Probability
posted by Heath .
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(1X^6)/(1X)
>
F(X) = X^3(1X^6)^3/(1X)^3
We now need to perform a series expansion. The series expansion of
1/(1X)^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(1X^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
So, answer A is correct!
The coefficient of X^8 of F(X) is clearly the coefficient of X^5 of A(X) which is
7!/(5!2!) = 21
Answer B is also correct!
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
Answer C is also correct!
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
Answer D is also correct!
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