By definition, if $0 \leq a<\frac{1}{2}$, then $A=0$ and if $\frac{1}{2} \leq a \leq 1$, then $A=1$. Similarly, if $0 \leq b<\frac{1}{2}$, then $B=0$ and if $\frac{1}{2} \leq b \leq 1$, then $B=1$. We keep track of our information on a set of axes, labelled $a$ and $b$. The area of the region of possible pairs $(a, b)$ is 1, since the region is a square with side length 1. Next, we determine the sets of points where $C=2A+2B$ and calculate the combined area of these regions. We consider the four sub-regions case by case. In each case, we will encounter lines of the form $a+b=Z$ for some number $Z$. We can rewrite these equations as $b=-a+Z$ which shows that this equation is the equation of the line with slope -1 and $b$-intercept $Z$. Since the slope is -1, the $a$-intercept is also $Z$. Case 1: $A=0$ and $B=0$ For $C$ to equal $2A+2B$, we need $C=0$. Since $C$ is obtained by rounding $c$, then we need $0 \leq c<\frac{1}{2}$. Since $c=2a+2b$ by definition, this is true when $0 \leq 2a+2b<\frac{1}{2}$ or $0 \leq a+b<\frac{1}{4}$. This is the set of points in this subregion above the line $a+b=0$ and below the line $a+b=\frac{1}{4}$. Case 2: $A=0$ and $B=1$ For $C$ to equal $2A+2B$, we need $C=2$. Since $C$ is obtained by rounding $c$, then we need $\frac{3}{2} \leq c<\frac{5}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{3}{2} \leq 2a+2b<\frac{5}{2}$ or $\frac{3}{4} \leq a+b<\frac{5}{4}$. This is the set of points in this subregion above the line $a+b=\frac{3}{4}$ and below the line $a+b=\frac{5}{4}$. Case 3: $A=1$ and $B=0$ For $C$ to equal $2A+2B$, we need $C=2$. Since $C$ is obtained by rounding $c$, then we need $\frac{3}{2} \leq c<\frac{5}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{3}{2} \leq 2a+2b<\frac{5}{2}$ or $\frac{3}{4} \leq a+b<\frac{5}{4}$. This is the set of points in this subregion above the line $a+b=\frac{3}{4}$ and below the line $a+b=\frac{5}{4}$. Case 4: $A=1$ and $B=1$ For $C$ to equal $2A+2B$, we need $C=4$. Since $C$ is obtained by rounding $c$, then we need $\frac{7}{2} \leq c<\frac{9}{2}$. Since $c=2a+2b$ by definition, this is true when $\frac{7}{2} \leq 2a+2b<\frac{9}{2}$ or $\frac{7}{4} \leq a+b<\frac{9}{4}$. This is the set of points in this subregion above the line $a+b=\frac{7}{4}$ and below the line $a+b=\frac{9}{4}$. We shade the appropriate set of points in each of the subregions. The shaded regions are the regions of points $(a, b)$ where $2A+2B=C$. To determine the required probability, we calculate the combined area of these regions and divide by the total area of the set of all possible points $(a, b)$. This total area is 1, so the probability will actually be equal to the combined area of the shaded regions. The region from Case 1 is a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$, so has area $\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{32}$. The region from Case 4 is also a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$. This is because the line $a+b=\frac{7}{4}$ intersects the top side of the square (the line $b=1$) when $a=\frac{3}{4}$ and the right side of the square (the line $a=1$) when $b=\frac{3}{4}$. The regions from Case 2 and Case 3 have identical shapes and so have the same area. We calculate the area of the region from Case 2 by subtracting the unshaded area from the area of the entire subregion (which is $\frac{1}{4}$). Each unshaded portion of this subregion is a triangle with height $\frac{1}{4}$ and base $\frac{1}{4}$. We can confirm this by calculating points of intersection as in Case 4. Therefore, the area of the shaded region in Case 2 is $\frac{1}{4}-2 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$. Therefore, the combined area of the shaded regions is $\frac{1}{32}+\frac{1}{32}+\frac{3}{16}+\frac{3}{16}=\frac{14}{32}=\frac{7}{16}$. Thus, the required probability is $\frac{7}{16}$.
Two numbers $a$ and $b$ with $0 \leq a \leq 1$ and $0 \leq b \leq 1$ are chosen at random. The number $c$ is defined by $c=2a+2b$. The numbers $a, b$ and $c$ are each rounded to the nearest integer to give $A, B$ and $C$, respectively. What is the probability that $2A+2B=C$?