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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition)>Unit 6

Lesson 5: Dependent events

# Dependent probability introduction

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.
• 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.
• What does the upside down U symbol at - mean?
• 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.
• At , why does Sal come to a conclusion that both the events need to be multiplied ? What is the explanation to multiply both the events ?
• 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.
• 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.
• 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)
• Statistically, is removing two green marbles simultaneously identical to removing one green marble and then removing another green marble?
• 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.
• at , why did Sal write the "0.30*\$1=0.30"??
• why did he multiply "the 30% chance of winnig" with "the 1\$ prize" ?
• I and willing to play the game if I can replace the marbles and all else stays the same.
``3/5 * 3/5 = 9/259/25 * 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.
• I know that his question is already asked, but it did not receive a good answer (at least for me). The question is this:
What is the fundamental reason that we must multiply those two probabilities? I know that we multiply because the second probability is only true when the first one is, but what I am asking is that how do we 'prove' that we must multiply, and not, say, add the two probabilities?
• Because we are looking at an intersection. When we want two events to both happen, then we multiply probabilities.

Imagine a Venn Diagram. We draw a circle for the first marble draw. Inside of it are the 60% of cases (or people, if we want to imagine 100 people playing the game) who draw green. Now, we ONLY care about that 60%. And of the 60%, 50% will then draw a green on their second draw. So we're looking at what was 60% of the players, and then taking 50% of that subset.
• 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.