CEMC Probelm Set 17/18 Two numbers a and b with 0 ≤ a ≤ 1 and 0 ≤ to b ≤ 1 are chosen at random. The number c is defined by c= 2a+2b. The number a, b and c are each rounded to the nearest integer to given A, B and C, respectively (For example, if a =0.432 and b =0.5 then c = 1.864 and so A=0 B=1 and C=2) What is the probability that 2A + 2B = c
 A: The probability that c is rational is 0. It therefore has probability 0 of being 2A+2B. Did you mean C?
A: I will assume that you meant $2(A+B)=C$.
Let us take the unit square and divide it into regions depending on where things get rounded.  We have a dividing line when $a=1/2$ and when $b=1/2$, which determines whether $A$ and $B$ are zero or one.  Further, we can draw the lines $2(a+b)=k$ where $k$ is $0.5, 1.5, 2.5, 3.5$ to divide the square into where $C$ will be rounded to each of $0, 1, 2, 3, 4$.  
Now, we need to find the regions of the square where our condition holds and calculate the area. The area is most easily calculated by noting that we can divide our square further into $32$ right triangles with short sides of length $1/2$ and that each region in our square will be a union of these squares, so we just have to add up the number of triangles in each region.  See here for a picture.
Since each of $A$ and $B$ have two possible values, we will have four cases:


*

*$A=B=0, C=0$, 

*$A=B=1, C=4$, 

*$A=0, B=1, C=2$, 

*$A=1,B=0, C=2$.


The first two regions have area $1/32$, and the last two have area $6/32$, for a total area of $14/32=7/16$
