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 = [tex] \frac{51\pi}{100\pi} =0.51 [/tex]
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 = [tex] \frac{33\pi}{100\pi} =0.33 [/tex]
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 = [tex] \frac{16\pi}{100\pi}=0.16[/tex]
