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I'm trying to solve a question that asks:

If 5 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained is between 30 and 40 inclusive.

My approach so far, was to create a random variable called $T$ whose range is from $10$ to $60$ and look for $P\{|T-35| \le 5\} $

I don't know what to do next. Can I proceed like this is a normal distribution or is five too small a sample to qualify for the Central Limit Theorem?

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    [Probability and Statistics for Engineers, by Ross] (http://goo.gl/PEUXa)2012-12-07

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I just made a spreadsheet to calculate it and get $\frac {41539796}{60466176}\approx 0.687$

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    @Imray: took a couple minutes. I made a column with 1 to 60 for the sum. The next ten columns were the number of ways to make that sum with 1 through 10 dice. In the column for 1 die, I put 1 in 1 through 6. Then in each successive column, a cell added the six cells above it and one column to the right (a little perturbation at the small end). Fill down/fill right, then sum the appropriate cells in the 10 dice column.2012-11-15
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This can be determined exactly with the use of generating functions (also known as probability generating functions) and a computer algebra system (like GP/PARI).

The probability is the sum of the coefficients of the terms in the polynomial $ \left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6\right)^{10} $ with degree 30 through 40, which is the sum of the coefficients of the terms in the polynomial $ (x+x^2+x^3+x^4+x^5+x^6)^{10} $ with degree 30 through 40, divided by $6^{10}$.

Using GP, this is calculated instantly as

sum(i=30,40,polcoeff(sum(j=1,6,x^j)^10,i))/(6^10)

which gives the result $\frac{10384949}{15116544} \approx 0.68699227812918084$.

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    I added in an extra expression to make it clearer. The polynomial $P=\frac{1}{6}x+\frac{1}{6}x^2+...+\frac{1}{6}x^6$ is the probability generating function of a fair 6-sided die: the coefficient on $x^i$ is the probability of face $i$ appearing on a single roll. Then, the distribution of sums of $n$ such dice is given via the coefficients of $P^n$. You can read more about it on the Wikipedia page I linked to in my answer.2012-12-04