# Probability

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Roll 3 dice. What is the probability of getting a sum of:
A) sum of 7: 15/216

B) sum of 8: 21/216

C) sum of 11: 27/216

D) sum of 12: 25/216

Are these correct?

How many ways are there to choose 3 numbers ranging from 1 to 6 such that the sum is N? Consider the function:

F(X) =

Sum over n1 from 1 to 6;

Sum over n2 from 1 to 6;

Sum over n3 from 1 to 6;

X^(n1 + n2 + n3)

Clearly the coefficient of X^N will give you the answer.

We can calculate F(X) as follows. Note that:

X^(n1 + n2 + n3) = X^(n1)*X^(n2)*X^(n3)

When summing over n3 from 1 to 6, the factors X^(n1)*X^(n2) don't change. You can take them outside that summation. So, what you see is that the summation factorizes into three summations which are identical.

Sum over n from 1 to 6 of X^n =

X(1-X^6)/(1-X)

-------->

F(X) = X^3(1-X^6)^3/(1-X)^3

We now need to perform a series expansion. The series expansion of
1/(1-X)^3 is given by:

A(X) =

Sum from k = 0 to infinity of

(k+2)!/[k!2!]X^k

We must multiply this by:

B(X) =

X^3(1-X^6)^3 =

X^3(1 - 3X^6 + 3 X^12 - X^18)=

X^3 - 3 X^9 + 3 X^15 - X^21

And we have:

F(X) = A(X)*B(X)

The coefficient of X^7 of F(X) is clearly the coefficient of X^4 of A(X), which is:

6!/(4!2!) = 15

The coefficient of X^8 of F(X) is clearly the coefficient of X^5 of A(X) which is

7!/(5!2!) = 21

The coefficient of X^11 of F(X) is clearly the coefficient of X^8 of A(X)minus 3 times the coefficient of X^2 of A(X), which is:

10!/(8!2!) - 3* 4!/(2!2!) = 27

The coefficient of X^12 of F(X) is clearly the coefficient of X^9 of A(X)minus 3 times the coefficient of X^3 of A(X), which is:

11!/(9!2!) - 3* 5!/(3!2!) = 25

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