Chapter CSB1.13, Problem 6E

To determine

The probability that a randomly thrown dot land in shaded region of each dartboard.

Answer to Problem 6E

Probability will be equal to 0.6511

Explanation of Solution

From the diagram total area is equal to,

  $\begin{array}{l}A=15\times 15\\ =225{\text{ m}}^2\end{array}$

Area of un shaded region is,

  $\begin{array}{l}{A}_{unshaded}=4\times \pi {\left(\frac{5}{2}\right)}^2\\ =4\times 3.14\times {2.5}^2\\ \text{=78}{\text{.5 m}}^2\end{array}$

Now, area of shaded region is,

  $\begin{array}{l}{A}_{shaded}=225-78.5\\ =146.5{\text{ m}}^2\end{array}$

Probability is given by ratio of favorable case and total number case.

Therefore, probability will be,

  $\begin{array}{l}p=\frac{A_{shaded}}{A}\\ =\frac{146.5}{225}\\ =0.6511\end{array}$