Probability in Length, Area, and Volume

It is given that there are 15 yellow squares of 2 square feet each and 18 gray squares of 3 square feet each. Therefore, by calculating the product of 15 and 2, as well as 18 and 3, the total areas of yellow and gray regions can be found. Yellow:& 15* 2=30 ft^2 Gray:& 18* 3=54ft^2 The sum of these values is the total area of the kitchen floor. Note that the whole area of the kitchen floor is the sample space of all the possible outcomes where the candy can fall. Total: 30+54=84 ft^2 Since the probability of a candy falling on gray region should be calculated, the area of success region is the area of gray region. Therefore, the area of success region can be substituted with 54 and the number of total region can be substituted with 84 into the Probability Formula.

The probability of the candy falling on the gray area of the floor is approximately 0.64 or 64 %.

First, the total areas of the forests, fields, and lakes should be calculated by multiplying the area of each geographical feature by the number of those features.

Object	Number	Area	Total Area
Forests	8	1200	8* 1200= 9600
Fields	24	750	24* 750= 18 000
Lakes	76	3	76* 3= 228

It is given that the total area of the country is 30 000 square kilometers. Since it is also given that Magdalena is not in a lake, the area of total region she could be in is the difference between the total area of the country and the total area of lakes. 30 000-228=29 772 km^2 Magdalena must be somewhere in a 29 772 square kilometer area. To calculate the probability of the event of her being lost in a forest, substitute 9600, the total area of land covered by forests, for the area of success region and 29 772 for the area of total region.

The probability of Magdalena being lost in a field can be calculated similarly by substituting 18 000, the total land area made up of fields, for the area of success region and the same number for the area of total region.

Therefore, the probability of Magdalena being lost in a forest is approximately 0.32, or 32 %, while the probability of her being lost in a field is about 0.60, or 60 %.

The geometric probability formula can be used to find the total length of the fence. P=Length of success region/Length of total region Since the probability of the event of turning and successfully finding a place in the grass is given, the length of success region is the length of parts without the fence. By substituting the length of total region with 170 and P with 310, this value can be calculated.

It can be concluded that the length of the parts without the fence is 51 feet. The parts of the path without the fence form the sample space of the event of turning successfully. By subtracting 51 from the length of the path, the total length of the fence can be found. Length of Fence 170-51=119 ft Therefore, the length of the fence is 119 feet.

The experiment uses a sphere, a three-dimensional object, so the geometric probability formula for volume should be used. P=Volume of success region/Volume of total region The sample space in this case is the whole region inside the sphere. The volume of total region is the same as the volume of the sphere. Let r be the radius of the sphere. Its volume is then determined by the following formula. \begin{gathered} V_\text{sphere}=\dfrac{4\pi}{3}r^3 \end{gathered} Outcomes in which the bubble is situated closer to the outside than the center of the sphere are said to be the success outcomes.

They can be calculated as the difference between the total volume of the sphere and the outcomes in which the bubble is situated closer to the center of the sphere. \begin{gathered} V_\text{success}=V_\text{sphere}-V_\text{center} \end{gathered} The bubble is closer to the center if it is in the sphere whose radius is half the radius of the whole sphere. The radius of this smaller sphere is r2. This means that the volume of that sphere is as follows.

Now the volume of the success outcomes can be found.

Finally, the found volumes can be substituted into the probability formula.

It can be concluded that the probability of the bubble being closer to the outside of the sphere is 78.

First, draw a graph that represents the possible times in which Magdalena and Tiffaniqua could meet. If Tiffaniqua arrives at 14:30 P.M., then Magdalena must arrive no later than 14:45 P.M.. If Tiffaniqua arrives at 14:31 P.M., Magdalena must arrive no later than 14:46 P.M., and so on.

Therefore, the 15-minute leeway shown on the graph represents their opportunity of meeting. The geometric probability formula for area will be used to calculate the probability of both girls being in that 15-minute span. P=Area of success region/Area of total region

Area of the Total Region

The total time available for the girls to meet or a sample space is represented by the square shown in the diagram. Its sides represent the 90-minute time span from 14:30 P.M. to 16:00 P.M..

The area of the total region can be calculated by substituting 90 for s into the area of a square formula.

Area of the Success Region

Analyzing the diagram, it can be noted that the area of the 15-minute time allowance is equal to the difference between the area of the square and the areas of the top and bottom triangles. Also, the areas of the top and bottom triangles are the same, so calculating only one of them would be enough.

As can be seen, these are right triangles whose legs represent 75 minutes. By using the area of a triangle formula, their areas can be calculated.

Now, the area of the 15-minute leeway zone can be found.

Therefore, the area of success region is 2475.

Calculating Probability

Finally, by substituting 2475 for the area of the success region and 8100 for the area of the total region, the probability of the event of the girls meeting can be calculated.

It can be concluded that the probability that Magdalena and Tiffaniqua will meet at the library is 1136.

P=Area of success region/Area of total region To use the formula, first the areas of the success region — the red region — and total regions should be found. For the purposes of the solution the circles on the board can be named.

First, the radii of all the circles can be found. It is given that the diameter of the small red circle C_1 at the center of the dartboard is 2 centimeters, so its radius is 22=1 centimeter. r_(C_1)=1 cm The diameter of the green circle C_2 is said to be 2 centimeters greater than the diameter of C_1. This implies that the diameter of C_2 is 2+2=4 centimeters and its radius is 2 centimeters. It is given that the length between C_2 and the next ring is 8 centimeters, so the radius of C_3 is obtained by adding 2 and 8. r_(C_2)&=2 cm r_(C_3)&=10 cm Next, by adding the width of the rectangles to the radius of C_3, it can be concluded that the radius of C_4 is 11.5 centimeters. r_(C_4)=11.5 cm The radius of C_5 is equal to the sum of 14 and radius of C_2, which is 2 centimeters. Finally, the radius of C_6 is 1.5 centimeters greater than r_(C_5). r_(C_5)&=16 cm r_(C_6)&=17.5 cm Now that all the radii are known, they can be substituted into the circle area formula. First, the area of C_1 can be found. Leave the areas in terms of pi to get a precise result at the end.

Similarly, the areas of the rest of the circles can be calculated.

Circle	Radius	π r^2	Area
C_1	1	π ( 1)^2	π
C_2	2	π ( 2)^2	4π
C_3	10	π ( 10)^2	100π
C_4	11.5	π ( 11.5)^2	132.25π
C_5	16	π ( 16)^2	256π
C_6	17.5	π ( 17.5)^2	306.25π

By calculating the difference between the areas of C_4 and C_3, the total area of smaller red and green rectangles can be found. A_(C_4)-A_(C_3)=132.25π-100π ⇓ A_(C_4)-A_(C_3)=32.25 π Since there are 20 rectangles and they all have the same size, the area of each rectangle can be calculated by dividing 23.25π cm^2 by 20. \begin{gathered} A_\text{small rectangle}=\dfrac{32.25\pi}{20}\\[0.5em] \Downarrow\\ A_\text{small rectangle}= 1.6125\pi \end{gathered} 10 of the 20 rectangles are red, so the total area of the red rectangles is 16.125π square centimeters. Finally, the total area of the greater green and red rectangles can be found as the difference between the areas of C_6 and C_5. A_(C_6)-A_(C_5)=306.25π - 256π ⇓ A_(C_6)-A_(C_5)=50.25π The area of each bigger rectangle can be obtained by dividing this value by 20. \begin{gathered} A_\text{big rectangle} =\dfrac{50.25\pi}{20} \\[0.5em] \Downarrow \\ A_\text{big rectangle}=2.5125\pi \end{gathered} Just as in case with smaller rectangles, there are 10 bigger red rectangles, so the total area of the bigger red rectangles is 25.125 square centimeters. The total red region on the board consists of 1 small circle, 10 small rectangles and 10 big rectangles. Therefore, the total area of the red region is the sum of these areas. \begin{gathered} A_\text{red}={\color{#0000FF}{\pi}}+{\color{#FD9000}{16.125\pi}}+{\color{#A800DD}{25.125\pi}}\\ \Downarrow\\ A_\text{red}=42.25\pi \end{gathered} The area of the success region was found to be 42.25 square centimeters. The area of the total region is the area of the largest circle C_6. With this information in mind, the probability of the dart hitting a red region can be found.

The probability of the dart hitting a red region is about 0.138, or 13.8 %.

\begin{gathered} A_\text{red and green}=32.25\pi\!+\!50.25\pi\!+\!4\pi \\ \Downarrow \\ A_\text{red and green}=86.5\pi \end{gathered} The total area of the success region is 86.5π square centimeters. The area of total region is the same as before, 306.25π square centimeters. Now enough information has been calculated to find the probability of hitting a red or green region.

The probability of hitting a red or green region is approximately 0.282, or 28.2 %.

The probability of hitting a black or white region is approximately 0.718, or 71.8 %.