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Lecture Comments (4)

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!

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

Determine if the two events are independent or dependent events:
Rolling a number cube four times
Independent
Determine if the two events are independent or dependent events:
Drawing a marble from a bag of marbles, and without replacing it, drawing another marble
Dependent
Susanna rolls a number cube twice. Find the probability of the event.
P(3,4)
  • P(3) = [1/6]
  • P(4) = [1/6]
  • P(3) × P(4)
[1/6] ×[1/6] = [1/36]
Susanna rolls a number cube twice. Find the probability of the event.
P(5, not 1)
  • P(5) = [1/6]
  • P(not 1) = [5/6]
  • P(5) × P(not 1)
[1/6] ×[5/6] = [5/36]
Susanna rolls a number cube twice. Find the probability of the event. P(not 2, not 1)
  • P(not 2) = [5/6]
  • P(not 1) = [5/6]
  • P(not 2) × P(not 1)
[5/6] ×[5/6] = [25/36]
There are 6 red, 5 blue, 2 green marbles in a bag. Find the probability for each when the first marble is not replaced back in the bag.
P(green, red)
  • P(green) = [2/13]
  • P(red) = [(6 ÷6)/(12 ÷6)] = [1/2]
  • P(green) × P(red)
  • [2/13] ×[1/2]
[2/26] or [1/13]
There are 7 red, 9 blue, 6 green marbles in a bag. Find the probability for each when the first marble is not replaced back in the bag.
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]
[3/33] or [1/11]
There are 2 red, 4 blue, 3 green marbles in a bag. Find the probability for each when the first marble is not replaced back in the bag.
P(green, red)
  • P(green) = [3/9] = [1/3]
  • P(red) = [2/8] = [1/4]
  • P(green) × P(red)
  • [1/3] ×[1/4]
[1/12]
There are 2 blue, 1 yellow, 6 black marbles in a bag. Find the probability for each when the first marble is replaced back in the bag. P(black, blue)
  • P(black) = [6/9] = [2/3]
  • P(blue) = [2/9]
  • P(black) × P(blue)
  • [2/3] ×[2/9]
[4/27]
There are 5 blue, 6 yellow, 9 black marbles in a bag. Find the probability for each when the first marble is replaced back in the bag. P(black, blue)
  • P(black) = [9/20]
  • P(blue) = [5/20] = [1/4]
  • P(black) × P(blue)
  • [9/20] ×[1/4]
[9/80]

*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

Transcription: Independent and Dependent Events

Welcome back to Educator.com.0000

For the next lesson, we are going to go over probability of compound events0002

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 for0058

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 multiply0289

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 for0402

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 affected0553

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 A0581

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 be0666

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 be0681

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 multiply1185

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 yellow1456

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 happening1505

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 means1537

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 afterwards1822

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 blue1842

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 151922

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 then1945

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