. Neither die is fair, but they are both weighted the same. The probability of rolling a certain number on one die is given in the table below:

# 1 2 3 4 5 6
probability 1/6 1/6 1/9 ? 2/9 ?

If the probability that the two dice both show the same numbers is (2/3)^4, we can express the probability of rolling 10 on these two dice as a/b where a and b are coprime positive integers. What is the value of a+b?

163.

To find the missing probabilities in the table, we can use the fact that the probabilities on each die must sum up to 1. Let's calculate the missing probabilities.

We know the following probabilities for each number on the die:

# 1 2 3 4 5 6
probability 1/6 1/6 1/9 ? 2/9 ?

To find the missing probability for number 4, we subtract the sum of the known probabilities from 1:

1/6 + 1/6 + 1/9 + ? + 2/9 + ? = 1

Simplifying the equation gives:

4/6 + 1/9 + ? = 1 - (1/6 + 2/9)
4/6 + 1/9 + ? = 9/18 - 3/18
4/6 + 1/9 + ? = 6/18

Combining like terms gives:

4/6 + 1/9 + ? = 1/3

To find the missing probability for number 4, we subtract the known probabilities from this sum:

? = 1/3 - (4/6 + 1/9)
? = 3/9 - (8/12 + 1/9)
? = 3/9 - (16/24 + 2/18)
? = 3/9 - (2/3 + 1/9)
? = 3/9 - 8/12 - 2/18
? = 3/9 - 8/12 - 4/18
? = 4/12 - 8/12 - 4/18
? = (4/12 - 8/12 - 4/18) * 18/18
? = (6 - 12 - 8) / 18
? = -14 / 18

Since probabilities cannot be negative, we conclude that the missing probability for number 4 is 0.

Now that we have the complete table of probabilities:

# 1 2 3 4 5 6
probability 1/6 1/6 1/9 0 2/9 ?

Next, we can calculate the probability of rolling the same number on both dice, which is given as (2/3)^4.

(2/3)^4 = (2/3) * (2/3) * (2/3) * (2/3) = 16/81

Therefore, the probability of rolling a 10 on these two dice is 16/81. Since the question asks for the sum of the numerator and denominator, a+b = 16+81 = 97.