The missing image for this question is attached below.
Area of circle = πr²
Question 1st)
The probability that the a point chosen at random will be in part A will be equal to = (Area of Part A) / (Total Area)
Area of Part A = Area of Big Circle - Area of Middle Circle
So, Area of Part A = 100π - 49π = 51π
Total Area = Area of Big Circle = 100π
Thus, the probability that the point is in Part A =
![(51\pi)/(100\pi) =0.51](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lsh9mo45ksa3pah83hkj9bg5268tj58053.png)
Question 2nd)
The probability that the a point chosen at random will be in part B will be equal to = (Area of Part B) / (Total Area)
Area of Part B = Area of Middle Circle - Area of Smaller Circle
So, Area of Part B = 49π - 16π = 33 π
Total Area = Area of Big Circle = 100π
Thus, the probability that the point is in Part B =
![(33\pi)/(100\pi) =0.33](https://img.qammunity.org/2019/formulas/mathematics/middle-school/mtjgp4h1ucm6i2esfe9ukmbaqow974cq2k.png)
Question 3rd:
The probability that the a point chosen at random will be in part C will be equal to = (Area of Part C) / (Total Area)
Area of Part C = Area of Small Circle
So, Area of Part C = 16π
Total Area = Area of Big Circle = 100π
Thus, the probability that the point is in Part C =
![(16\pi)/(100\pi)=0.16](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ybx9icsenfrhi4wzxyxdx76gr5cd7ali59.png)