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### Course: AP®︎/College Statistics>Unit 7

Lesson 4: Independent versus dependent events and the multiplication rule

# 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" ?
• For the question at , it won't be reasonable to earn 30 cents by losing 35. Economically speaking, it is a loss. Now as per the question in , despite having better odds as per the information below
``P(green at first pick)=3/5``

We don't change the sample space size after replacement. All outcomes remain the same.
``P(green at second pick)=3/5P(green both times with replacement)=3*3/5*5=9/25=0.36``

The price is \$0.35, and the prize is \$0.36. This means we only gain a cent.

You can give it a shot if you don't care about the gain amount
But in case 2, that is, 1 cent gain is better than 5 cent loss, and you are more likely to play if you could replace marbles.

But if I were to choose, both options would not appeal to me.
• This option is appealing if you have many chances at playing the game. If you play the game a thousand times, on average, you will make 10 dollars. If you play it a million times, you would make 10 thousand.
• I did the calculations on his question, "is it worth it to play the game?" before watching the whole video, and I came up with a different answer than him, and I am confused how my answer is wrong.

I also got a 3/10 chance of winning the game (30%), so i figured that means that for every 10 games you play you will on average win 3 and lose 7. Thus you will win \$3 and lose \$2.45, an overall win of 55cents per 10 games.

Can anyone explain why my answer is wrong?
• You pay the \$0.35 whether you win or lose. So you don't win \$3 on average, you win \$3·(1-0.35)=\$1.95.

Since you expect to win \$1.95 and lose \$2.45, you expect to lose \$0.50 every 10 games, or \$0.05 per game, as Sal calculated.
• 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?
``3/5 * 3/5 = 9/259/25 * 1 = \$0.36\$0.36 - \$0.35 = \$0.01 = ¢1``