Find the probability of rolling a total of 8 using 3 die.
I want to do this without having to write down every option individually.
Thanks for the help.
The total number of combos is $\displaystyle 6^{3}=216$
$\displaystyle \left(\sum_{k=1}^{6}x^{k}\right)^{3}=(x+x^{2}+x^{3 }+x^{4}+x^{5}+x^{6})^{3}$
Can you expand this?. Look at the coefficient of the x term whose exponent is 8. That's the number of ways it's possible to sum to 8 with 3 dice.
The expansion will start with x^3 and the first coefficient is 3. So, we have $\displaystyle x^{3}+3x^{4}$, thus far.
Now, to find the next coefficient, multiply the coefficient of the x^4 term by the exponent 4 and divide by one less than the preceding exponent. (3)(4)/2=6
$\displaystyle x^{3}+3x^{4}+6x^{5}$
(6)(5)/3=10:
$\displaystyle x^{3}+3x^{4}+6x^{5}+10x^{6}$
Proceed in that manner a few more times and we get $\displaystyle 21x^{8}$
The probability of rolling a sum of 8 with 3 dice is $\displaystyle \frac{21}{216}=\frac{7}{72}$