Chapter 13.3, Problem 3CYP

To determine

To calculate; to find the probability that archer will hit center.

Answer to Problem 3CYP

Solution: $A.P\left(skydiverlandsintheblueregion\right)=\frac{5}{9}or55.56\%$

  $B.P\left(skydiverlandsinthewhiteregion\right)=\frac{1}{3}or33.33\%$

Explanation of Solution

Area of red circle

  $\begin{array}{l}r=\frac{d}{2}=\frac{2}{2}=1yard\\ Are{a}_{redcircle}=\pi r^2=\pi {\left(1\right)}^2\end{array}$

Area of red and white circle

  $\begin{array}{l}r=1+1=2yards\\ Are{a}_{redandwhitecircle}=\pi r^2=\pi {\left(2\right)}^2\end{array}$

Area of white region $=Area-Are{a}_{red}=\pi {\left(2\right)}^2-\pi {\left(1\right)}^2=3\pi$

Total area of target

  $r=1+1+1=3yards$

  $Are{a}_{total}=\pi r^2=\pi {\left(3\right)}^2$

Area of blue region $=Are{a}_{total}-Are{a}_{redandwhite}=\pi {\left(3\right)}^2-\pi {\left(2\right)}^2=5\pi$

  $\begin{array}{l}A.P\left(skydiverlandsintheblueregion\right)=\frac{Areaofblueregion}{totalarea}\\ =\frac{5\pi }{9\pi }=\frac{5}{9}=55.56\%\end{array}$

  $$

  $\begin{array}{l}B.P\left(skydiverlandsinthewhiteregion\right)=\frac{Areaofwhiteregion}{totalarea}\\ =\frac{3\pi }{9\pi }=\frac{1}{3}or33.33\%\end{array}$