geometric probability assignment

Geometric Probability

Geometric probability is the visual representation of probability. Many of the probabilities are continuous and for these probabilities, it is easy to represent as a geometric probability. The area of the geometric figure is used to compute the probabilities. The geometric probability is the probable area divided by the entire area.

Let us learn more about geometric probability and the representation of geometric probability with the help of examples, FAQs.

What Is Geometric Probability?

Geometric probability is the representation of probability geometrically. The probabilities which are non-discrete or are continuous are represented across a number line or across a two-dimensional plane. The experiment having numerous probability outcomes are often represented as a geometric probability.

Probability is the mathematical way of calculating how likely a particular event may happen. Probability values always lie between 0 and 1,and sometimes they are also expressed as a percentage. The geometric probability is the representation of the probability as a form of a geometric figure such that the happening of an event is shaded as part of the area of the entire figure. We can assume geometric probability as a dartboard and the probability of hitting a particular area of the dartboard.

For calculation of geometric probability, we are only required to know the expected area where we want to hit with reference to the total area. Geometric probability is obtained by dividing the expected area by the total area.

Geometric Probability = Probable Area/Total Area

Geometric probability is also used to sometimes represent the different outcome probabilities at the same time. A few quick examples of geometric probability are as follows.

• The probability of the arrival of a train across a time period is a continuous probability, which can be represented as a geometric probability.
• In a class, the probability of a number of students gaining 65% or more marks can be represented as a geometric probability. This can be represented as a normal distribution, and the probability region can be shaded.
• Among a group of people, 30% drink tea, 40% drink coffee, and the remaining 30% drink soft drinks. This can be represented as a shaded region in a pi diagram.
• The probability of incidence of covid-19 among the population can be graphically represented as a geometric probability.

Representation of Geometric Probability

The geometric probabilities can be represented in a single dimension, two-dimensions, or n-dimensions, and can be represented as an area of the shaded probable region, with reference to the area of the entire region.

Number Line: The continuous probabilities are marked as a point on the number lines, and the length of the line from the starting point divided by the entire length of the line gives the probability answer.

Two Dimensional Plane: The geometric probability can be represented in the two-dimensional graph as a normal curve, as a pi diagram, or any other geometric shape. For the representation of the probability of a single event we use the normal curve, and represent the probability as the part of the area of the normal curve. And for representing multiple probabilities we use a Pi diagram or any other diagram. Further, we can represent geometric probability in higher dimensions also.

Percentage Format: Geometri probability is numerically represented as a percentage. The percentage value represents the happening of the event or the area of the geometric figure. 100% refers to the entire shaded image and it means the sure happening of the event. A 50% is represented as a half-shaded region and is equal to half of the chances of happening of an event.

☛ Related Topics

• Discrete Probability Distribution
• Probability Theory
• Events in Probability
• Theoretical Probability
• Probability Distribution
• Probability Tree Diagram

Examples on Geometric Probability

Example 1: Find the geometric probability of selecting a girl from a class, if the girls of the class are represented by a circle of radius 3.5 cms, and the entire class of students is represented by a larger square of side 8 cms.

Given that the girls represent the circle if radius 3.5cms, and the entire class of students represent a square of side 8 cms.

Area of the circle = πr 2 = 22/7 × 3.5 × 3.5 = 38.5

Area of the square = s 2 = 8 × 8 = 64

Probability of selecting the girl = Area of Circle/Area of Square

Probability of selecting the girl = 38.5/64

Therefore the probability of selecting the girl from the class of students is 38.5/64.

Example 2: Find the geometric probability of the number of students who can speak English, if the people are represented s sectors in a circle, and the English-speaking people represent a sector of 120°.

The number of people who speak English is represented by a sector of angle 120°, and the total number of students of the class are represented by the entire circle, which is an angle of 360°.

Probability of people speaking English = Area of sector/Area of circle = Angle of the sector/Complete angle

Probability of people speaking English = 120°/360° = 1/3

Therefore the probability of people speaking English language is 1/3.

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Practice Questions on Geometric Probability

Faqs on geometric probability, what is geometric probability in maths.

Geometric probability is the geometric or visual representation of probability in one, two or n-dimensional framework. The experiment having numerous probability outcomes are often represented as a geometric probability. For calculation of geometric probability, we are only required to know the expected area where we want to hit with reference to the total area. Geometric probability is obtained by dividing the expected area by the total area.

Where Do We Use Geometric Probability?

The geometric probabilities are used if the range of probabilities is huge and it is difficult to represent it with a simple value. The geometric quantities of length, area, angle are used to express the probabilities of events.

What Are The Formulas Of Geometric Probability?

The formula of geometric probability is the expected area divided by the total area. We have Geometric Probability = Probable Area/Total Area . Also sometimes in place of area, we can also use the length, angle, or any other geometric quantity.

What Are The Examples Of Geometric Probability?

The examples of geometric probability include events that have a range of probabilities such as the probability of arrival of a bus across a time period. Also simple probabilities with multiple events can be represented as a geometric probability. If a party has people drinking tea, coffee, soft drink, milkshake, water, soda lime, then the probabilities can be easily represented using geometric probability.

What Is The Difference Between Geometric Probability And Discrete Probability?

The geometric probability is different from discrete probability, as the discrete probability has countable events which can be easily computed. The probabilities having continuous probabilities cannot be easily calculated and here we can use geometric probability. Probabilities involving dice, playing cards, coins are all simple examples of discrete probability, but probabilities involving student with a performance range, probabilities with time range can be represented using geometric probability.

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Applied probability.

A framework for understanding the world around us, from sports to science.

• Sharky Kesa
• Mahindra Jain
• A Former Brilliant Member
• Samarpit Swain

Geometric probability is a tool to deal with the problem of infinite outcomes by measuring the number of outcomes geometrically, in terms of length, area, or volume. In basic probability, we usually encounter problems that are "discrete" (e.g. the outcome of a dice roll; see probability by outcomes for more). However, some of the most interesting problems involve "continuous" variables (e.g., the arrival time of your bus).

Dealing with continuous variables can be tricky, but geometric probability provides a useful approach by allowing us to transform probability problems into geometry problems. If this sounds surprising, take a look at the following problem:

Your bus is coming at a random time between 12 pm and 1 pm. If you show up at 12:30 pm, how likely are you to catch the bus?

Intuitively, the answer seems to be \(\frac{1} {2} \). We can show this geometrically by considering a point chosen randomly on a 1-dimensional number line: the length of the number line between 12:30 pm and 1 pm is equal to the length from 12 pm to 12:30 pm.

While this example is fairly straightforward, many complicated problems can be solved simply by using geometric probability. On this page, we will start with 1D examples, which are the simplest and easy to understand and then work our way up to 2D, 3D, and higher dimensions.

Introduction

1-dimensional geometric probability, 2-dimensional geometric probability, 3-dimensional geometric probability, applications, extra challenges.

One of the main ideas in probability is to count the number of equally likely "desired" outcomes, and then divide that by the number of equally likely total outcomes:

\[ P(X) = \frac{\mbox{desired outcomes}}{\mbox{total outcomes}} .\]

However, when a variable is continuous , it becomes impossible to "count" the outcomes in the traditional sense. For example, if \(X\) is a random real number between 0 and 1, it could be \(0.2\) or \(0.53\) or \(0.434662565465465\) or even something irrational like \(\frac{\pi}{4}.\) It is clear that there are infinite outcomes if we count in the traditional sense.

Let's look more at the situation where \(X\) is a random real number, as mentioned in the Introduction section.

\(X\) is a random real number between 0 and 3. What is the probability \(X\) is closer to 0 than it is to 1? Since there are infinitely many possible outcomes for the value of \(X,\) we will take the equally likely outcomes as random points along the number line from 0 to 3. It’s easy to see that \(X\) will be closer to 0 than it is to 1 if \(X<0.5.\) Probability line Now, we can use the measures (lengths, in this 1D case) of our possible outcomes and apply the usual probability formula. Here, \[P(X\text{ is closer to 0 than to 1}) = \frac{\mbox{length of segment where }0<X<0.5}{\mbox{length of segment where }0<X<3} = \frac{0.5}{3} = \frac{1}{6} \approx 17\%.\ _\square\]

To reiterate, the core idea in one-dimensional (1D) geometric probability is translating a probability question into a geometry problem on a number line, where we measure outcomes with length . To make sure you've got this concept down, try this problem related to rounding errors:

A number is uniformly chosen from \( [0.15, 0.25] \). It was rounded to two decimal places and then to one decimal place. The probability that the final value is \( 0.2 \) is \( X \% \). What is \(X?\)

Assumption: Use rounding "half away from zero". That is, if the number is equally far from the two closest numbers, choose the one away from zero. For example, 2.5 is equally far from 2 and 3, so round 2.5 to 3.

The reason as to why this works is a more advanced topic, which deals with the idea of measure theory . Measure theory gives a rigorous framework for probability theory, including probabilities on finite sets. Measure theory is also the key idea behind integration in calculus, and can be used to find integrals of functions that seem non-integrable using “standard” methods. These two ideas are not unrelated, as at a fundamental level, probability theory is just a special case of integration.

We will do a few more examples on working with geometric probabilities in higher dimensions to get a better feel for how to work with the concept. It is often helpful to use a figure to help with understanding and solving these types of problems.

Many probability problems include more than one variable, so 1D geometric probability won't be enough. For problems with two variables, it is often helpful to transform them into 2D geometric probability questions, where the outcomes are measured by area :

\[P(X) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}}.\]

This is most easily understood when the problem at hand is explicitly a 2D geometry problem:

A dart is thrown at a circular dartboard such that it will land randomly over the area of the dartboard. What is the probability that it lands closer to the center than to the edge? The set of outcomes are all of the points on the dartboard, which make up an area of \(\pi r^2\) where \(r\) is the radius of the circle. The points that are closer to the center than to the edge are those that lie within the circle of radius \(\frac{r}{2}\) around the center, so the area of the "success" outcomes is \(\pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}.\) Thus, \[P(\text{closer to center than edge}) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}} = \frac{\frac{\pi r^2}{4}}{\pi r^2} = \frac{1}{4} = 25\%.\ _\square\]

A square \( S \) has side length 30. A standard 20-sided die is rolled, and a square \( T \) is constructed inside \( S \) with side length equal to the roll. Then, a dart is thrown and lands randomly somewhere inside square \( S \). What is the probability that the dart also lands inside square \( T ?\) Suppose the die rolls \( i \). Then the probability that the dart will land inside square \( T \) is the ratio of the area of square \( T \) to the area of square \( S \). This is \( \frac{i^2}{900} \). For each \( i \), the probability that the die will roll \( i \) is \( \frac{1}{20}, \) so the probability that the dart lands inside \( T \) will be \[ \sum\limits_{i=1}^{20} \frac{1}{20}\cdot \frac{i^2}{900} = \frac{1}{20} \cdot \frac{20 \times 21 \times 41}{900 \times 6} = \frac{287}{1800} . \ _\square\]

The difficulty associated with geometric probability usually comes from one of two areas: the first is finding a good way to model the problem geometrically, and the second is in trying to determine the areas/volumes of particular regions in order to calculate the relative probabilities. As in finite probability, it is sometimes simpler to find the probability of the complement.

To make sure you've got down the basic ideas of 2D geometric probability, try this similar question. Note that many 2D geometry problems, such as the one below, use the ideas of composite figures. If you are not familiar with that concept, you may want to take a look at composite figures first.

The dartboard above is made up of three concentric circles with radii \(1, 3,\) and \(5.\) Assuming that a dart thrown will land randomly on the dartboard, what is the probability that it lands in the green region?

However, one of the most powerful uses of geometric probability is applying it to problems that are not inherently geometric. Identifying when and how to use geometric probability is never obvious, but a good sign is that you are dealing with probabilities in a situation with continuous variables. Let's take a look at a modified example of the bus problem mentioned at the beginning of this wiki.

Both the bus and you get to the bus stop at random times between 12 pm and 1 pm. When the bus arrives, it waits for 5 minutes before leaving. When you arrive, you wait for 20 minutes before leaving if the bus doesn't come. What is the probability that you catch the bus? We have two continuous variables here: \(b,\) the time in minutes past 12 pm that the bus arrives, and \(y,\) the time in minutes past 12 pm that you arrive. Since there are 2 independent variables, we will convert this into a 2-dimensional geometry problem. Specifically, we can think of the set of all outcomes as the points in a square: Then, we need to determine the region of "success"; that is, the points where we catch the bus. Since the bus will wait for 5 minutes, you need to arrive within 5 minutes of the bus' arrival, or \(y \le b+5:\) However, you only wait for 20 minutes, so you can't arrive more than 20 minutes before the bus, so \(y \ge b-20:\) Combining our two conditions, we have a region of success as shown below: Now, we just need to find the area of this success region. A simple method is to find the area of the non-success region, and then subtract that from the total area: Thus, the probability of catching the bus is \[P(\text{catching the bus}) = \dfrac{\text{area of desired outcomes}}{\text{area of total outcomes}} = \frac{60^2 - \frac{55^2}{2}-\frac{40^2}{2}}{60^2} = \frac{103}{288} \approx 36\%.\ _\square\]

Now that we have changed our problem into a geometric one, we can easily answer other questions about the situation such as the following:

1) What is the probability that the bus does not have to wait for you? 2) What is the probability that you had to wait less than 10 minutes, given that you were able to catch the bus? 3) What is the probability that the bus came and went before you, given that you were not able to board the bus?

To practice these ideas, let's try a similar question:

Dave and Kathy both arrive at Pizza Palace at two random times between 10:00 p.m. and midnight. They agree to wait exactly 15 minutes for each other to arrive before leaving. What is the probability that Dave and Kathy see each other?

If the probability is \( \frac ab\) for coprime positive integers, give the answer as \(a+b\).

You have many chocolate bars of unit length and start breaking each of them into 3 pieces by randomly choosing two points on the bar. What are the average lengths of the shortest, medium, and longest pieces?

If the product of these averages can be expressed as \( \frac pq\), where \(p\) and \(q\) are coprime positive integers, give your answer as \(p+ q\).

At this point, you can probably guess where this is headed! 3D geometric probability is when we are dealing with 3 continuous variables, and we measure the volume of the various outcomes; that is,

\[P(X) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}}.\]

To get started, let's look at an example which is analogous to the first problem we solved in the 2D geometric probability section.

An atom is inside a sphere and it is equally likely to be anywhere within the sphere. What is the probability that it lands closer to the center of the sphere than the outside? The set of outcomes are all of the points in the sphere, which make up a volume of \(\frac{4\pi}{3} r^3\) where \(r\) is the radius of the sphere. The points that are closer to the center than to the edge are those that lie within the sphere of radius \(\frac{r}{2}\) around the center, so the volume of the "success" outcomes is \(\frac{4\pi}{3} \left(\frac{r}{2}\right)^3 = \frac{\pi}{6}r^3.\) Thus, \[P(\text{closer to center than outside}) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}} = \frac{\frac{\pi}{6}r^3}{\frac{4\pi}{3} r^3} = \frac{1}{8} = 12.5\%.\ _\square\]

Of course, not all problems will be so explicitly geometric in nature. As usual, one of the signs that we might want to apply geometric probability is that we are dealing with continuous variables. Let's see how we can approach the following example:

Alex, Bob, and Charlie each randomly pick a real number between 0 and 1. What is the probability that the sum of the squares of their numbers does not exceed 1? First, if we let their 3 numbers be \(x,\) \(y,\) and \(z,\) it is easy to see that the outcomes can be represented as points within a unit cube \(\big(\)the cube that encloses the region \(x,y,z \in [0,1]\big),\) which has volume \(1^3 = 1.\) Then, the region where the sum of the squares of their numbers does not exceed 1 is given by \(x^2+y^2+z^2 \le 1,\) which (without restriction) is the sphere of radius 1 centered around the origin \((0,0,0).\) However, since \(x,y,z\ge 0,\) exactly \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\) of this sphere (one "octant") lies within the unit cube of possible outcomes. Hence, the volume of this "success" region is \(\frac{1}{8} \cdot \left(\frac{4\pi}{3} \cdot 1^3\right) = \frac{\pi}{6}.\) Thus, \[P(\text{sum of squares does not exceed 1}) = \dfrac{\text{volume of desired outcomes}}{\text{volume of total outcomes}} = \frac{\frac{\pi}{6}}{1} = \frac{\pi}{6} = 52\%.\ _\square\]

If 3 real numbers are chosen randomly and uniformly from \([0,1],\) what is the probability that the square of one of the numbers will be greater than the sum of the squares of the other two numbers?

In addition to being a useful mathematical problem-solving tool, geometric probability can also be applied in other scientific fields. Let's start with an example from mechanics, using the ideas of velocity and acceleration .

We are playing shuffleboard on the table below, where the lengths of the regions are labelled below (in meters). You push the puck with initial velocity \(v,\) where \(v\) is randomly chosen between 5 and 15 meters/second. Since the table is rough, the puck decelerates at a constant rate of 5 m/s\(^2.\) What is the probability that you slide the puck off of the table? (You may assume that the puck is negligibly small.) In this problem, we have only one variable--the initial speed of the puck--so this is going to be a 1-dimensional geometry problem. Recall the kinematics formula \( v_{f} ^{2} = v_{i}^{2} + 2 a s \). The final speed \(v_ {f }\) is zero because the puck comes to a rest. The initial speed \( v_{i} = v\). The distance traveled \(s\) will decide how many points we get. After plugging in the values, we get \[s = \frac{ v^{ 2 } }{ 10 }, \] so \(s>8+4+2+1=15\) occurs when \(v>\sqrt{150}.\) If we think of \(v\) as being a point on a number line between 5 and 15, then we can find our probability as \[\frac{15-\sqrt{150}}{15-5} \approx 28\%.\ _\square\]

Awesome! Who would have expected that geometric probability would allow us to solve a physics problem? Let's check out a few more examples.

A block performs simple harmonic motion with time period \(T\) and maximum speed \(v\). The speed of the block is measured at a random time. What is the probability that the measured speed is more than \(\frac v2?\) Let's plot the block's position (light gray) and velocity (gray). As we are interested in relative values, the specific vertical scaling of the plot is not important. Green time intervals represent times at which we would do a positive measure, where the measured speed, or absolute value of velocity, would exceed \(\frac v2\). These intervals are \(\left(k \pi + \frac{\pi}6, k \pi + 5 \cdot \frac{\pi}6\right)\) for any whole number \(k\), since the start of the first green interval from the vertical axis is at time \(t\) for which \(-\sin{t} = -\frac12\), so \(t = \frac{\pi}6\). Since the motion is periodic and the measure can take place at any time with equal probability, we can argue that the probability of a positive measure is \[\frac{N \times 2 \left(5\cdot \frac{\pi}6 -\frac{\pi}6\right)}{N \times 2\pi} = \frac{2}{3},\] where the system makes \(N\) periods. \(_\square\)

Johannes Kepler worked out that all planets revolve around the sun in elliptical orbits with the sun at one focus. He also deduced that planets revolve around the sun with constant areal velocity. Let's model a small star system in which a planet revolves around the star in an elliptical orbit. It has semi-major axis \(a\) and semi-minor axis \(b\). During one revolution, the minimum speed of the planet is \(u\) and the maximum speed is \(v\). In a complete revolution, an instant of time is randomly and uniformly chosen. What is the probability that the distance between the planet and the star at that instant is more than \(b?\) Kepler observed that planets don't orbit the sun at a uniform speed, but rather move faster when they are closer to the sun and more slowly when they are farther away. He specifically determined that the orbital speed of a planet is such that the line drawn from the sun to the planet sweeps out equal areas in equal intervals of time. This means that the time the planet spends at distant positions is proportional to the area the line sweeps at those positions. Our probability is then \[ \mathbb{P}\big[(\text{measured distance}) > b\big] = \frac{(\text{time spent at distance}) > b}{(\text{time period})} = \frac{A}{A_0} = \frac{A}{\pi \, a \, b}, \] where \(A_0\) is the area of the ellipse and \(A\) is the unknown area that the sun-planet radius vector sweeps out while the planet is far away. The problem then reduces to finding this area: Let's draw the orbit in Cartesian coordinates with \(x\) and \(y\) in units of semi-axis \(a\). Let axis ratio \(\mu =\frac ba\), with \(\mu = 0.6\) in the picture above. The sun is at the left focus \(S\) with coordinates \(\big(-\sqrt{a^2 - b^2}, 0\big).\) Draw a circle with radius \(b\) that is centered at the sun, and find an intersection with the ellipse at point \(P\) with coordinates \(\left( -a \, \sqrt{\frac{a - b}{a + b}}, \sqrt{\frac{2 \, b^3}{a + b}} \right).\) This follows by writing down the equations of the ellipse and circle, respectively: \[\begin{align} \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 &= 1 \\ (x + c)^2 + y^2 &= b^2 \end{align} \] with \(c = \sqrt{a^2 - b^2}\), and by finding point \(P\) at which the curves meet. We will naturally divide the calculation of the area by integrating the upper branch of the ellipse from the \(x\)-coordinate of \(P\), and by adding the area of triangle \(P S Q\). The picture indicates area division by different shades of green: \[ \begin{align} A &= 2 \times \Bigg( \frac{1}{2} \cdot \left(c - a \, \sqrt{\frac{a - b}{a + b}}\right) \cdot \sqrt{\frac{2 \, b^3}{a + b}} + b \, \int_{x_1}^{x_2} \sqrt{1 - \left(\frac{x}{a}\right)^2} \, dx \Bigg)\\ &= \frac{1}{2} + \frac{1}{\pi} \cdot \left( \sqrt{2 \, \mu \, (1 - \mu)} + \arcsin{\sqrt{\frac{1 - \mu}{1 + \mu}}} \right). \end{align} \] We integrated from \( x_1 = -a \, \sqrt{\frac{a - b}{a + b}} \) to \( x_2 = a \). Below is the plot of the resulting probability in terms of axis ratio \(\mu\). With \(\mu = 0.6\) the probability of finding the planet farther than \(b\) is about 89%. \(_\square\) With the eccentricity of the orbit close to zero, the ellipse is heavily prolonged and the chance of finding the planet in the vicinity of the sun is small, because i) the major part of the orbit is farther from sun, and ii) when the planet does approach the sun, its speed is large, and hence its visitation time is short. On the other hand, when the eccentricity is close to one, the ellipse is close to a circle and the orbital velocity of the planet is approximately constant, which means that the planet spends about half the time at distances less than \(b\), and half the time farther away. This can be seen in the animation below where the axis ratio \( \mu \) oscillates between \(0.2\) and \(0.99\). Caveat: Imagine the ideal case \( b = a \) of a perfect circle. If a planet orbits its sun at a constant distance \( b,\) we surely could never measure its distance at more than \( b \). But the result above suggests the chance should be even. How would you explain this?

There are many great problems in geometric probability. If you'd like some extra challenges, check out these problems. If you'd like to contribute to this wiki, you can add a solution to one of the examples!

Two numbers are chosen randomly and uniformly from \([-a, a]\). What is the probability that the absolute value of the smaller number is greater than two times the absolute value of the larger number? Does the final answer depend on the value of \(a?\) Let \(X\) and \(Y\) represent the two random numbers. Chances of \(X\) being smaller than \(Y\) are even, so we can focus on the case \(X < Y\). Our probability is then \[ \mathbb{P} \big( |X| > 2 \cdot |Y| \big) = \frac{a^2}{2} : \frac{(2 a)^2}{2} = \frac{1}{4}. \] The probability expression involves two absolute values, so it splits to four cases, depending on the signs of the variates: i) \( (X < 0) \land (Y > 0)\) ii) \( (X < 0) \land (Y < 0)\) iii) \( (X > 0) \land (Y > 0)\) iv) \( (X > 0) \land (Y < 0)\). Solutions of the last two, together with the starting assumption, are empty sets. The union of the solutions of the first two cases is shaded in the picture below. Mind that we divided by half of the area of the square \(2 \cdot a^2\), since we started by reducing the probability space to the region above the line \(X = Y\). Note also that the answer doesn't involve \(a\), so that it would stay the same as long as we picked the two variates from some common interval. \(_\square\)

Two points are randomly and uniformly selected on the circumference of a circle. The center of the circle and the two points are joined together. What is the probability that we obtain each of the following? \(\begin{array}{rrl} &\text{i)} &\text{a line segment} \\ &\text{ii)} &\text{an acute-angled triangle} \\ &\text{iii)} &\text{a right-angled triangle} \\ &\text{iv)} &\text{an obtuse-angled triangle} \end{array}\) We may suppose that the two points \(A\) and \(B\) are selected independently so we can, without loss of generality, select and place \(B\) relatively to \(A\). i) Answer: \(0\). Why? Line segment would occur if \(B\) was placed at exact angle \(\pi\) to \(A\). But a continuous variable has zero chance of taking a specific value (what is the width of a point?). ii) Answer: \(\frac{1}{2}\). Why? Triangle would be acute if \(B\) was placed at some angle in the interval \(\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]\). Values \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) (such triangle would be right-angled ) and \(0\) ( degenerate ) should be excluded, yet again, some specific discrete values don't change the probability. iii) Answer: \(0\). Why? See above. iv) Answer: \(\frac{1}{2}\). Why? Triangle can be either acute or obtuse, with even chances. \(_\square\)

Two points are randomly and uniformly selected from the interior of a circle. The center of the circle and the two points are joined together. What is the probability that we obtain each of the following? \(\begin{array}{rrl} &\text{i)} &\text{a line segment} \\ &\text{ii)} &\text{an acute-angled triangle} \\ &\text{iii)} &\text{a right-angled triangle} \\ &\text{iv)} &\text{an obtuse-angled triangle} \end{array}\)

Two points are chosen randomly and uniformly along a stick of length \(1.\) The stick is cut at those points to form \(3\) pieces. What is the probability that these pieces can form a triangle?

Two real numbers \(a\) and \(b\) are randomly chosen from the range \((0,1).\) The probability that \(\frac{a}{b}\) rounded to the nearest integer is odd is equal to \(P.\) What is \(\lfloor1000P\rfloor?\)

Large tablecloth has parallel vertical lines unit apart. Thin wooden stick of length \( \frac{3}{2} \) is twirled and tossed on the table. If \(P\) is the chance that the stick crosses two lines when it lands and comes to rest, how much is \(10^{10} \times P\), rounded to nearest whole number?

How large is the table exactly? Assume infinite expanse, infinitely thin lines, etc. The picture below depicts some sticks that are colored by the number of crossings.

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Bernoulli Trials:

Conditions for bernoulli trials:, geometric probability definition,  geometric distribution formula, geometric distribution:, mean of the geometric distribution:, the variance of the geometric distribution:, cumulative geometric probability, practice questions:, geometric probability – explanation & examples.

Geometric probability or geometric distribution refers to calculating the probability of first success in a sequence of Bernoulli trials.

Before reading this article, it might be helpful to refresh the following topics: 1. Basic probability theory 2. Independent events 3. Mean and Variance

After reading this article, you should understand

1. What is meant by geometric probability and geometric distribution. 2. What are Bernoulli Trials. 3. How to calculate geometric probability. 4. How to find the mean and variance of the geometric distribution.

What is Geometric Probability

To understand geometric probability, we first need to understand what constitutes a Bernoulli trial.

Bernoulli trials deal with experiments with two possible outcomes: ” success ” and ” failure ”. For example, when we toss a coin , we either get heads or tails. We can arbitrarily label getting either heads or tails as success and the other as a failure. Another example is the rolling of a six-sided fair die . Although we have six possible outcomes in this case, however, we can convert it into a Bernoulli trial by considering the probability of one particular outcome, say 2. So if we roll a die and get a 2, it is a success, and if we get any number other than 2, it is a failure.

For an experiment to qualify as a Bernoulli trial, it must satisfy two conditions:

• There are only two outcomes.
• The probability of success (and failure) remains the same for each trial.

For example, let’s say we draw a card from a deck of 52 cards. Let us say that it is a success if we get a king; otherwise, it is a failure. Each time we draw a card, if we also replace it in the deck, then the probability of each draw remains the same. However, if we draw a card from the deck and do not replace it before drawing the next card, then the probability of each subsequent draw depends upon previous draws, and the probabilities keep changing in each draw. In such a case, the experiment does not qualify as a Bernoulli trial.

Now that we have understood what a Bernoulli trial means, the concept of geometric probability becomes simple.

The probability of getting the first success in a sequence of Bernoulli trials is referred to as geometric probability .

How to find Geometric Probability:

It is best to consider an example to understand this concept.

Example 1: Let us suppose we are rolling a six-sided fair die many times. What is the probability of the following events:

• Getting a 2 on the first attempt.
• The first 2 appear on the third attempt.
• The first 2 appear on the 10th attempt.

1. Let us define “getting a 2” as success and “not getting a 2” as a failure . So,

$P(\textrm{Getting a 2 in a single roll}) = \frac16$ $P(\textrm{Not getting a 2 roll}) = 1 – \frac16 = \frac56$

Since each time we roll a die, the probability of a success or a failure remains the same, so it is an example of a Bernoulli trial . Also, since we are interested in the probability of getting a first 2 in the first, third, or 10th attempt. This is a case of geometric probability.

Now, the probability of getting the first $2$ in the first attempt is, of course, $\frac16$. If we get the first $2$ in the third attempt, it means we did not get a $2$ in the first AND second attempt, so

$P(\textrm{Getting the first 2 in 3rd attempt})$

$= P(\textrm{Not getting 2 in 1st attempt})\; \textrm{AND} \;P(\textrm{Not getting 2 in 2nd attempt})\; \textrm{AND}$

$P(\textrm{Getting a 2 in 3rd attempt})$.

Note that the probability of any given attempt is independent of what has happened in the previous attempts or what might happen in future attempts. Recall from basic probability theory that if two events, say $E1$ and $E2$ are independent, then the probability of the event $E1 \;\textrm{AND}\; E2 = P(E1) \times P(E2)$. Hence,

$= P(\textrm{Not getting 2 in 1st attempt}) \times P(\textrm{Not getting 2 in 2nd attempt})$

$\times \;P(\textrm{Getting a 2 in 3rd attempt})$.

$P(\textrm{Getting the first 2 in 3rd attempt}) = (1 – \frac16) \times (1 – \frac16) \times \frac16 = 0.115$

$P(\textrm{Getting the first 2 in 10th attempt})$

$= P(\textrm{Not getting 2 in 1st attempt}) \times \cdots \times P(\textrm{Not getting 2 in 9th attempt})$

$\times\; P(\textrm{Getting 2 in 10th attempt})$.

$P(\textrm{Getting the first 2 in 10th attempt}) = (1 – \frac56)^{9} \times \frac16 $.

Let us consider a Bernoulli trial with a probability of success equal to $p$ and probability of failure equal to $1-p$, then the probability of getting first success in the kth attempt is given as

$P(\textrm{First success in kth attempt}) = (1-p)^{k-1}p$.

Example 2 : A card is drawn randomly from a deck of $52$ cards and replaced. The process is continued until a king is drawn. Find the probability that the first king is drawn in the 5th attempt.

Let us suppose, if a king is drawn, we call it a success. If any other card apart from the king comes up, we call it a failure. Since the cards are being replaced at each draw, so we are dealing with a Bernoulli trial. Also, we are interested in the first success, so we are dealing with geometric probability. There are 4 kings in a deck of 52 cards, so the probability of success is $4/52 = 1/13$. Using the formula described above $P(\textrm{First king in 5th draw}) = (1 – \frac{1}{13})^{5-1} \times \frac{1}{13} = 0.056$.

Let $X$ is a random variable. Informally, a random variable is a collection of numbers, where each number has some probability. If we define $P(X=k) = (1-p)^{k-1}p$, then $X$ is called a geometric random variable and the function $f(k) = (1-p)^{k-1}p$ is called the geometric distribution. Note that $f(k)$ is only defined for discrete values of $k$ and hence is a discrete function. Below, we plot geometric distribution for various values of probability of success.

When we are performing Bernoulli trials, we might get success on the first attempt, or we might get success on the 10th attempt, or maybe we won’t get success until the 100,000th attempt( although the probability of such a case would be very low). We might be interested in the question, “How many attempts, on average, are required to get the first success?”. Such a number is called the mean or the expected value of a distribution. For geometric distribution, the expected value can be calculated using the formula

$E(X) = \sum^{\infty}_{k=1}(1 – p)^{k-1} \times p \times k$.

We omit the proof, but it can be shown that $E(X) = \frac1p$ if $X$ is a geometric random variable and $p$ is the probability of success.

Variance is a measure of the spread of the distribution. It tells us how much the distribution deviates from the mean/expected value. In some applications, we might be interested in the expected value and the variance of the geometric distribution. Again, we omit the proof and state the formula for the variance $VAR(X) = \frac{1 – p}{p^2}$.

Let us see a few examples.

Example3: A six-sided fair die is rolled many times until we get a 3. How many rolls, on average, are required to get the first 3.

Solution: We are obviously dealing with a geometric distribution here with $p=\frac16$. Since we are interested in the first success on average, so we can use the formula for the expected value of the geometric random variable. Hence,

$\textrm{Number of rolls, on average, to get first 3} = E(X) = \frac1p = \frac{1}{1/6} = 6$.

So, on average, it will take us six attempts to get the first 3.

Example4: In a community, for every 100 persons, 2 are infected with COVID-19. A doctor performs ideal PCR tests (with 100% detection accuracy) to detect COVID patients each day by randomly selecting persons from the community. Each test is assumed to be independent of the other. How many tests, on average, would the doctor perform to get the first positive result?

Let us define a positive test as a success (ironically). The probability of success is $2/100 = 1/50$. Since each test is independent, so it is a Bernoulli trial. Since we are interested in first success, so it is a geometric distribution. Using the formula for the expected value of a geometric distribution

$\textrm{Number of tests, on average, to get first positive} = \frac{1}{p} = 50$.

So, on average, after every $50$ test, the doctor will get a positive result.

Sometimes, we are not exactly interested in whether the first success will appear in the 4th or 5th attempt but instead if the first success will appear WITHIN 4th or 5th attempt. In other words, instead of asking for $P(X=k)$, we are asking for $P(X \leq k)$. One possible method is to note that

$P(X \leq k) = P(X=1) + P(X=2) + \cdots + P(X=k)$.

However, there is another simpler method to find $P(X \leq k)$. We note from basic probability that

$P(X \leq k) = 1 – P(X>k)$.

If we ask what the probability that the first success appears after the 4th attempt, i.e., $X>4$, it is the same as asking what the probability that the first four attempts are not a success is. So,

$P(X>k) = P(\textrm{1st attempt is failure}) \times P(\textrm{2nd attempt is failure}) \times \cdots P(\textrm{kth attempt is failure})$.

$P(X>k) = (1-p)^k$.

Accordingly,

$P(X \leq k) = 1 – (1-p)^k$.

Example5: A card is drawn randomly from a deck of 52 cards and replaced. The process is continued until a king is drawn. Find the probability that the first king is drawn WITHIN first 5 attempts.

$P(\textrm{First King not in first 5 attempts}) = (1 – \frac{1}{13})^{5} = 0.67$

$P(\textrm{First King within first 5 attempts}) = 1 -(\textrm{First King not in first 5 attempts}) = 0.33$

Question No.1: Let a factory is producing PlayStation-4 consoles. The probability of producing a defective console is $p=0.001$. A tester is testing randomly selected items from the production line. Find the probability of the following events: 1. The first defective console is detected in the 4th test. 2. The first defective console is detected in the 100th test. 3. The first defective console is not detected in the first 100 tests. 4. The first defective console is detected in the first 100 tests.

Answers: 1) 0.00099 2) 0.000904 3) 0.904 3) 0.0952

Question No.2 : A traveling salesperson is selling an item. The probability that he/she will successfully sell the item to a random customer is $0.1$. On average, how many customers he will try until he gets the first sale?

Answer : 10

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Geometric Probability

What is probability?

Probability  is the likelihood of a desired outcome against all possible outcomes.

An archery target for recreational target practice is often just a large circle with four other circles inside it. As you get closer to the center, the points you win increase: 1 , 3 , 5 , 7 , and 9 for the bullseye. You can use geometric probability to calculate the chance of hitting a bullseye.

If something can never happen, its probability is 0 , which could also be considered 0 percent .

If something may or may not happen, its probability is between 0 and 1 , or from 0 percent to 100 percent .

If something is absolutely certain to happen, its probability is 1 , or 100 percent .

The desired outcome of an event, like the roll of two dice, is the probability you want; subtracted from  1 , it leaves the probability of unfavorable outcomes.

Together, what you want to have happen and everything that could happen will add to  1 , or to  100 percent .

We can illustrate this with dice:

[insert three pictures of sets of normal dice, or possibly make a video showing the three probabilities?]

The probability of getting a total of 1 from two, six-sided, normally marked dice is 0 , or 0 percent .

The probability of getting a 7 from two, six-sided, normally marked dice is 0.166666667 , or 16.66 percent .

The probability of getting a number ranging from 2 to 12 from two, six-sided, normally marked dice is 1 , or 100 percent .

Geometric probability

Here is a square with a smaller square inside it, tucked in the corner:

[insert drawing of square, like a tic-tac-toe board, with one corner staked out as a smaller, shaded square; smaller square is 1/9 of the larger one]

We got the little square by dividing height and width into three equal sections, so the little square is one square out of nine squares. Geometric probability is the chance of hitting the little shaded square, say with a dart or arrow, out of hitting anywhere on the square.

The shaded square is  1 9 \frac{1}{9} 9 1 ​ of the whole area, so the probability of hitting it is  1 9 \frac{1}{9} 9 1 ​ , while the chance of not hitting it is  8 9 \frac{8}{9} 9 8 ​ . Together, these two probabilities add to  9 9 \frac{9}{9} 9 9 ​ , or 100% . To find the geometric probability of an event, we use this formula, where  P  is the probability:

Imagine the big square is  9 cm wide and  9 cm tall. That makes the little square  3 cm × 3 cm , or  9 c m 2 9c{m}^{2} 9 c m 2 . The large square is  81 c m 2 81c{m}^{2} 81 c m 2 , so the geometric probability of hitting the little square is:

As a decimal,  1 9 \frac{1}{9} 9 1 ​  is (interestingly, we think)  0.111111 , which is far closer to  0  than to  1 . If you were given a small beanbag and invited to toss the beanbag onto the big square, you do not have a very good chance of hitting the little shaded square.

Geometric probability of Greenland's flag

The flag of Greenland is interesting, both for its meaning (white is snow and ice; red is the sun) and its mathematics.

And here is a diagram showing its proportions:

We can see that the entire flag is  18  units wide x  12  units tall. It is designed in abstract units so you could make it in cm, m, inches, or feet, so long as you kept all the ratios and proportions the same.

To see if you get the hang of geometric probability, try these three questions:

What is the total area of Greenland's flag?

The flag is  18 × 12  units, for an area of  216 u n i t s 2 216unit{s}^{2} 216 u ni t s 2 .

The circle is 8 units in diameter ( 4 units in radius). If the area of a circle is  π r 2 \pi {r}^{2} π r 2 , what is the circle's area?

The circle's area is calculated by A = π r 2 A=\pi {r}^{2} A = π r 2 .

What is the geometric probability of a Greenland butterfly, the  Arctic fritillary , accidentally flying into the circle?

P = 23.27% , or less than a 1-in-4 chance.

Let's make it more interesting. Notice that half of the circle is white, and half is red. What is the geometric probability that the butterfly will bump into  only  the red part of the circle?

You already know the geometric probability of hitting the circle:  0.23270 , so since the red part is half of the circle, that makes the red area half of the whole circle, or  0.11635 , which is around  11.63 percent .

Ready to try one more?

What is the geometric probability that our little Greenland butterfly will bump into  any  red area of Greenland's flag? If you look carefully, the half-circle of red could flip downward, to make the entire bottom half of the flag all read. So:

P = 0.5 or a 50% chance So now, if you ever go to Greenland, you will not only know the flag, you will know the geometric probability of its elegant, simple design.

Geometric probability of a bullseye

Suppose you have an archery target that is a circle  60 cm  in diameter, with a bullseye  5 cm  in diameter. What is the geometric probability of hitting the bullseye? Let's start with our formula for the area of a circle for the archery target:

Now, let's take a look at the bullseye:

Now we can calculate the geometric probability:

You need real skill to hit a bullseye, because you only have a  2.7%  chance of doing it at random!

Lesson summary

Now that you have explored this lesson (and a little geography), you are able to explain the concept of probability ranging from 0 to 1 , recall and state a definition of geometric probability, and use the formula for geometric probability to calculate the likelihood of an outcome involving flat shapes.

Geometric Probability

Related Topics: More Lessons for Grade 8 Math Geometry Help Math Worksheets

The following are more probability problems that involve geometrical shapes.

ABCD is a square. M is the midpoint of BC and N is the midpoint of CD . A point is selected at random in the square. Calculate the probability that it lies in the triangle MCN .

Let 2 x be the length of the square.

Area of square = 2 x × 2 x = 4 x 2

The figure shows a circle with centre O and radius 8 cm. Ð* BOD* = 72˚. The radius of the smaller circle is 4 cm. A point is selected at random inside the larger circle BCDE .

Calculate the probability that the point lies a) inside the sector BODC . b) inside the smaller circle c) neither in the sector BODC nor in the smaller circle.

Geometric Probability Example: Find the probability of a dart landing in the light purple region.

Geometric Probability Using Area Examples: (1) A circle with radius 2 lies inside a square with side length 6. A dart lands randomly inside the square. What is the probability that the dart lands inside the circle? Give the exact probability and the probability as a percent rounded to the nearest tenth. (2) A point is chosen at random on this figure. What is the probability that the point is in the yellow region? (3) A square is inscribed in a circle. What is the probability that a point chosen at random inside the circle will be inside the square? (4) A circle is inscribed in an equilateral triangle. What is the probability that a point chosen at random inside the triangle will be inside the circle? (5) Find the probability that a point chosen inside the circle will be inside the shaded region.

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4.4: Geometric Distribution

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There are three main characteristics of a geometric experiment.

• There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.
• In theory, the number of trials could go on forever. There must be at least one trial.
• The probability, \(p\), of a success and the probability, \(q\), of a failure is the same for each trial. \(p + q = 1\) and \(q = 1 − p\). For example, the probability of rolling a three when you throw one fair die is \(\dfrac{1}{6}\). This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is \(q = \dfrac{5}{6}\), the probability of a failure. The probability of getting a three on the fifth roll is \(\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\left(\dfrac{1}{6}\right) = 0.0804\).

\(X =\) the number of independent trials until the first success.

You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is \(p = 0.57\). What is the probability that it takes five games until you lose? Let \(X =\) the number of games you play until you lose (includes the losing game). Then \(X\) takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is \(P(x = 5)\).

Example \(\PageIndex{1}\)

Exercise \(\pageindex{1}\).

You throw darts at a board until you hit the center area. Your probability of hitting the center area is \(p = 0.17\). You want to find the probability that it takes eight throws until you hit the center. What values does \(X\) take on?

\(1, 2, 3, 4, \dotsc, n\). It can go on indefinitely.

Example \(\PageIndex{2}\)

A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions?

Let \(X\) = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. \(X\) takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find \(P(x \geq 3)\). ("At least" translates to a "greater than or equal to" symbol).

Exercise \(\PageIndex{2}\)

An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically?

\(P(x \leq 10)\)

Example \(\PageIndex{3}\)

Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?

This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).

• Let \(X =\) the number of ____________ you must ask ____________ one says yes.
• What values does \(X\) take on?
• What are \(p\) and \(q\)?
• The probability question is \(P(\)_______\()\).
• Let \(X =\) the number of students you must ask until one says yes.
• 1, 2, 3, …, (total number of students)
• \(p = 0.55; q = 0.45\)
• \(P(x = 4)\)

Exercise \(\PageIndex{3}\)

You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are \(p\) and \(q\)?

\(p = 0.1\)

\(q = 0.9\)

Notation for the Geometric: \(G =\) Geometric Probability Distribution Function

\(X \sim G(p)\)

Read this as "\(X\) is a random variable with a geometric distribution." The parameter is \(p\); \(p =\) the probability of a success for each trial.

Example \(\PageIndex{4}\)

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?

Let \(X\) = the number of computer components tested until the first defect is found.

\(X\) takes on the values 1, 2, 3, ... where \(p = 0.02\). \(X \sim G(0.02)\)

Find \(P(x = 7)\). \(P(x = 7) = 0.0177\).

To find the probability that \(x = 7\),

• Enter 2 nd , DISTR
• Scroll down and select geometpdf(
• Press ENTER
• Enter 0.02, 7); press ENTER to see the result: \(P(x = 7) = 0.0177\)

To find the probability that \(x \leq 7\), follow the same instructions EXCEPT select E: geometcdf as the distribution function.

The probability that the seventh component is the first defect is 0.0177.

The graph of \(X \sim G(0.02)\) is:

The y -axis contains the probability of \(x\), where \(X =\) the number of computer components tested.

The number of components that you would expect to test until you find the first defective one is the mean, \(\mu = 50\).

The formula for the mean is

\[\mu = \dfrac{1}{\text{p}} = \dfrac{1}{0.02} = 50\]

The formula for the variance is

\[\sigma^{2} = \left(\dfrac{1}{p}\right)\left(\dfrac{1}{p} - 1 \right) = \left(\dfrac{1}{0.02}\right)\left(\dfrac{1}{0.02} - 1 \right) = 2,450\]

The standard deviation is

\[\sigma = \sqrt{\left(\dfrac{1}{p}\right)\left(\dfrac{1}{p} - 1\right)} = \sqrt{\left(\dfrac{1}{0.02}\right)\left(\dfrac{1}{0.02} - 1\right)} = 49.5\]

Exercise \(\PageIndex{4}\)

The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.

\(P(x = 9) = 0.0092\)

Example \(\PageIndex{5}\)

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let \(X =\) the number of people you ask until one says he or she has pancreatic cancer. Then \(X\) is a discrete random variable with a geometric distribution: \(X \sim G\left(\dfrac{1}{78}\right)\) or \(X \sim G(0.0128)\).

• What is the probability of that you ask ten people before one says he or she has pancreatic cancer?
• What is the probability that you must ask 20 people?
• Find the (i) mean and (ii) standard deviation of \(X\).
• \(P(x = 10) = \text{geometpdf}(0.0128, 10) = 0.0114\)
• \(P(x = 20) = \text{geometpdf}(0.0128, 20) = 0.01\)
• Mean \(= \mu = \dfrac{1}{p} = \dfrac{1}{0.0128} = 78\)
• Standard Deviation \(= \sigma = \sqrt{\dfrac{1-p}{p^{2}}} = \sqrt{\dfrac{1-0.0128}{0.0128^{2}}} \approx 77.6234\)

Exercise \(\PageIndex{5}\)

The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let \(X =\) the number of Afghani women you ask until one says that she is literate.

• What is the probability distribution of \(X\)?
• What is the probability that you ask five women before one says she is literate?
• What is the probability that you must ask ten women?
• \(X \sim G(0.12)\)
• \(P(x = 5) = \text{geometpdf}(0.12, 5) = 0.0720\)
• \(P(x = 10) = \text{geometpdf}(0.12, 10) = 0.0380\)
• Mean \(= \mu = \dfrac{1}{p} = \dfrac{1}{0.12} \approx 3333\)
• Standard Deviation \(= \sigma = \dfrac{1-p}{p^{2}} = \dfrac{1-0.12}{0.12^{2}} \approx 7.8174\)
• “Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at www.pewsocialtrends.org/files...-to-change.pdf (accessed May 15, 2013).
• “Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/...pen-to-change/ (accessed May 15, 2013).
• “Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/...last&sort=desc (accessed May 15, 2013).
• Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/N...eshman2011.pdf (accessed May 15, 2013).
• “Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at ec.europa.eu/europeaid/where/...summary_en.pdf (accessed May 15, 2013).
• “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publicat...k/geos/af.html (accessed May 15, 2013).
• “UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/vide...y-centers.html (accessed May 15, 2013).

There are three characteristics of a geometric experiment:

• There are one or more Bernoulli trials with all failures except the last one, which is a success.
• The probability, \(p\), of a success and the probability, \(q\), of a failure are the same for each trial.

In a geometric experiment, define the discrete random variable \(X\) as the number of independent trials until the first success. We say that \(X\) has a geometric distribution and write \(X \sim G(p)\) where \(p\) is the probability of success in a single trial. The mean of the geometric distribution \(X \sim G(p)\) is \(\mu = \dfrac{1-p}{p^{2}} = \sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}\).

Formula Review

\(X \sim G(p)\) means that the discrete random variable \(X\) has a geometric probability distribution with probability of success in a single trial \(p\).

\(X =\) the number of independent trials until the first success

\(X\) takes on the values \(x = 1, 2, 3, \dotsc\)

\(p =\) the probability of a success for any trial

\(q =\) the probability of a failure for any trial \(p + q = 1\)

\(q = 1 – p\)

The mean is \(\mu = \dfrac{1}{p}\).

The standard deviation is \(\sigma = \dfrac{1-p}{p^{2}} = \sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}\).

Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask.

Exercise 4.5.6

In words, define the random variable \(X\).

\(X =\) the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

Exercise 4.5.7

\(X \sim\) _____(_____,_____)

Exercise 4.5.8

What values does the random variable \(X\) take on?

Exercise 4.5.9

Construct the probability distribution function (PDF). Stop at \(x = 6\).

Exercise 4.5.10

On average (\(\mu\)), how many freshmen would you expect to have to ask until you found one who replies "yes?"

Exercise 4.5.11

What is the probability that you will need to ask fewer than three freshmen?

1 ”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/...pi_data_value- last&sort=desc (accessed May 15, 2013).

\[\sqrt{\dfrac{1}{p}\left(\dfrac{1}{p} - 1\right)}\]

• The probability, \(p\), of a success and the probability, \(q\), of a failure do not change from trial to trial.

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Probability

9th -  12th  , 3rd -  5th  , conditional probability, 11th -  12th  , basic probability.

Geometric Probability Assignment

10th - 12th grade, mathematics.

12 questions

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Find the probability of a point chosen at random being selected in the non-shaded area. (Within the circle of radius 14)

Three concentric circles (same center) have radii of 1unit, 2 units, and 4 units. What is the probability that a point selected at random within the largest circle will be in the ring between the smallest and next largest circles?

ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. A point is selected at random in the square. Calculate the probability that it lies in the triangle MCN.

A circle with radius 2 lies inside a square with side length 6. A dart lands randomly inside the square. What is the probability the dart lands inside the circle? Give the exact probability

4 π 36 \frac{4\pi}{36} 3 6 4 π ​

2 π 36 \frac{2\pi}{36} 3 6 2 π ​

4 π 12 \frac{4\pi}{12} 1 2 4 π ​

8 π 36 \frac{8\pi}{36} 3 6 8 π ​

Find the probability that a randomly chosen point in the figure lies in the shaded region. The shape is a regular pentagon.

Find the probability of the shaded region as a % rounded to the nearest tenth.

Find the probability of a point being in the shaded region. Express your answer as a percentage. (Please take a moment to appreciate Ms. Khraishi's hard work drawing that picture. YES, that is a regular hexagon...)

Find the probability of a point falling within the shaded region. Express as a percentage.

Find the probability of a point falling in the shaded region. Express as a percentage.

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GEOMETRIC PROBABILITY

Probability :

A probability is a number from 0 to 1 that represents the chance that an event will occur. Assuming that all outcomes are equally likely, an event with a probability of 0 cannot occur. An event with a probability of 1 is certain to occur,and an event with a probability of 0.5 is just as likely to occur as not.

We already know, how to evaluate probabilities by counting the number of favorable outcomes and dividing that number by the total number of possible outcomes.

In this section, we will use a related process in which the division involves geometric measures such as length or area. This process is called geometric probability. 

Geometric Probability

Probability and Length :

In the diagram shown below, AB is a segment that contains the segment CD. 

If a point K on AB is chosen at random, then the probability that it is on CD is as follows :   

Probability and Area :

In the diagram shown below, J is a region that contains region M.

If a point K in J is chosen at random, then the probability that it is in region M is as follows :

Finding a Geometric Probability

Example 1 :

Find the probability that a point chosen at random on RS is on TU.

P(Point is on TU)  =  2/10

P(Point is on TU)  = 1/5

So, the probability can be written as 1/5 or 0.2 or 20%.

Example 2 :

Mr. Johnson works for a temporary employment agency. He lives on the west side of town and prefer to work there. The work assignments are spread evenly throughout the rectangular region shown below. Find the probability that an assignment chosen at random for Mr. Johnson is on the west side of town.

The west side of town is approximately triangular. 

Its area is

=  1/2  ⋅ 2.25  ⋅ 1.5

=  1.69 square miles

The area of the rectangular region is

=  1.5 • 4

=  6 square miles

So, the probability that the assignment is on the west side of town is

P(Assignment is on west side)   ≈  1.69/6

P(Assignment is on west side)   ≈  0.28

So, the probability that the work assignment is on the west side is about 28%.

Using Areas to Find a Geometric Probability

Example 3 :

A dart is tossed and hits the dart board shown below. The dart is equally likely to land on any point on the dart board. Find the probability that the dart lands in the red region.

Find the ratio of the area of the red region to the area of the dart board.

P(Dart lands in red region)  =   π(2) 2 /16 2

P(Dart lands in red region)  =  4 π/256

Use calculator. 

P(Dart lands in red region)   ≈   0.05

So, the probability that the dart lands in the red region is about 0.05, or 5%.

Using a Segment to Find a Geometric Probability

Example 4 :

Mr. Jacob is visiting San Francisco and taking a trolley ride to a store on Market Street. He is supposed to meet a friend at the store at 3:00 P.M. The trolleys run every 10 minutes and the trip to the store is 8 minutes. Mr. Jacob arrives at the trolley stop at 2:48 P.M. What is the probability that Mr. Jacob will arrive at the store by 3:00 P.M.?

To begin, find the greatest amount of time Mr. Jacob can afford to wait for the trolley and still get to the store by 3:00 P.M. Because the ride takes 8 minutes, he needs to catch the trolley no later than 8 minutes before 3:00 P.M., or in other words by 2:52 P.M. So, Mr. Jacob can afford to wait 4 minutes (2:52 º 2:48 = 4 min). He can use a line segment to model the probability that the trolley will come within 4 minutes.

The required probability is

P(Get to store by 3:00)  =  4/10

P(Get to store by 3:00)  =  2/5

So,  the probability that Mr. Jacob will arrive at the store by 3:00 P.M. is 2/5 or 0.4 or 40%. 

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Visualization of basic probability assignment

• Foundation, algebraic, and analytical methods in soft computing
• Published: 19 September 2022
• Volume 26 , pages 11951–11959, ( 2022 )

Cite this article

• Hongfeng Long 1 ,
• Zhenming Peng   ORCID: orcid.org/0000-0002-4148-3331 1 &
• Yong Deng   ORCID: orcid.org/0000-0001-9286-2123 2  

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Applying geometry to the analysis and interpretation of basic probability assignment (BPA) is a unique research direction in evidence theory. Among them, the visualization of BPA helps to intuitively analyze the geometric properties and characteristics of BPA, which is an important research content in this direction, but there are currently few related studies. In response to this problem, we proposed a new BPA visualization method based on the vector representation of the BPA to illustrate the image of BPA directly. First, the basic point and the uncertain vectors could be obtained by the given BPA, and then we connected these components to construct the image of BPA. Through the image of BPA, we can effectively analyze the interaction effect of focal elements in BPA and observe the potential characteristics of BPA directly. For example, the uncertainty of focal elements can be expressed by geometric area. Moreover, the geometric meanings of parameters in the vector representation of the BPA can be explained. Finally, numerical examples are illustrated to demonstrate the advantages and related applications of the proposed method.

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Acknowledgements

The work is partially supported by National Natural Science Foundation of China (Grant Nos. 61973332 and 61775030), and is also partially supported by Natural Science Foundation of Sichuan Province of China (Grant No. 2022NSFSC40574).

The work was partially supported by National Natural Science Foundation of China (Grant No.61973332 and No.61775030).

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Hongfeng Long & Zhenming Peng

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All authors contributed to the study conception and design. HFL performed all experiments and wrote the manuscript. ZMP and YD contributed to the conception of the study and provided critical revisions. All authors read and approved the final manuscript.

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Long, H., Peng, Z. & Deng, Y. Visualization of basic probability assignment. Soft Comput 26 , 11951–11959 (2022). https://doi.org/10.1007/s00500-022-07412-1

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Accepted : 11 July 2022

Published : 19 September 2022

Issue Date : November 2022

DOI : https://doi.org/10.1007/s00500-022-07412-1

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