For more information, please see full course syllabus of Basic Math

For more information, please see full course syllabus of Basic Math

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Independent and Dependent Events

#### Related Links

- Independent events: When the outcome of the second event does not depend on the outcome of the first event
- If A and B are independent events, then P(A, B) = P(A) × P(B)
- Dependent Events: When the outcome of the second event depends on the outcome of the first event
- If A and B are dependent events, then P(A, B) = P(A) × P(B after A)

### Independent and Dependent Events

Rolling a number cube four times

Drawing a marble from a bag of marbles, and without replacing it, drawing another marble

P(3,4)

- P(3) = [1/6]
- P(4) = [1/6]
- P(3) × P(4)

P(5, not 1)

- P(5) = [1/6]
- P(not 1) = [5/6]
- P(5) × P(not 1)

- P(not 2) = [5/6]
- P(not 1) = [5/6]
- P(not 2) × P(not 1)

P(green, red)

- P(green) = [2/13]
- P(red) = [(6 ÷6)/(12 ÷6)] = [1/2]
- P(green) × P(red)
- [2/13] ×[1/2]

P(green, red)

- P(green) = [(6 ÷2)/(22 ÷2)] = [3/11]
- P(red) = [(7 ÷7)/(21 ÷7)] = [1/3]
- P(green) × P(red)
- [3/11] ×[1/3]

P(green, red)

- P(green) = [3/9] = [1/3]
- P(red) = [2/8] = [1/4]
- P(green) × P(red)
- [1/3] ×[1/4]

- P(black) = [6/9] = [2/3]
- P(blue) = [2/9]
- P(black) × P(blue)
- [2/3] ×[2/9]

- P(black) = [9/20]
- P(blue) = [5/20] = [1/4]
- P(black) × P(blue)
- [9/20] ×[1/4]

*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.

Answer

### Independent and Dependent Events

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
- Independent Events 0:11
- Definition
- Example 1: Independent Event
- Example 2: Two Independent Events
- Dependent Events 9:09
- Definition
- Example: Dependent Events
- Extra Example 1: Determine If the Two Events are Independent or Dependent Events 13:38
- Extra Example 2: Find the Probability of Each Pair of Events 18:11
- Extra Example 3: Use the Spinner to Find Each Probability 21:42
- Extra Example 4: Find the Probability of Each Pair of Events 25:49

### Basic Math Online Course

### Transcription: Independent and Dependent Events

*Welcome back to Educator.com.*0000

*For the next lesson, we are going to go over probability of compound events*0002

*and those events being independent and dependent.*0007

*Before we go over these events, let's first review over probability.*0012

*Probability is talking about the chances of something happening.*0017

*What is the probability of picking a card from a deck?*0024

*Or what is the probability of rolling a 2 if you roll a die?*0030

*Probability talks about your chances of something occurring.*0035

*To find probability, we are looking at a ratio; a ratio is like a fraction.*0040

*It is comparing the top number with the bottom number.*0055

*Probability is talking about your desired outcome or the outcome that you are looking for*0058

*over the total possible number of outcomes or just total for short.*0072

*It is desired outcome over the total--is the probability.*0081

*Probability, once you have this, you can leave it as a fraction.*0085

*You can change it to a decimal; it is just fraction to decimal.*0089

*You just divide the top number with the bottom number.*0094

*Or you can change it to a percent.*0098

*Usually probability is left as a fraction, desired outcome over the total.*0101

*If I have a die, a number cube... we are going to talk about number cubes in our examples later.*0112

*A number cube has different numbers of dots on each side.*0122

*There is 6 sides; each side has a different number, 1 through 6.*0131

*If I said what is the probability, what are the chances of rolling a 1?*0135

*My desired outcome then, I am going to say my probability of rolling a 1.*0142

*That is how we write it out, probability of rolling a 1.*0148

*That is my desired outcome; how many sides has a 1?--only 1 side.*0154

*My desired outcome, there is only 1; over... how many total sides are there?*0162

*How many total possible outcomes are there?*0170

*There is 6 sides; my total is 6.*0173

*The probability of rolling a 1 is 1/6.*0176

*Same thing if I said probability of rolling a 3.*0181

*Be careful, this number is not going to be the number on top.*0186

*How many sides have a 3?--only 1 side.*0190

*My desired outcome would be just 1 because there is only 1 possible side that is going to be a 3.*0196

*It is 1 over... how many total sides are there?--6.*0205

*The chances of rolling a 1 is the same as the chance of rolling a 3.*0210

*That is probability.*0214

*When we talk about two events, each of this, this is one event.*0218

*You are rolling a 1; this is another event, rolling a 3.*0224

*Each of those are events.*0228

*We are talking about when we have two events or two or more events.*0231

*When we have two events, when there is two things going on,*0236

*those two events can be either independent events or dependent events.*0240

*That is what we are talking about now.*0246

*Independent events is when the outcome of the second event does not depend on the outcome of the first event.*0248

*Think about what the word independent means.*0256

*It doesn't depend on it; it is not affected by anyone else, anything else.*0259

*The first event happens; the second event happens; they don't affect each other.*0266

*When you have two events that are independent, then we write each of those events as A and B.*0273

*It is probability of A, that is the first event.*0280

*The probability of B, that is the second event.*0285

*When we have two events that are independent, all we have to do is multiply*0289

*the probability of that first event with the probability of the second event.*0294

*Let's say I have a bag of marbles.*0301

*In this bag, I have 1, 2, 3 red marbles.*0309

*Let's say I have 2, 3, 4 blue marbles.*0316

*And let's say I have 2 green; I don't have green.*0323

*So I am going to just use black for green; I will put G for green.*0328

*3 red marbles, 4 blue, and 2 green marbles; I have a bag of marbles.*0333

*Just talking about one event, let's just say what is the probability of picking a red marble?*0339

*That is one event because I am going to pick up one marble.*0346

*The probability of picking a red, that red is my desired outcome.*0350

*That is what I am asking for.*0358

*How many reds do I have?--I have 3 reds.*0361

*That number, the desired outcome, is going to go on top.*0366

*3 over... how many total number of marbles do I have?*0369

*I have 1, 2, 3, 4, 5, 6, 7, 8, 9.*0375

*If you have a fraction, you always have to simplify.*0381

*I can simplify this by 3; divide each by 3.*0384

*It is going to become 1/3; the probability of picking a red marble is 1/3.*0390

*That is only when you have one event.*0399

*Talking about compound events, two events, if I ask for*0402

*the probability of picking a red and afterwards picking a blue...*0408

*Probability of picking a red, we already found that; that is 1/3.*0420

*It is going to be probability of picking red times the probability of picking the blue.*0425

*The only way both of these events, picking the red and then picking another marble the blue,*0440

*the only way these two events are going to be independent is if after we pick the red marble,*0447

*after this first event, after you pick the first marble, you have to place it back into the bag.*0455

*You are going to pick one out; put it back in.*0461

*Then pick the second one.*0464

*It will be independent because then picking this red or picking this one, it won't affect this one.*0466

*Probability of picking a red marble, we know that is 3/9 or 1/3.*0474

*Times probability of picking a blue.*0484

*Blue is my desired outcome; I have 4.*0489

*Over a total number of 4, 5, 6, 7, 8, 9.*0494

*Again after you pick the first marble, we put it back in the bag.*0503

*Now it is just original number of marbles,*0510

*the same number of marbles when we picked the blue one, when we picked the second one.*0515

*This is 1 times 4 is 4; over... 3 times 9 is 27.*0519

*This can't be simplified; this is our answer.*0527

*The probability of picking a red and then after replacing it, picking a blue, would be 4/27.*0533

*This is independent events.*0545

*When we have two events and the second outcome is affected*0553

*by the first outcome, then we have dependent events.*0560

*The second event depends on the first event.*0567

*Finding the probability of two dependent events is a little bit different.*0573

*Same thing here; when we have probability of the first event A*0581

*and then the probability of the second event B, we are still going to multiply them.*0586

*It will be the probability of A times the probability of B after A because remember this second event is affected.*0594

*It depends on the first event A.*0605

*Back to the bag of marbles; again 3 red, 4 blue, and 2 green marbles.*0612

*Same bag of marbles; but now the way it becomes dependent events.*0631

*I want to find the probability of picking a red and then my second event will be picking a blue.*0637

*But the difference is here after we pick the first marble, after we find the probability of picking a red,*0648

*we are not going to put the marble back in the bag.*0658

*We are going to take it out; we are going to leave it out.*0661

*Then the second event, the probability of picking a blue, is going to be*0666

*slightly different because the total number of marbles is different.*0670

*There is less marbles; that is why these would be dependent events.*0675

*Because the probability of picking a blue is not going to be*0681

*the same as if we were to place the marble back in.*0685

*This will be probability of red times probability of blue.*0692

*Again we are not going to replace it back in.*0699

*The probability of picking a red, how many reds do I have?*0703

*My desired outcome is 3; desired outcome goes on top.*0709

*3 over... total number of marbles is 9; you can simplify this to become 1/3.*0715

*Let's say that... let me just change this to 1/3.*0729

*Because this red is no longer there, we took it out.*0738

*That is the first event.*0743

*For the second event, since this marble was not replaced back in, it is left out.*0745

*This is going to be different.*0755

*Probability of picking a blue, my desired outcome is number of blue.*0756

*How many blues do I have?--4.*0760

*My total number of marbles is going to be different.*0765

*It is going to be 1, 2, 3, 4, 5, 6, 7, 8.*0768

*It is going to be 1 less than the total here.*0772

*This was originally 9 before we simplified.*0775

*Now it is going to be 8; there is 1 less marble in the bag.*0779

*Now we multiply these numbers.*0784

*It is going to be... 1 times 4 is 4; over... 3 times 8 is 24.*0787

*This can be simplified; 4 goes into both numbers.*0795

*Divide each number by 4; this is 1; this is 6.*0799

*The probability of picking a red and then picking a second marble blue without replacing marbles is going to be 1/6.*0806

*Let's go over some examples.*0824

*Determine if the two events are independent or dependent events.*0825

*The first one, rolling a number cube twice.*0831

*Remember for independent or dependent events, we have to have two events; at least two.*0835

*Here rolling a number cube twice.*0843

*The first event would be the first time you roll the number cube.*0846

*The second event is going to be the second time you roll the number cube.*0850

*Does the second event depend on the outcome of the first event?*0857

*Meaning if we roll a number cube, if we roll a die,*0865

*we get either 1 through 6, a number from 1 to 6.*0870

*If you roll it again the second time, does it change or is it affected?*0876

*If I roll a 2 the first time, does that mean I can't roll a 2 the second time?*0883

*The first time we roll a number cube, all the numbers...*0892

*let's say I want to find the probability of picking a 5.*0896

*How many 5s are there?--how many sides on the number cube is a 5?*0899

*There is only 1 side; it will be 1/6.*0904

*That would be the probability of my first roll.*0909

*Then for my second roll, what is the probability of picking a 5 or picking any number?*0914

*It is also 1; do the number of sides change?--no, still the same.*0922

*This roll and this roll, my second roll, they don't affect each other.*0930

*They have nothing to do with each other.*0936

*In this case, this would be independent.*0938

*The second one, drawing a card from a deck of cards and without replacing it, drawing another card.*0946

*There are 52 cards in a deck.*0954

*If I pull a number out or take a card, my total number of cards is going to be 52.*0958

*If I don't put it back in, then for my second draw,*0969

*when I draw my second card, my total is going to be different.*0976

*My probability will be different because it is always desired outcome over the total.*0981

*For my second draw, there is less cards in the deck.*0986

*In this case, this would be dependent because the second draw depends on the first draw.*0991

*The outcome of the second is affected by the first; dependent.*1003

*Picking two students in your class to be class representatives.*1015

*Imagine your class; there is let's say 30 students in the class.*1022

*You pick the first student.*1031

*Let's say you are picking the president and vice-president as class representatives.*1036

*If you pick the first student to be your president,*1041

*how many students do you have left to pick from when you pick the vice-president?*1045

*The total number of students, does it change?*1052

*It does change because you already picked one student and that same person can't be both.*1055

*You pick one student to be the president of your class.*1060

*Then for the vice-president, you have one less student to pick from.*1066

*You have 29 students; so this is dependent; this is dependent.*1074

*These two events would be considered dependent events.*1085

*Samantha rolls a number cube twice; find the probability of each pair of events.*1094

*Here rolling twice, two events, this is the first event; this is the second event.*1100

*We want to know the probability of rolling a 2 and then probability of rolling a 5 afterwards.*1108

*Probability of picking a 2; a number cube... let's say 1 here; 2 here; let's say 5 here.*1120

*There is 6 total sides; how many sides have 2?*1148

*Only 1 side does; my desired outcome is the 2.*1154

*But how many 2s are there?--only 1.*1159

*It is 1 out of a total 6.*1161

*What is the probability for my second roll, for my second event, probability of rolling a 5?*1169

*Again there is only 1 side with a 5.*1174

*1 over... still number of sides is the same, 6.*1177

*The probability of rolling both of those, I just have to multiply*1185

*probability of 2 times the probability of 5 occurring.*1189

*It is going to be 1/6 times 1/6.*1195

*1 times 1; 6 times 6 is 36.*1202

*The probability of rolling a 2 and then rolling a 5 afterwards is 1/36.*1207

*This one here, the probability of rolling a number that is not a 3.*1216

*Probability of not 3; that is my desired outcome; not 3.*1225

*How many numbers are not 3?*1236

*We have 6 of them; only 1 is a 3; the rest aren't.*1240

*There is 5 sides that are not 3; that is going to be 5 on the top.*1245

*Over... how many do I have total?--6.*1252

*That is my first roll; that is my first event.*1260

*My second event, my second roll, is probability of rolling a 6.*1262

*Again there is only 1 side with a 6; that is 1/6.*1267

*Probability of the first one times the probability of the second one.*1274

*Probability of that is 5/6 times probability of the second one 1/6.*1281

*5 times 1 is 5; over 36; that can't be simplified; that is my answer.*1288

*Here we have a spinner that we are going to use to find the probability of each.*1304

*The first one, I only have one event, only 1 spin.*1310

*I am looking for the probability of rolling a black; there are no blacks.*1317

*It is red, orange, yellow, green, blue, purple, light purple, and then another orange.*1324

*Probability of rolling a black, my desired outcome is black.*1332

*Do I have any black?--no; 0 on top.*1337

*Over... how many total do I have?--1, 2, 3, 4, 5, 6, 7, 8; over 8.*1343

*This 0/8 is always 0.*1354

*If have a 0 on top, that is going to make the whole thing 0.*1358

*Here there is no probability of picking a black; that is what the 0 means.*1364

*For the second one, we want to know the probability of spinning a red.*1372

*And then if we do a second spin, because there is two events...*1378

*First spin lands on red; second spin lands on green.*1384

*Probability of red; how many sections of red do I have?*1388

*I only have 1; 1 over... total number of sections, 8.*1397

*What about probability of green?*1406

*This would be considered independent events because if I spin the first time, I land on red.*1410

*That is not going to affect what my second spin is going to land on.*1420

*These would be independent; the probability of green is 1 out of 8.*1425

*I multiply them together; 1/8 times 1/8.*1433

*1 times 1 is 1; 8 times 8 is 64.*1440

*The probability of landing on red and then spinning again landing on green is 1/64.*1448

*The next one, the probability of landing on any color that is not yellow*1456

*and then for the second spin, landing on blue.*1464

*Probability of not yellow; how many are not yellow?*1468

*There is 8; there is only 1 yellow; 7 are not yellow; 7/8.*1475

*The probability of blue; there is 1 blue; 1/8.*1484

*We are going to multiply them together; 7/8 times 1/8 is 7/64.*1492

*If you notice these two numbers, the chances of this happening is greater than the chances of this happening*1505

*because here the chances of the spinner landing on a color that is not yellow is actually pretty high.*1519

*7/8, that is pretty high because there is so many spaces that are not yellow.*1529

*If this fraction is greater than this fraction, that means*1537

*the probability of this happening is greater than the chances of that happening.*1541

*For our fourth example, we have a bag of marbles; draw my bag of marbles.*1550

*I have 5 red marbles; 1, 2, 3, 4, 5.*1560

*I have 4 blue; 1, 2, 3, 4.*1566

*I have 6 green; I don't have green color.*1572

*I am going to use black G for green; 1, 2, 3, 4, 5, 6.*1575

*We are going to find the probability for each when the first marble is not replaced back in the bag.*1586

*Here we have two events; two things are happening.*1593

*We are going to pick two marbles.*1596

*After you pick the first marble, we are not going to put it back in the bag.*1601

*It is not going to be replaced back in.*1606

*We pick one; that one stays out of the bag.*1609

*Then we are going to pick our second marble; that is my two events.*1612

*Remember the second event, because after we pick the first marble, we are not going to replace it back in.*1618

*That is going to affect the probability of that second marble.*1626

*Both of these would be considered dependent events because the second one is affected by that first event.*1633

*Let's first talk about this event, the probability of picking a green.*1645

*That is our first pick, green.*1652

*Probability, we look at the desired outcome over the total possible number of outcomes.*1656

*How many green marbles do I have?--I have 1, 2, 3, 4, 5, 6.*1661

*6 is going to be my top number.*1670

*Over... how many marbles do I have total, in all?*1672

*1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.*1676

*That is going to go right there; the probability of picking a green is 6/15.*1683

*It is still a fraction; I still have to simplify it.*1691

*What number goes into both 6 and 15?--they share a factor of 3.*1694

*This is going to be 2/5; the probability of picking a green is 2/5.*1702

*Now I have to pick my second marble; that is going to be red.*1710

*Because that first marble was not replaced back in the bag, this marble, one of the green, is now gone.*1719

*It is no longer in there.*1727

*Probability of picking the red, how many reds to I have?*1732

*I have 5 red marbles; 5 on top.*1735

*Over... how many marbles do I have now?*1739

*After that 1 is gone, I have 14 left.*1743

*It was 15 and then minus the 1 that we have already picked.*1748

*Now it is 5/14; this fraction I can't simplify.*1753

*That is the probability of picking the red.*1758

*Now to find the probability of both happening, I take this one and I take this one; I multiply it together.*1761

*It will be probability of the green times the probability of picking the red.*1769

*This is dependent; it is the red after green; 2/5 times 5/14.*1780

*2 times 5 is 10; 5 times 14; 14 times 5.*1792

*We do this; that is 20; 5 times 1 plus the 2 is 70.*1799

*Fraction, I have to simplify it; what number goes into both top and bottom?*1809

*10; I divide by 10 for each; I get 1/7 as my answer.*1814

*The probability of picking a green and then afterwards*1822

*without replacing it back in, picking a red marble, is 1/7.*1825

*I need to write this for this second problem.*1834

*Now we want to find the probability of picking a blue*1842

*and then afterwards without replacing it back in, pick another blue.*1845

*Again two dependent events.*1851

*Probability of picking the first blue, the first event, what is my desired outcome?*1855

*How many blue marbles do I have here in the bag?*1864

*I have 4 blue over a total of 15 marbles.*1866

*The probability of picking a blue marble is 4/15.*1874

*I can't simplify it; so that is the probability.*1878

*For my second pick, because again it is not being replaced in the bag.*1881

*This one is no longer in the bag; I have 1 less marble.*1891

*For my second pick, I want to look at how many blue marbles I have left.*1896

*I have 3 left; I had 4 but 1 is gone; now I have 3.*1904

*Over... I don't have 15 anymore; I have 14 now.*1913

*Probability of picking the first blue was 4 out of 15*1922

*because I had all my blue, just 4 of them, out of a total of 15 marbles.*1925

*For my second pick, I am also wanting to pick another blue one.*1933

*I only have 3 left because the first one wasn't replaced back in.*1937

*Out of a total of 14 marbles left.*1941

*Now I am going to take the first event and then*1945

*multiply it to the probability of the second event happening.*1951

*It is 4/15 times 3/14.*1956

*4 times 3 is 12; over... 15 times 14; you are going to multiply it.*1962

*This is 20; that is 4 times 1 is 4; plus 2 is 6.*1971

*I leave the space alone; 1 times 5 is 5; 1 times 1 is 1.*1978

*I am going to add them down; 0; this is 11; this is 2; 210.*1983

*I know I can simplify this fraction because this number is an even number and this number is an even number.*1998

*Let's divide each of these by 2; 12 divided by 2 is 6.*2007

*Over... this, if I take the 200 and I divide it by 2, I get 100.*2020

*This divided by 2 is 100; this divided by 2 is 5.*2032

*If I divide the whole thing, it will be 105.*2037

*It looks like 6/105; I also have another factor.*2046

*I can divide this again by... I know 3 goes into that one and 3 also goes into this one.*2051

*6 divided by 3 is... write it down here... 2.*2059

*Over... 105 divided by 3; let me show you that one.*2065

*3 goes into 10 three times; that gives you a 9.*2073

*Subtract it; I get 1 left over; bring down the 5.*2077

*3 goes into 15 five times; that gives you 15.*2082

*We subtract it; I have no remainders; my answer is 35.*2088

*Can I simplify this further?--no, I can't because this is not an even number.*2095

*This is my answer.*2100

*Again finding the probability of two events, you have to find the probability of each event occurring.*2102

*Then you are going to multiply them together, whether it is independent or dependent events.*2111

*That is it for this lesson; thank you for watching Educator.com.*2116

1 answer

Last reply by: Professor Pyo

Sat Jan 4, 2014 1:14 AM

Post by KUMARI Mahendran on January 3, 2014

But if the first choice was blue, and it is not replaced, then there is one less blue ball in the bag.

1 answer

Last reply by: Peter Fraser

Sun Jan 20, 2013 11:14 AM

Post by Anwar Alasmari on January 11, 2013

In Example IV, I think that the P(blue, blue) has to be 4/15 * 4/14, not 3/14. because the second choice I won't know if the frist choice will be blue or not. Please advice!