Chapter 13.3, Problem 32PPS

To determine

To calculate:

The probability of following equation:

  ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$

Answer to Problem 32PPS

The probability of ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$ is $0.50$ .

Explanation of Solution

Given information:

The probability ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$ in the system $1\le x\le 6,y\le x,y\ge 1$ is represented by portion of shaded region that lies outside of circle.

Calculation:

  

The area of triangle minus that area of triangle in the circle divided by the area of triangle is equal to the probability that

  ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$

The area of the triangle is:

  $\begin{array}{l}=\frac{1}{2}\times l\times h\\ =\frac{1}{2}\times 5\times 5\\ =12.5{\text{units}}^2\end{array}$

The triangle is the right triangle with congruent side lengths, the triangle has angles ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ . The area of triangle lies inside the circle is:

  $\begin{array}{l}=\frac{{45}^{\circ }}{{360}^{\circ }}\left(16\pi \right)\\ =2\pi {\text{units}}^2\end{array}$

The probability that ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$ :

  $\begin{array}{l}P\left({\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16\right)=\frac{\text{area}\text{entire}\text{triangle}-\text{area}\text{of}\text{triangle}\text{within}\text{circle}}{\text{area}\text{entire}\text{triangle}}\\ P\left({\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16\right)=\frac{12.5{\text{units}}^2-2\pi {\text{units}}^2}{12.5{\text{units}}^2}\\ P\left({\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16\right)=0.50\end{array}$

The probability of ${\left(x-1\right)}^2+{\left(y-1\right)}^2\ge 16$ is $0.50$ .