Win that stuffed toy

We can find the successful area and the total area for the center of the disk, which will give us the answer.

The successful area for the center of the disk consists of a region inside each of the 9 little squares. These regions are squares with side-length $10-2\cdot 2.5=5$ , and hence have total area $9\cdot 25=225$ .

The total area for the center of the disk consists of a region inside the big square, since the entire disk must land inside the square. This side-length is, similarly, $30-2\cdot 2.5=25$ , and so the total area is $625$ .

The probability is $\dfrac{225}{625}=\dfrac{9}{25}$ , and the answer is $9+25=\boxed{34}$ .

View the diagram here

Hence, the probability is $\frac{9 \times (5 \times 5)}{25 \times 25} = \frac{9}{25}$

Sample area = $(30 - 2 * 2.5) * (30 - 2 * 2.5) = 25 * 25$

Event area = $9 * (10 - 2 * 2.5) * (10 - 2 * 2.5) = 9 * 5 * 5$

So probability = $\frac{9}{25}$

Answer: $9 + 25 = \boxed{34}$

For each $10\times10$ square, the center of the disk must be in a $5\times5$ region. The center of the disk can only be in a $25\times25$ region, so the answer is $\dfrac9{25}$ .

Let the centre of the disk be $O$ .

Within one of the squares, the area of the $O$ can move to win the stuffed toy is $(10-2.5-2.5)(10-2.5-2.5)=25$ . Then among all the squares, the area of the $O$ can move is $9\times25=225$ .

The total area of $O$ can move is $(30-2.5-2.5)(30-2.5-2.5)=625$ .

Hence, the answer is $\frac{225}{625}=\frac{9}{25}$ . $a+b=34$ .

We will consider the center of the circle.. for the circle to lie within the $3 \times 3$ grids surrounded by walls , its center always has to be atleast 2.5cm away from each of the boundry walls..giving us total usable area as $30-2 \times 2.5)^2$ =625 ... and for the circle to lie within an $10 \times 10$ grid its center has to be atleast 2.5cm away from each of the $10 \times 10$ grid lines..giving us total required area as $(10-2 \times 2.5)^2=25$ now like this there are 9 grids..total probabality becomes $(25 \times 9)\625=(9\25)$