Let r be the radius
Area of the shaded part (which is 2 complete circles)
= 2 x πr²
= 2πr²
Area of the unshaded part (which is a square that has a length 2r)
= 2r x 2r
= 4r²
Probability that the point is in the unshaded area
[tex] \dfrac{\text{area of square}}{\text{area of square + 2 circles}} [/tex]
[tex] = \dfrac{4\pi r^{2}}{2\pi r^{2} + 4\pi r^{2}} = \dfrac{2r^{2}(2)}{2r^{2}(\pi + 2)} = \dfrac{2}{\pi + 2} = 0.389 [/tex]
Answer: 0.4 (nearest tenth)
