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 0 answersPost by Jamal Salim on October 27, 2015in Example 2 you can find the probability direct using 72/360 = 1/5 because (Pi)r2 will cancel each other 2 answersLast reply by: Jamal SalimTue Oct 27, 2015 8:57 PMPost by Mirza Baig on December 6, 2013I think the example 3 where you find the side of a square and you said that it was 10 radical 2 but in my opinion i think that's wrong because we know that  squares sides and hypotenuse are same because  of Pythagorean theoremand so add 10+10 = 20,20 is the side of a square so the area of a square would be 400cm^2.

### Geometric Probability

• Length Probability Postulate: If a point on AB is chosen at random and C is between A and B, then the probability that the point is on AC is AC/AB
• Area Probability Postulate: If a point in region A is chosen at random, then the probability that the point is in region B, which is the interior of region A, is the area of B to the area of A
• Area of a Sector of a Circle: A sector of a circle is a region of a circle bounded by a central angle and its intercepted arc. Area = (central angle)/(360 degrees) × πr2

### Geometric Probability

Find the probability that a point is on AB .
[AB/AC] = [6/(6 + 10)] = [3/8]

AB = AC , if a point is in ∆ABC,find the probability that it is in ∆ABD.
• ∆ABD ≅∆ACD
• [Area of ∆ ABD/Area of ∆ ACD] = [1/2]
The probability is [1/2]

Determine whether the following statement is true or false.
If a point is in circle A, then it is also in pentagon BCDEF.
False

Circle A, m∠BAC = 60o, AB = 5, find the area of sector ABC.
• Area = [(m∠BAC)/360]*πr2
• Area = [60/360]*3.14*52
Area = 94.2

Rhombus ABCD, if a point is in rhombus ABCD, find the probability that it is also in ∆ABE.
• [Area of ∆ ABE/Area of rhombus ABCD] = [([1/2]*AE*BE)/([1/2]*AC*BD)] = [AE*BE/AC*BD] = [AE*BE/(2AE)*(2BE)] = [1/4]
The probability is [1/4]

Trapezoid ABCD,BE ⊥AD , if a point is in trapezoid ABCD, find the probability that it is in ∆ABE.
• [Area of ∆ ABE/Area of trapezoid ABCD] = [([1/2]*AE*BE)/([1/2]*(AD + BC)*BE)] = [AE/(AD + BC)] = [2/((2 + 10) + 7)] = [2/19]
[2/19]

Square ABCD, find the probability that a point is in the square but outside the circle.
• [Area of the circle/Area of the square] = [(πr2)/((2r)2)] = [(π)/4] = [3.14/4] = 0.785
• The probability that the point is in the shaded region = 1 − 0.785 = 0.215
0.215

Rectangle ABCD, circle E and circle F, a point is in the rectangle, find the probability that it is in the shaded region.
• r1 = [1/2]AB = 4
• 2r1 + 2r2 = BC
• r2 = [1/2]BC − r1 = [1/2]*12 − 4 = 2
• Area of circle E = πr12 = 3.14*4*4 = 50.24
• Area of circle F = πr22 = 3.14*2*2 = 12.56
• Area of rectangle ABCD = 8*12 = 96
• Area of shaded region = 96 − 50.25 − 12.56 = 33.2
• The probability that the point is in the shaded region: [33.2/96] = 0.346
0.346

Determine whether the following statement is true or false.
If a point is in the circle, then it is also in square ABCD.
True
Determine whether the following statement is true or false.
For two parallel lines, if a point is on one line, then the probability that the point is also on the other line is 0.
True

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Geometric Probability

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Length Probability Postulate 0:05
• Length Probability Postulate
• Are Probability Postulate 2:34
• Are Probability Postulate
• Are of a Sector of a Circle 4:11
• Are of a Sector of a Circle Formula
• Are of a Sector of a Circle Example
• Extra Example 1: Length Probability 11:07
• Extra Example 2: Area Probability 12:14
• Extra Example 3: Area Probability 17:17
• Extra Example 4: Area of a Sector of a Circle 26:23

### Transcription: Geometric Probability

Welcome back to Educator.com.0000

For the next lesson, we are going to go over geometric probability.0001

The first thing that we are going to go over is the Length Probability Postulate.0008

It is when we are using segments for probability.0013

If a point on segment AB is chosen at random, and point C is between A and B, then the probability that the point is on AC0018

is going to be (and this is a ratio) segment AC, over AB.0033

Now, if you remember probability, probability measures the part over the whole.0041

You can also think of the top number as the desired outcome, over the total outcome, the total possible number of all of the different types of possible outcomes.0052

So then, it is a desired outcome, what you are looking for, over the total--over the whole thing.0077

So, it is just part over whole; this is the most basic way you can remember probability, part over whole.0083

Here, the same thing applies to the Length Probability Postulate; you are looking at the part.0091

You are looking at what the desired outcome is, which is the point being on AC, over the whole thing; that is AB--that is the whole thing.0099

It is AC to AB--always part over whole.0110

So, let's say that this right here is 5; CB is 5; the probability of a point landing on AC...what is AC?0116

That is the desired outcome; that is the top number, which is 5, over the whole thing (it is not 5; it is not the other half);0128

it is the whole thing, which is AB, and that is 10; so when I simplify this, this becomes 1/2.0139

The probability of landing on AC is 1/2.0148

And for the Area Probability Postulate, when you are talking about the probability of something to do with area, you are looking at space.0156

So, you are looking to see, for example, maybe a dart hitting the dartboard; that is area, because it is space that you are looking at.0166

If you look to see, maybe, a spinner (we are going to have both of those for our examples) landing on a certain space, that is area.0178

So, that has to do with this probability; and this postulate says that if a point in region A0191

(this rectangle is region A) is chosen at random, then the probability that the point is in region B0199

(which is inside region A) is going to be the area of region B, over...remember: the whole thing is the area of region A.0207

Region A is the whole thing; the area of the whole thing is the total,0231

and the top one will be the area of the desired outcome, or the part that we were just looking at; and that is region B.0238

The area of a sector of a circle: now, a sector is this little piece right here.0254

This is the center; the sector is the area of this piece, so it is bounded by the central angle0264

(this is a central angle right here; this angle is a central angle, so if I need the θ, that is the central angle) and its intercepted arc.0273

This is the intercepted arc; so those are the boundaries, this angle and that.0283

This whole thing is called a sector; now, I like to refer to the sector as a pizza slice.0289

Think of this whole thing as a pizza; this is a slice of pizza; so a sector is a slice of pizza.0300

We are finding the area of that slice; to find the area of this, it would be this formula:0309

the central angle (which is this angle right here, this central angle) over 360...0318

now, why is it over 360?--because going all the way around a full circle, including that, is 360; so it is like the part over the whole,0326

the central angle over the whole thing, which is 360...times the area of the circle.0337

Now, another way (an easier way, I think) would be (to figure out how to find the area of this):0346

instead of looking at this formula, I like to use proportions.0355

So, what we can do to find the area of this pizza slice right here: remember: a proportion is a ratio equaling another ratio;0359

so, we are going to look at the probability (probabilities are ratios, something to something, which is part to whole)0373

of the measures to the areas, because we are looking both: we are looking at measure, and we are looking at area.0387

So, for the measures, the part over the whole for the angle measures is going to be the central angle, over the whole thing, which is 360.0403

And the probability of the area...isn't that part over whole, also?...so the part will be the area of the sector.0422

That is the area of the sector; so let's just call that A for area of the sector...over the whole thing, which is0429

(the area of this whole circle is going to be) πr2.0440

Again, the ratio (or probability) of the part to the whole is the angle measure to the whole thing.0445

And the ratio of the areas is going to be the area of the sector, over the area of the circle, because this is part to whole.0459

We are going to make them equal to each other; that is our proportion; and you are just basically going to solve.0467

Let's say that this angle measure is 40, and the radius, r, is 6.0475

If this is 40, that is the part; that is 40 degrees, over...what is the whole thing?...360 degrees, is equal to the area of the sector;0497

that is what we are looking for; that is the area of the sector, over the area of the whole thing; that is the circle, so it is πr2.0512

So, that is π(6)2; and if you are going to solve this out, remember how you solve proportions.0524

You do cross-multiplying; so then, the area of the sector, times 360, times a, equals 40 degrees times π(6)2.0531

And when you solve this out, you divide this by 360, because we are solving for the a.0558

Now, if you look at this, this is exactly the same thing as this right here: the central angle,0568

the angle of that right there, divided by the 360, the whole circle, times the area of the circle--that is πr2.0579

It is the same exact thing; if you want to just use this, that is fine--it is the same exact thing.0591

But this way, you just know that you are looking at the part, the angle measure, over the whole, the circle's angle measure.0597

That is equal to the area of the sector (that is the part), over the area of the whole circle.0608

It is part over whole, for angle measures, equals part over whole, for the areas.0612

And that is just a way for you to be able to solve this out without having to memorize this formula.0620

And then, we solve this out; and you can just do that on your calculator; I have a calculator here on my screen.0627

I get that my area is 12.57; and again, that is the area of this sector, the pizza slice; and that is units of area, squared.0642

That is the area of a sector.0663

Let's go ahead and do some more examples: What is the probability that a point is on XY?0668

Again, for probability, we are looking at part to whole; so the desired outcome, the part that we are looking for, is XY;0675

that is going to be my numerator--that is the top part of my ratio--so XY is from 0 to 2; that is 2 units.0685

It is XY; again, we are looking at XY over the whole thing, which is XZ, so that is 2 over...the whole thing, from X to Z, is 10.0698

If I simplify this, this becomes 1/5, because 2 goes into both; it is a factor of both 2 and 10.0715

I can divide this by 2 and divide that by 2, and I get 1/5; that is the probability that a point will land on XY.0723

Find the probability of the spinner landing on orange, this space right here; here is that spinner.0736

This angle measure is 72, and the radius is 4; so, if this is 4, then we know that any segment from the center to the circle is going to be 4.0749

OK, so then, I want to use that proportion: the angle measure, over the measure of the circle, the total angle measure,0767

is equal to the area of the sector, over the whole thing (is going to be the area of the circle; and that just means "circle").0794

This is my proportion: the angle measure...any time I am dealing with the part (since it is always part over whole),0810

it is always going to be about the sector, this piece right here, the orange; and then, any time I am talking about the whole,0821

it is going to be the whole circle...(that is that) is 72 degrees, over the whole thing (is 360), is equal to0828

the area of the sector (and again, that is what I am looking for, so I can just say A for area of the sector),0842

over the area of the circle (that is the whole thing); and that is πr2.0850

My radius is 4, squared; so then, I can go ahead and cross-multiply.0859

360A = 72(π)(42); then, to solve for A, divide the 360; divide this whole thing by 360.0870

And then, from there, you can just use your calculator: 72 times π times the 42...and then divide 360; you get 10.05.0891

And we have inches here for units, so it is inches squared; that is the area of this orange.0919

Now, to find the probability...we found the area of this orange; and be careful, because,0926

if they ask you for the area of this base right here, then that would be our answer; but they are asking for a probability0933

of landing on orange; and any time you are looking at probability, you are always looking at part over whole.0939

And again, since we are talking about area, it is the area of the orange, over the area of the whole thing.0947

I found the area of a sector; now, to find the area of the whole thing, the area of the circle is πr2.0958

And all you have to do is...we know that r is 4, so 16 times π is 50.27 inches squared; that is the area of the circle.0970

And then, the probability is going to just be (I'll write it on this side) 10.05/50.27.0990

You can change this to a decimal, so you can go ahead and divide this; or maybe you can just leave it like that,1007

You can definitely have probability as a decimal; you can just go ahead and take this and divide it by this number; and that would be your answer.1020

This is the probability, part over whole, the area of the orange over the area of the circle.1028

The circle is circumscribed about a square; if a dart is thrown at the circle, what is the probability that it lands in the circle, but outside the square?1040

We want to know what the probability is of landing in the gray area: it said "in the circle, but outside the square"; that is all the gray area.1055

That is the probability: they are not asking for the area of that part; they are asking for the probability of landing on that part.1066

So then, we have to make sure that we are going to do the part over the whole.1073

First, I have to find the area of that gray area, because that is my desired outcome; that is my part.1082

The desired outcome is the area of the gray, over the whole thing, which would be the area of the circle, because that is the whole thing.1087

So then, my part is going to be, again, area of gray over the area of the circle.1097

To find the area of the gray region, we have to first find the area of the circle and subtract the area of the square.1117

The area of the gray is going to be the circle, minus the square.1132

The area of the circle is πr2, minus...the area of this is going to be side squared.1154

We know that the radius is 10, because, from the center of the circle to the point on the circle, it is π(10)2;1168

minus...do we know what the side is?...we actually don't, because this is from the center to the vertex of this square.1180

So, let me make a right triangle: I know that this angle right here (let me just draw the triangle out again--that doesn't look good;1191

this is more accurate)...this is that triangle here: this is 10, and I want to know either this or this.1208

Let's say that we are going to call that x.1219

Now, this is a right angle; we know that this is a 45-degree angle, because it is half the square; in squares, everything is regular.1223

So, to find the other sides of a 45-45-90 degree triangle, since we know that it is a special right triangle, we are going to use that shortcut.1236

If this is n, then this is n, and this is n√2; and in this case, I should label this n, because that is the side opposite the 45, which is n.1250

The side opposite this 45 is n, and then the side opposite the 90 is 10.1268

Here is the shortcut; I am given the 90-degree side, which is this right here, so I am going to make those equal to each other,1273

because this is n, and this is n√2, which is 10; so n√2 = 10.1281

Divide the √2 to both sides: I am going to solve for n; n = 10/√2...what do I do here?1292

Well, this square root is in the denominator, so I have to rationalize it; when I do that, this becomes 10√2/2; simplify this out; this becomes 5√2.1303

So again, what did I do? I took this...because I have this right here, the hypotenuse of this right triangle, I want to find this side right here.1324

I am going to use special right triangles, since this is a 45-45-90 degree triangle: n, n...the side opposite the 90 is n√2.1332

That is the side that I am given, so I am going to make that equal to n√2: n√2 = 10.1343

Solve for n by dividing the √2; let me rationalize the denominator, because I can't have a radical in the denominator.1350

So then, this becomes 10√2, over...√2 times √2 is just 2; simplify that out, and I get 5√2.1360

That means that n is 5√2; this side is 5√2; this side is 5√2.1370

Well, if this is 5√2, then what is this whole thing? We labeled that as s.1382

So, if this is 5√2, then this is 5√2; so you basically have to just multiply it by 2, because this is half of this whole side.1389

My side is 5√2 times 2, which is 10√2.1400

So, to find the area of the square, I am going to do 10√2 times 10√2, base times height (or side squared): 10√2, squared.1412

And then, I am going to use my calculator: this is 3.14 times 100, which is 314, minus 10√2 times 10√2;1427

that is 100 times 2; that is 200 (if you want, you can just double-check on your calculator).1445

This right here is (I'll just show you really quickly) 10√2 times 10√2.1452

10 times 10 is 100; √2 times √2 is times 2, so it is 100 times 2, which is 200.1464

Then, this is going to be 114; so the area of the gray is 114, because I took the area of the circle,1478

which was πr2, 314, and then I found the area of the square, which is 200, 10√2 times 10√2.1491

And then, I got 114; now, that is just the area of the gray; we are looking for the probability that it lands in the gray.1510

That is the area of the gray, over the area of the whole thing, which is the circle.1523

We take 114 over the area of the circle (where is the area of the circle?), which is 314; and that is the probability.1528

Now, we know that both of these numbers are even, so I can simplify it.1546

So then, if you were to cut this in half, this is going to be 57; if you cut this in half, this is going to be 157.1552

And that would be the probability, 57/157.1564

So again, the probability is going to be the area of the gray over the area of the whole thing, which is the circle.1573

The fourth example: we have a hexagon, and I am just going to go ahead and write that this is a regular hexagon with side length of 4 centimeters.1586

It is inscribed in a circle; what is the probability of a random point being in the hexagon?1599

"Inscribed": now, I don't have a diagram to show you, so I am going to have to draw it out.1606

"Inscribed" means that it is inside, so the hexagon is inside the circle; but all of the vertices of the hexagon are going to be on the circle.1611

They have to be intersecting; so let me first draw a circle, and the regular hexagon ("hexagon" means 6 sides).1627

I am going to try to draw this as regular as I possibly can, something like that, so it will look like it is inscribed...something like that.1641

What else do we have? Side length is 4 centimeters; what is the probability of a random point being in the hexagon?1666

OK, so then, again, we are giving a probability, so it is part over whole.1679

What is the part? The part will be the hexagon; inside the hexagon is the desired outcome--that is what we are looking for.1686

So, it is going to be the area of the hexagon, over...the whole thing is going to be the area of the circle.1693

Let's see, now: to find the area of this hexagon...remember: to find the area of a regular polygon,1711

if we were to take this hexagon and then break this up into triangles, we have 1, 2, 3, 4, 5, 6 triangles.1727

Each triangle is going to be 1/2 base times height; and then, we have 6 of them...times 6.1743

Now, if we take the base (the base is right here), and we multiply this base with this 6, isn't that the same thing as the perimeter?1756

The base, with the 6, is going to be the perimeter; the height, this right here, we call the apothem.1766

I am going to draw arrows to show that the base and the 6 together became the perimeter, and the height became the apothem.1785

1/2 just stayed as 1/2; this is the formula for the area of a regular polygon: it is 1/2 times the perimeter of the polygon, times the apothem.1792

The apothem is from the center, the segment going, not to a vertex, but to the center of the side; so it is perpendicular.1808

Now, we don't know what the apothem is, so I am going to have to look for it.1824

Now, remember: you always want to use right triangles, if you possibly can; we can, because the apothem is perpendicular to the side.1833

So, if I just maybe draw it bigger to show: this is the apothem.1843

If the whole side measures 4 (see how this is 4), then this half is going to be 2.1853

And then, I want to look for this angle measure, because I don't have this side.1863

If I have this side, then I can use the Pythagorean theorem, because then I have a2 + b2 = c2.1868

But I don't, so instead, I want to see: this is a circle; the whole thing, all the way around, is 360 degrees.1876

If I break this up into parts, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.1886

I basically just look to see how much this triangle is from the whole 360; if the whole thing measures 360,1901

remember how we said that this is actually one of the 6 triangles; so 360 divided by 6 is 60.1913

Then, since this whole thing measures 60 degrees, what is this half right here?1927

This half is 30 degrees, so this is 30 and this is 60; and again, all I did was just...1932

I know that the whole thing, the full circle, measures 360; since I know that this right here is 60,1942

because I have 6 triangles, so it is like 6 triangles sharing 360 degrees;1950

then this angle right here is 60 degrees, which means that this half right here is 30.1956

A 30-60-90 triangle is a special right triangle; if this is n, the side opposite the 60 is n√3; the side opposite the 90 is 2n.1964

This is the special right triangle; what do I have--what am I given?1981

The side opposite the 30 is 2; so I am going to make those two equal to each other: n = 2.1988

That means that the side opposite the 60 is going to be 2√3; the side opposite the 90 is going to be 2 times n, which is 4.1999

That means that a, which is the side opposite my 60, is 2√3; isn't that my apothem, my a?2012

So, I know that that took a while; but it is just going over the area of a regular polygon.2020

It is 1/2 perimeter (which is 6 times 4, because there are 6 sides, and each side is 4...so perimeter is 24)...my apothem is 2√3.2033

And then, I can just cut this in half; so 24/2 is 12, times 2 is 24, √3.2053

To make that into a decimal, 24 times √3...we get 41.57 units squared...oh, we have centimeters, so this is centimeters squared.2065

And again, this is the area of the hexagon; so this is the hexagon, and then I want to find the area of the circle.2094

Now, if it were just the area of the hexagon that we were looking for, then this would be the answer.2110

But again, we are looking at probability: what is the probability of a random point being in the hexagon?2113

It is the area of the hexagon, over the area of the full circle.2124

So then, I look at it: here is the hexagon; here is the circle; the area is πr2.2131

π...do I know r?...r would be this length right here; this is the radius, because this is the center of the circle; this is a point on the circle.2144

That side is opposite the 90, and that is 4; so this is 4 squared; this is 16π; 16 times π is 50.27 centimeters squared.2168

So then, hexagon over circle is 41.57, over 50.27; so let me just do that on a calculator: 41.57/50.27...and I get (so the probability is) 0.83.2199

Now, one thing to mention here: when you have a decimal, when you change your probability fraction into a decimal,2239

you have to make sure that it is less than 1, because, if you are looking at a part over the whole,2247

it is going to be a proper fraction; the part is going to be smaller that the whole.2257

If the whole is everything--it is the whole thing--well, then, the part can only be a fraction of it.2262

So, the only time you can get anything greater than this...2269

I'm sorry: the biggest number you can get for probability, when you change it to a decimal, is 1, because,2276

when you look at the fraction, it can just be the whole thing over the whole thing.2284

And when you have whole over whole, well, that is just going to equal 1, because it is the same number over itself.2289

So, make sure that your probability decimal is not greater than 1, unless you are talking about the whole thing.2296

Then, it is going to be 100%, all of it, which is 1; but otherwise, if the part is smaller than the whole, then your decimal has to be less than 1.2304

That is it for this lesson; thank you for watching Educator.com.2318