We assume that the cylinder of a revolver has 6 chambers. The procedure returns a structure of each trial and vector of   times the trigger was pulled.

RR:=proc(N)  # N is the number of trials

local roll, i, a, L, j, b, M, Q, A;

roll:=rand(1..6);

for i to N do

a:=roll(); L:=[[a]];  # a is the chamber number in which the bullet

for j do

b:=roll(); if b=a then L:=[op(L), b]; break else  L:=[op(L), b]; fi;

od;

M[i]:=L;

od;

Q:=[seq(M[i], i=1..N)];

A:=Vector([seq(nops(Q[i])-1, i=1..nops(Q))]);

print(op(Q));

print(A);

end proc:

Example:

RR(10);

SolvingSystem:=proc(system, n)
 if n=1 then method s1 fi;
 if n=2 then method s2 fi;
 if n=3 then metod s3 fi;
 result;
end proc:

add(add(int00[i,j], j=1..3), i=1..3);

                           0

You have to write  exp(-10*x)  instead of  exp^(-10*x) 

Should be:

l4:=animate(textplot3d, [[0, 0, 0.9*tend,'direction'], color=red, align={BELOW}, font=[TIMES,ROMAN,14]], tend=0..50, frames=150):

 Coordinates of the end of a vector are the coordinates of the basis  plus the coordinates of the vector.

See  help on commands  ?plots[matrixplot]  and  ?table

code1.mw

Your error: in the last lines should be the sign  :=  instead of  =

Let  x  is serial number 5, y  is serial number 6, z  is serial number 7.

L:=[]:
for x from 1 to 2 do
for y from 0 to 9 do
for z from 0 to 9 do
 if z=5-y then L:=[op(L), [x,y,z]]: fi:
od: od: od:
L; nops(L); combinat[numbperm]([9,8,7,7])*%;
 

 [[1, 0, 5], [1, 1, 4], [1, 2, 3], [1, 3, 2], [1, 4, 1], [1, 5, 0],

        [2, 0, 5], [2, 1, 4], [2, 2, 3], [2, 3, 2], [2, 4, 1],

        [2, 5, 0]]

                                  12

                                 144

Replace the last line by the line

solve({seq(lhs(U)[i] = rhs(U)[i], i = 1 .. 2)}, {Gx, Gy})

Solve command does not solve vector equations. Index  i  not equals to 3, because the last coordinate of lhs(U) does not contain  Gx  and  Gy .

Of course, we assume that  lhs(U)[3]=0 , otherwise there are no solutions.

Assignment   i:='i'  is incorrect, since then  Term1:=i-1  has not a specific value, and therefore the command  add  can not be executed.

algsubs command  is more powerful than applyrule command.

Example:

a:=x[2]+y-b+x[1]-c+x[5]+d+x[4]+x[3]+expand(sum(x[i], i=1..5)^2);
applyrule(sum(x[i], i=1..5)=Sum(x[i], i=1..5),a);
algsubs(sum(x[i], i=1..5)=Sum(x[i], i=1..5), a);

u:=sin(w*t-theta)+sin(w*t-theta-2*Pi/3)-sin(w*t-theta+2*Pi/3);
algsubs(w*t-theta=x, u);
expand(%);
simplify(applyrule(a::algebraic*sin(x)+b::algebraic*cos(x)=sqrt(a^2+b^2)*sin(x+arctan(b,a)),%));
subs(x=w*t-theta, %);

is here  http://www.mapleprimes.com/questions/145061-Based-On-2-Sets-Of-Vectors-In-R4-How 

Read help  ?LinearAlgebra[LinearSolve]

One vector equation  x1*v1+x2*v2+x3*v3=x4*w1+x5*w2+x6*w3  is equivalent to the homogeneous system of four scalar equations with six unknowns from  x1  to  x6 . Since the rank of the leading matrix is greater  or equal to 3, the dimension of the solution space will be equal to 2 or 3. Linear systems in Maple are solved by  LinearAlgebra[LinearSolve]  command.

without changing the code of the procedure:

a:=1: b:=2:

w(a,b); 

         99