Mellie

Haha, hopefully not.

13 1/3*10

40/3*10

400/3

133 1/3

9*2=18

18/3=6

Hi   Why?

Actually, Melody was correct, but still thank you so much for helping me!!

$$The total area of the dartboard is $4\times4=16$ square inches. The area of the shaded region is the area of the $3\times3$ region after subtracting the area of the $2\times2$ region, or $9-4=5$ square inches. The shaded region is $5$ square inches out of the possible $16$ square inches the dart could hit, so the probability of the dart hitting the shaded region is $\boxed{\frac{5}{16}}$.$$

Thank you geno, but this was incorrect. Sorry!!!!!!

$$The point $(x,y)$ satisfies $x < y$ if and only if it belongs to the shaded triangle bounded by the lines $x=y$, $y=1$, and $x=0$, the area of which is 1/2. The ratio of the area of the triangle to the area of the rectangle is $\frac{1/2}{4} = \boxed{\frac{1}{8}}$. [asy] draw((-1,0)--(5,0),Arrow); draw((0,-1)--(0,2),Arrow); for (int i=1; i<5; ++i) { draw((i,-0.3)--(i,0.3)); } fill((0,0)--(0,1)--(1,1)--cycle,gray(0.7)); draw((-0.3,1)--(0.3,1)); draw((4,0)--(4,1)--(0,1),linewidth(0.7)); draw((-0.5,-0.5)--(1.8,1.8),dashed); [/asy]$$

Thank you so much geno!!!