Probability in Length, Area, and Volume

Let's start with the rectangles. To determine how much of each rectangle is shaded, let's number each of the parts in every rectangle until there are no more parts to number.

For A, we see that 1 of 3 parts are shaded, which gives a ratio of 13. For D, we have 3 of 8 shaded parts, so the shaded part covers 38 of the whole area. Finally, for E, there are 2 shaded parts out of 8, which reduces to 14. i.& 1/2 → ? [1em] ii.& 1/3 → A [1em] iii.& 1/4 → E [1em] iv.& 3/8 → D [1em] v.& 3/4 → ? Finally, we will identify the ratios that correspond to B and C. For C, we can see that the shaded part forms a semicircle. Therefore, this corresponds to a ratio of 12. Since we only have one option left, we can pair v with 34. To prove this, let's divide the circle into quarters and then count how many of them are shaded.

There are indeed 3 shaded parts out a total of 4 quarters. Now we can complete the pairing. i.& 1/2 → C [1em] ii.& 1/3 → A [1em] iii.& 1/4 → E [1em] iv.& 3/8 → D [1em] v.& 3/4 → B

To determine the probability of the ship landing in Alabama, we must divide the length of Alabama's shoreline by the total length of the Gulf of Mexico's shoreline. P=length of Alabama's shoreline/length of Gulf of Mexico's shoreline Let's first add the length of the shorelines. 367+397+44+53+770=1631 Finally, we divide the length of Alabama's shoreline by the length of the Gulf of Mexico's shoreline. P=53/1631≈ 3 % There is a 3 % probability of randomly landing in Alabama.

As in part A, we must divide the length of the given state's shoreline by the total length of the Gulf of Mexico's shoreline. We already calculated the length of the shoreline in part A. With this information, we can determine the probability of a ship landing in Texas. P=367/1631≈ 23 %

The probability of the tennis ball landing on the plot of land where the watermelons grow is the ratio of the area of this piece of land to the area of the entire garden.

Examining the diagram, we see that the watermelons occupy a square piece of land with a side length of 5 meters. The garden itself is also a square, with a side length of 10 meters. Now we can determine the areas that we need. 5* 5&=25 m^2 10* 10&=100 m^2 Finally, we can determine the probability of the tennis ball landing where the watermelons grow. area of watermelons/area of entire garden=25/100 =0.25 Therefore, the probability is 0.25, or 25 %.

As in the previous exercise, we must first find the area of the land where the strawberries are grown. We can see that the strawberry region can be divided into two parts, a square and an isosceles triangle that is also a right triangle.

We have already determined the area of the whole garden. To find the area of the part with strawberries, we should add the area of the square and the area of the right isosceles triangle. \begin{aligned} {\color{#0000FF}{A_\text{square}}}+{\color{#009600}{A_\text{triangle}}}&= \\[0.2cm] {\color{#0000FF}{5\cdot 5}}+{\color{#009600}{\dfrac{1}{2}(5)(5)}}&=37.5 \end{aligned} Now we have everything we need to determine the probability of the tennis ball landing in the area were strawberries are grown. area of strawberries/area of entire garden=37.5/100=0.375 Therefore, the probability is 0.375 or 37.5 %.

To determine the probability that the point falls somewhere along AC, we must divide the length of this segment by the length of AB. Let's identify the lengths of these segments.

Now we can determine the probability of the point falling on AC. P(X is on AC)=1/9

As in the previous exercise, we need to identify the length of the given segment CE. We already know that AB is 9 units.

With this information, we can determine the probability of the point falling on CE. P(X is on CE)=7/9

To determine the probability that a point randomly chosen lies in the shaded region, we have to divide the area of the shaded region by the area of the circle. The area of the shaded region can be calculated as the difference of the area of the circle and areas of the white triangles.

Area of the Circle A_C

The circle has a diameter of 8 inches, which means it has a radius of 4 inches. With this information, we can calculate the area of the circle.

Areas of the Triangles A_S and A_L

To calculate the areas of the triangles, we multiply the triangle's base and height and then divide the product by 2. A=1/2bh Notice that the smaller triangle has legs that correspond to the circle's radius. As for the big triangle, it has a base and height that correspond to the diameter and radius, respectively. Let's add this information to the diagram.

Now we can determine the areas of each triangle. Small:& A_S=1/2( 4)( 4)=8 in.^2 [1em] Large:& A_L=1/2( 8)( 4)=16 in.^2

Area of Shaded Region A_(SR)

The area of the shaded region is the circle's area less the areas of the two triangles. A_(SR)=16π -8-16=16π -24

Probability

Now that we know the area of the shaded region and the area of the circle, we can determine the probability of picking a point that is on the shaded area by dividing the area of the shaded region by the area of the circle.

There is a 52 % probability that the point lies in the shaded region.