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## Precalculus

### Course: Precalculus > Unit 8

Lesson 2: Multiplication rule for probabilities- Compound probability of independent events
- Independent events example: test taking
- General multiplication rule example: independent events
- Dependent probability introduction
- General multiplication rule example: dependent events
- Probability with general multiplication rule
- Interpreting general multiplication rule
- Interpret probabilities of compound events

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# Dependent probability introduction

AP.STATS:

VAR‑4 (EU)

, VAR‑4.D (LO)

, VAR‑4.D.2 (EK)

CCSS.Math: Let's get you started with a great explanation of dependent probability using a scenario involving a casino game. Created by Sal Khan.

## Want to join the conversation?

- Hmm. I'm not very clear on the logic he used in determining if you would want to play the game. I said it would make sense. If you play the game 100 times for example, you win 30 times. so you get $30. This also means you lose 70 times to you play 70 x 0.35 = $24.5. In the end you seem to be gaining money.(17 votes)
- Well, you lose 0.35 EVERY time, because it costs this much to play. So when you win, you only really win 0.65, not the full $1. (You had to pay 0.35 for the chance to get that $1) Hence, 30 out of 100 times (to use your example), you win 0.65, and 30x0.65=19.5. The other 70 times, you lose 0.35, and 70x-0.35= -24.5, so over those 100 plays, you are losing $5.(101 votes)

- What does the upside down U symbol at2:32-2:35mean?(29 votes)
- In this example, Sal is asking "what is the probability of both the first AND second being green". The upside down U symbol in this case stands for the AND.

The symbol typically stands for "intersection" and is used in set theory to refer to common numbers or letters in sets. For example, the "intersection" of {1, 3, 5, 7} and {4, 5, 6, 7} is {5, 7} because those are the numbers that you can find in both sets.(49 votes)

- At04:00, why does Sal come to a conclusion that both the events need to be multiplied ? What is the explanation to multiply both the events ?(31 votes)
- To get the probability of both events being true. If you are asking why you multiply, it is because, for example, if there is a 1/2 probability of the 1st being green and a 1/3 probability of the 2nd being green, the probability of the 2nd being green and the 1st is green is 1/2 of the time the 2nd is green (1/3) since an of means multiplication, the probability of both being green is 1/2 x 1/3.(11 votes)

- I'm a high school senior in India and we are studying probability at school. My question concerns conditional probability. We have defined probability to be the formula- P(A/B)= P(A int B)/P(B). However, when solving many problems we don't use the definition directly and instead use the vague notion of assuming the occurrence of the "given" event. Even though this makes some intuitive sense, it is rather vague and not at all rigorous. So I would be grateful is someone could provide me with a better explanation or working rule, and connect that to the aforementioned formula.(15 votes)
- Ill try explain this using an example:

Given a box of 50 marbles

20 marbles are blue and 30 marbles are white.

There are 5 smooth, and 15 rough blue marbles.

While there are 12 smooth and 18 rough white marbles.

Let event A: "Draw blue marble"

Let event B: "Draw rough marble"

What is the probability of drawing a blue marble?

1. P(A) = 20/50

What is the probability of drawing a rough marble?

2. P(B) = (15+18)/50 = 33/50

What is the probability of drawing a rough and blue marble?

3. P(A int B) = 15/50

Given as you take a marble, you feel a rough marble, what is the probability that it is a blue marble?

4. P(A | B)

= P(A int B) / P(B)

= [15/50] / [33/50]

Given that the chosen marble is blue, what is the probability that the marble is rough?

5. P(B | A) = P(B int A) / P(B)

using commutative property P(B int A) = P(A int B)

= P(A int B) / P(A)

= [15/50] / [20/50]

As you can see, the sample space has considerable changed once we have a condition. That is the main point.

I hope this simple example helps for understanding this on a small scale.

(P.S. anyone is welcome to correct me as I am only human and prone to make mistakes)(18 votes)

- Statistically, is removing two green marbles
**simultaneously**identical to removing one green marble and then removing another green marble?(12 votes)- It is still a 30% chance (or 3/10.) think of it this way. We have 5 different marbles: g1, g2, g3, r1, and r2, 'g' standing for green, and 'r' standing for red. There are ten different ways to pull two of these out of the bag simultaneously.

3 different pairs that are only green: g1 and g2, g1 and g3, g2 and g3. To visualize this imagine the three marbles arranged in a triangle formation. (kind like they are in the video) Then attempt to draw lines between them. You will find you can only draw three lines between them (note: in all pairings the order does not matter (e.g. g1 and g2 is the same as g2 and g1). This is because we only care about the quantity of red/green marbles (e.g. in the previous example, there are still two green marbles in each.))

Next, there are 6 different pairs that include a red marble. g1, 2, and 3 with r1 and g1, 2, and 3 with r2. Picture this as a grid with two columns (the two red marbles) and 3 rows (the three green marbles) filling all of the cells will give you six different parings.

Lastly, there is one situation where only reds are pulled out. r1 and r2.

Thus, if there are 10 possible outcomes, and 3 of those fulfill our conditions, then placing 3 over ten we get the probability is equal to 3/10, or 30%. This is because in the end, the situations are the same. If you viewed the above listed pairings of marbles drawn out simultaneously as two marbles drawn out one after the other then you would get the same probability.(8 votes)

- at6:30, why did Sal write the "0.30*$1=0.30"??(6 votes)
- why did he multiply "the 30% chance of winnig" with "the 1$ prize" ?(3 votes)

- Hi I have a question about independent probability i assume, but it is not specifically related to this video. If there is a better place to ask such a question, please do point me towards it.

Here is the question I am trying to solve:

What is the probability that, in six throws of a die, there will be exactly one each of “1” “2” “3” “4” “5” and “6”?

.00187220

.01176210

.01543210

.01432110

My thought process is that these are independent events and in each roll the probability of getting a number like 1 or 2 is 1/6. So, for six rolls, the probability should be just (1/6)^6, but that doesn't seem to be right. Can you please help explain how I can go about solving this? Thanks.(2 votes)- What you have would be the correct way to find the probability of rolling the same number six times in a row. But in this problem, they are asking for the odds of rolling no repeats in six rolls, which is different.

In the first roll, there is a 100% chance you will get one a number 1 through 6, and will not repeat. On the second, , odds are 5/6 that you will not repeat....On the sixth roll, you are looking at 1/6 odds that you won't repeat. Try 1 ·5/6 ·4/6 ·3/6 ·2/6 ·1/6, and one of those choices should match.(7 votes)

- I and willing to play the game if I can replace the marbles and all else stays the same.
`3/5 * 3/5 = 9/25`

9/25 * 1 = $0.36

$0.36 - $0.35 = $0.01 = ¢1

I would want to play this altered version of the game, because if I play it many times, my average gaine per game would be ¢1.(4 votes) - Since both marbles need to be green, I assume that that is the reason that Sal ignored the what would happen if the first marble wasn't green.(3 votes)
- Correct. He only wants to know the probability of winning, and the only way that will happen is if the first marble is green.(3 votes)

- ?? Such a real-life example:

There's a raffle, 1 ticket = 15usd, 100 participants, 3 prize places. 1st = 100 usd, 2nd = 75usd, 3rd = 50usd. Is it economically rational to participate?

!! I solved it this way:

1/100=1% 1/99=1.01% 1/98=1.02%

1% * 100usd = 1usd (Expected Value_E.V.)

1.01% * 75usd = 0.758usd E.V.

1.02% * 50usd = 0.51usd E.V.

----------------

Total E.V. = 1 + 0.758 + 0.51 = 2.27usd

Since 2.27<15, CONCLUSION: it's not rational.

Is it correct?(3 votes)- You reached the correct conclusion, but the math is a bit off.

You actually have an equal chance of winning any of the 3 prizes (1/100) so the EV of a ticket is simply ($100+$75+$50)/100 = $2.25(2 votes)

## Video transcript

Let's imagine
ourselves in some type of a strange casino
with very strange games. And you walk up to a
table, and on that table there is an empty bag. And the guy who
runs the table says, look, I've got
some marbles here, 3 green marbles,
2 orange marbles. And I'm going to
stick them in the bag. And he literally sticks
them into the empty bag to show you that it's
truly 3 green marbles and 2 orange marbles. And he says, the game
that I want you to play, or if you choose
to play, is you're going to look away-- stick
your hand in this bag, the bag is not transparent--
feel around the marbles. All the marbles feel
exactly the same. And if you're able to
pick 2 green marbles, if you're able to take 1 marble
out of the bag, it's green. You put it down on the table. Then put your hand back in the
bag, and take another marble. And if that one's also
green, then you're going to win the prize. You're going to win $1
if you get 2 greens. If you get 2 greens,
you're going to win $1. You say, well, this sounds
like an interesting game. How much does it cost to play? And the guy tells you,
it is $0.35 to play, so obviously a fairly
low stakes casino. So my question to you is, would
you want to play this game? And don't put the
fun factor into it. Just economically, does
it make sense for you to actually play this game? Well, let's think through the
probabilities a little bit. So, first of all,
what's the probability that the first marble
you pick is green? Actually, let me just
write, first green, probability first green. Well, the total possible
outcomes-- there's 5 marbles here,
all equally likely. So there's 5 possible outcomes. 3 of them satisfy your event
that the first is green. So there's a 3/5 probability
that the first is green. So you have a 3/5 chance,
3/5 probability I should say, that after that first
pick you're kind of still in the game. Now, what we really care
about is your probability of winning the game. You want the first to be
green, and the second green. Well, let's think about
this a little bit. What is the probability that
the first is green-- first, I'll just write g for green--
and the second is green? Now, you might be
tempted to say, oh, well, maybe the second being green
is the same probability. It's 3/5. I can just multiply
3/5 times 3/5, and I'll get 9/25, seems like
a pretty straightforward thing. But the realization
here is what you do with that first green marble. You don't take that first
green marble out, look at it, and put it back in the bag. So when you take
that second pick, the number of green
marbles that are in the bag depends on what you
got on the first pick. Remember, we take
the marble out. If it's a green
marble, whatever marble it is, at whatever
after the first pick, we leave it on the table. We are not replacing it. So there's not any
replacement here. So these are not
independent events. Let me make this
clear, not independent. Or in particular, the second
pick is dependent on the first. Dependent on the first pick. If the first pick
is green, then you don't have 3 green
marbles in a bag of 5. If the first pick
is green, you now have 2 green marbles
in a bag of 4. So the way that we
would refer to this, the probability of both
of these happening, yes, it's definitely
equal to the probability of the first green times-- now,
this is kind of the new idea-- the probability of the second
green given-- this little line right over here, just a
straight up vertical line, just means given-- given that
the first was green. Now, what is the probability
that the second marble is green given that the first
marble was green? Well, we drew the
scenario right over here. If the first marble
is green, there are 4 possible outcomes,
not 5 anymore, and 2 of them satisfy your criteria. So 2 of them satisfy
your criteria. So the probability of the
first marble green being green and the second
marble being green, is going to be the probability
that your first is green, so it's going to be 3/5,
times the probability that the second is green,
given that the first was green. Now you have 1 less
marble in the bag, and we're assuming that
the first pick was green, so you'll only have
2 green marbles left. And so what does this give
us for our total probability? Let's see, 3/5 times 2/4. Well, 2/4 is the
same thing as 1/2. This is going to be
equal to 3/5 times 1/2, which is equal to 3/10. Or we could write that as 0.30,
or we could say there's a 30% chance of picking
2 green marbles, when we are not replacing. So given that, let me ask
you the question again. Would you want to
play this game? Well, if you played this
game many, many, many, many, many times, on average you have
a 30% chance of winning $1. And we haven't covered this yet,
but so your expected value is really going to
be 30% times $1-- this gives you a little
bit of a preview-- which is going to be $0.30
30% chance of winning $1. You would expect on average,
if you were to play this many, many, many times, that playing
the game is going to give you $0.30. Now, would you want to
give someone $0.35 to get, on average, $0.30? No, you would not want
to play this game. Now, one thing I will
let you think about is, would you want to play
this game if you could replace the green marble,
the first pick. After the first pick, if you
could replace the green marble, would you want to play
the game in that scenario?