Do you really need probability?

Shaded area indicates that $|x|+2|y| <1$ beacause it consists of four equations

$x+2y<1 ; -x+2y <1 ; x-2y<1 ; -x-2y<1$ .

The area of the rhombus is $\frac{1}{2} \times (1+1) \times (1/2+1/2)=1$ .Total area of the domain is $2^2=4$

So required probability is $\dfrac{1}{4}$

Notice that since $|x|, |y| \ge 0$ , all four quadrants are symmetric, so we can solve a simpler problem: $x,y \in [0,1]$ , probability that $x+2y < 1$ .

So $P(y < -\frac{1}{2} x + \frac{1}{2}) = \frac{1}{1-0} \int_{0}^{1} (-\frac{1}{2} x + \frac{1}{2}) = | -\frac{1}{4}x^2 + \frac{1}{2}x |_{0}^{1} = \frac{1}{4} \implies a+b =1+4= \boxed{5}$

$$