Chapter 13.3, Problem 33PPS

To determine

To calculate:

The probability that bear is equally likely to swim in the pool will be in incline region:

Answer to Problem 33PPS

The probability that bear is equally likely to swim in the pool will be in incline region is $0.31$

Explanation of Solution

Given information:

  

Calculation:

The probability that certain animal swims in the region $2$ of pool by dividing the volume of the region $2$ by the volume of the given pool:

The volume of Region $1$ , $V_1$ is given by the product of the width, depth and height of region $1$ :

  $\begin{array}{l}{V}_1=\left(7\text{ft}\right)\left(20\text{ft}\right)\left(20\text{ft}\right)\\ V_1=2800{\text{ft}}^3\end{array}$

The volume of Region $3$ , $V_3$ is given by the product of the width, depth and height of region $1$ :

  $\begin{array}{l}{V}_1=\left(\text{20ft}\right)\left(30\text{ft}\right)\left(30\text{ft}\right)\\ V_1=12000{\text{ft}}^3\end{array}$

The volume of Region $2$ , $V_2$ is given by the product of the width, depth and height of region $2$ :

  $\begin{array}{l}{V}_1=\frac{1}{2}\left(\text{25ft}\right)\left(\text{7+20 ft}\right)\left(20\text{ft}\right)\\ V_1=6750{\text{ft}}^3\end{array}$

The volumes of the regions to get the total volume of pool, V.

  $\begin{array}{l}V=V_1+V_2+V_3\\ V=2800{\text{ft}}^3+6750{\text{ft}}^3+12000{\text{ft}}^3\\ V=21550{\text{ft}}^3\end{array}$

The probability that certain animal swims in the region $2$ of pool by dividing the volume of region $2$ by the volume of the given pool:

  $P\left(A\right)$ : P (Animal swims in region 2)

  $\begin{array}{l}P\left(A\right)=\frac{V_2}{V}\\ P\left(A\right)=\frac{6750{\text{ft}}^3}{21550{\text{ft}}^3}\\ P\left(A\right)=0.31\end{array}$