Unlocking Geometric Probability: the Duet of Probability and Geometry

Object	Number	Area	Total Area
Forests	$8$	$1200$	$9600 ∣ ∣ ∣ ∣ 8 \cdot 1200 =$
Fields	$24$	$750$	$18000 ∣ ∣ ∣ ∣ 24 \cdot 750 =$
Lakes	$76$	$3$	$228 ∣ ∣ ∣ ∣ 76 \cdot 3 =$

Object

Number

Area

Total Area

Forests

8

1200

9600 ∣ ∣ ∣ ∣ 8 \cdot 1200 =

Fields

24

750

18000 ∣ ∣ ∣ ∣ 24 \cdot 750 =

Lakes

76

3

228 ∣ ∣ ∣ ∣ 76 \cdot 3 =

Circle	Radius	$π r^{2}$	Area
$C_{1}$	$1$	$π (1)^{2}$	$π$
$C_{2}$	$2$	$π (2)^{2}$	$4 π$
$C_{3}$	$10$	$π (10)^{2}$	$100 π$
$C_{4}$	$11.5$	$π (11.5)^{2}$	$132.25 π$
$C_{5}$	$16$	$π (16)^{2}$	$256 π$
$C_{6}$	$17.5$	$π (17.5)^{2}$	$306.25 π$

Circle

Radius

π r^{2}

Area

C_{1}

1

π (1)^{2}

π

C_{2}

2

π (2)^{2}

4 π

C_{3}

10

π (10)^{2}

100 π

C_{4}

11.5

π (11.5)^{2}

132.25 π

C_{5}

16

π (16)^{2}

256 π

C_{6}

17.5

π (17.5)^{2}

306.25 π

Participants in a lottery spin the following wheel to play. They win if a green sector stops under the arrow.

What is the angle of these sectors if we know they are congruent and the probability of winning two times in a row is

1 % ?

A probability of 1 % can be written as the following ratio. 1 %=1/100 Let's label the probability of winning once as p. Since we spin the wheel twice, the probability of winning twice becomes p^2. This expression is said to be equal to 1100, which allows us to write an equation containing p.

The probability of winning once is 110, meaning that the green sectors must cover one-tenth of the wheel. If we divide 360^(∘) by 10, we get the sum of the central angles of the winning sectors. 360^(∘)/10 = 36^(∘) The four sectors together make up 36^(∘). Since all of them have the same angle measures, we get that each of them has a central angle of 36^(∘)4=9^(∘).

A blue sphere is inscribed in the red cube. What is the probability that a point chosen at random in the cube is also inside the sphere? Answer with a fraction in its simplest form.

Probability is calculated as the quotient of the favorable outcomes and the number of possible outcomes. P=Number of favorable outcomes/Number of possible outcomes In order to determine the probability of the point being inside the sphere, we can divide the volume occupied by the sphere by the volume of the cube.

Volume of the Cube

To calculate the volume of the cube, we should cube the length of its side. Notice that we have not been given any information about the side length. If we label it 2x, we get the following volume. V_C=(2x)^3= 8x^3

Volume of the Sphere

Since the sphere is inscribed in the cube, its diameter must be as long as the cube's side.

Since a circle's radius is half the length of its diameter,the sphere's radius is x. Now we can calculate the volume of the sphere.

Probability

Now that we know the volumes of the sphere and of the cube, we can determine the probability of the random point being located inside the cube by dividing the volume of the sphere by the volume of the cube.

The local library has nine small classrooms available for group studies. On average, a classroom is occupied for four hours every day during the ten hours the library is open. Calculate the probability that at least one classroom is occupied at any given time. Round to the nearest percent.

Let's label the event that at least one classroom is occupied as A. The complement to this event is that all classrooms are vacant. We will label this as A^c. According to the complement rule, the probabilities of these events add up to 1. P(A)+P(A^c)=1 From the exercise, we know that on average, a classroom is occupied for 4 hours during the 10 hours that the library is open. This means it is vacant for 6 hours every day. P(vacant)=6/10 Since there are nine classrooms, we can multiply this probability nine times to obtain the probability that all of the classrooms are vacant. P(A^c)=(6/10)^9 Since we know the value of P(A^c), we can now solve the equation we wrote in the beginning for P(A).

The probability that at least one classroom is occupied at any point in the day is 99 %.

Dominika has two straws, one

8

inches long and the other

6

inches long. She picks a straw at random and cuts it into two parts without looking where she is cutting. What is the probability that the uncut straw and the two pieces from the straw that was cut can form a triangle? Answer with a fraction in its simplest form.

There are two events to consider, picking the straw and then cutting the straw that was picked.

Picking the Correct Straw

According to the Triangle Inequality Theorem, the longest side of a triangle must always be shorter than the sum of the remaining sides. Therefore, since one straw is longer than the other, we must cut the longer straw in order for us to be able to form a triangle at all. This means we have a probability of 50 % of picking the correct straw. P(longer straw)=1/2

Cutting the Longer Straw

As already discussed, we must cut the longer straw to form a triangle. Let's label the length of one of the cut pieces x. This means the remaining piece must be (8-x) inches long. Depending on which piece is smaller, we can write two inequalities with respect to the Triangle Inequality Theorem. x is smaller than (8-x) [-1em] x+6> 8-x [1em] (8-x) is smaller than x [-1em] (8-x)+6> x Let's solve both of these inequalities for x.

Let's also solve the second inequality.

As we can see, x has to be greater than 1 but less than 7. This gives us the following acceptable area for cutting the longer straw.

As we can see, the 8-straw can be cut anywhere within the 6-inch span in order to be able to form a triangle. P(successful cut area)=6/8=3/4

Probability

Now we can calculate the probability of being able to form a triangle with the straws by multiplying the probability of picking the correct straw by the probability of cutting the straw in the correct place. P(forming a triangle)=1/2* 3/4=3/8

Tiffaniqua and Ali meet at a street corner every Friday between noon and $1 P . M .$ so that they can eat lunch together. They have an agreement that if either one of them has to wait for more than $15$ minutes after arriving, they will eat lunch on their own. What is the probability that Ali and Tiffaniqua will eat lunch together on a given Friday if Ali arrives at $12 : 20 ?$

Neither of them will wait past

1 P . M .,

no matter when they arrive. Answer with a fraction in its simplest form.

According to the exercise, Tiffaniqua and Ali will only wait for 15 minutes after arriving at the street corner. If Tiffaniqua has not yet showed up when Ali arrives and Ali abides by the rules, Ali will go eat lunch by himself when the clock shows 12:35. Therefore, the sector formed by 12:20 and 12:35 shows a favorable event.

However, Tiffaniqua could also have showed up before Ali did. If she showed up at 12:05, Ali would meet her just before she was about to leave. With this information, we can expand the sector of the clock that shows the event of Ali and Tiffaniqua eating lunch together.

As we can see, there is a 30 minute interval where Tiffaniqua could have showed up which would result in them going to lunch. Therefore, we have a probability of 3060= 12 of them having lunch.

Probability in Length, Area, and Volume

Catch-Up and Review

Calculating the Probability of Bumping Into a Person

Geometric Probability

Probability and Length

Probability and Area

Probability and Volume

Probability of a Candy Falling on a Gray Square

Hint

Solution

Calculating the Probability of Being Lost in a Forest

Hint

Solution

Calculating the Probability of Turning Successfully

Hint

Solution

Calculating the Probability of Different Positions of a Bubble

Hint

Solution

Calculating the Probability of the Girls Meeting at the Library

Hint

Solution

Area of the Total Region

Area of the Success Region

Calculating Probability

Calculating Probability of Hitting Different Regions on a Dartboard

Hint

Solution

Finding Probability Using the Values of Area

Hint

Solution

Volume of the Cube

Volume of the Sphere

Probability

Picking the Correct Straw

Cutting the Longer Straw

Probability

Probability in Length, Area, and Volume

Recommended exercises

	10 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions