Chapter 13.4, Problem 13PPS

To determine

To calculate:

The average and expected values of shaded region by comparison:

Answer to Problem 13PPS

The average value is close to expected value.

Explanation of Solution

Given information:

  

Calculation:

The dashboard consists of thirteen regions of three different colors. The innermost circle is red hundred point circle, there are six white wedge shaped regions are worth $50$ points and there are six blue wedge shaped regions are worth $25$ points.

The center red circle has radius of $0.5$ units, the circular dashboard has radius of $5$ units.

The area of circle:

  $A=\pi r^2$

Where,

  $r$ : The radius of circle

The probability of red region of circle being hit by dart:

  $P\left(R\right)$ : P (Red)

  $A$ : Area of red region

  $R$ : Area of dashboard

  $\begin{array}{l}P\left(R\right)=\frac{A}{R}\\ P\left(R\right)=\frac{\pi {\left(0.5\right)}^2}{\pi {\left(5\right)}^2}\\ P\left(R\right)=\frac{0.25}{25}\\ P\left(R\right)=0.01\end{array}$

The probability of white region of circle being hit by dart:

  $P\left(W\right)$ : P (White)

  $D$ : Area of dartboard $-$ red area$P$ : Area of dashboard

  $\begin{array}{l}P\left(W\right)=\frac{D}{P}\\ P\left(W\right)=\frac{\frac{\pi {\left(5\right)}^2-\pi {\left(0.5\right)}^2}{2}}{\pi {\left(5\right)}^2}\\ P\left(W\right)=\frac{24.75}{2}\\ P\left(W\right)=0.495\end{array}$

The probability of blue region of circle being hit by dart:

  $P\left(B\right)$ : P (Blue)

  $D$ : Area of dartboard $-$ red area$P$ : Area of dashboard

  $\begin{array}{l}P\left(W\right)=\frac{D/2}{P}\\ P\left(W\right)=\frac{\frac{\pi {\left(5\right)}^2-\pi {\left(0.5\right)}^2}{2}}{\pi {\left(5\right)}^2}\\ P\left(W\right)=\frac{\frac{24.75}{2}}{25}\\ P\left(W\right)=0.495\end{array}$

The expected values of the dashboard adding the products of the probabilities and corresponding point values:

  $E$ : Expected value

  $\begin{array}{l}E=P\left(R\right)\left(100\right)+P\left(W\right)\left(100\right)+P\left(B\right)\left(100\right)\\ E=\left(0.01\right)\left(100\right)+\left(0.495\right)\left(100\right)+\left(0.495\right)\left(100\right)\\ E=1+24.75+12.375\\ E=38.1\end{array}$

  $AVG$ : Average value

  $\begin{array}{l}AVG=P\left(R\right)\left(100\right)+P\left(W\right)\left(50\right)+P\left(B\right)\left(25\right)\\ AVG=\left(\frac{0}{50}\right)\left(100\right)+\left(\frac{26}{50}\right)\left(100\right)+\left(\frac{24}{50}\right)\left(25\right)\\ AVG=0+26+12\\ AVG=\left|38\right|\end{array}$

The average value of simulation was found to be $38$ , while the expected value of dashboard was found to be $38.1$ . Hence, the average value is close to expected value.