Understanding the Concept of Independent Events: Venn Diagrams Explained

  • B
  • Thread starter Addez123
  • Start date
In summary: You have it backwards. We were given that A and B were independent, and drew the diagram accordingly.
  • #1
Addez123
199
21
If A and B are independent then the probability of both happening at once should be 0.
If we drew a ven diagram it'd be just two circles who don't intersect.

I'm guessing I got wrong definition of independent, can someone explain please?
 
Physics news on Phys.org
  • #2
Addez123 said:
If A and B are independent then the probability of both happening at once should be 0.
If we drew a ven diagram it'd be just two circles who don't intersect.

I'm guessing I got wrong definition of independent, can someone explain please?

By definition, two events ##A## and ##B## are independent if the occurrence of one does not affect the probability of occurrence of the other. The concept of independence extends to more than two events, taking pairwise independence for every pair of events inside a finite set of events.

In order to understand why ##P(A\cap B) = P(A)P(B)## you can write this expression using conditional probabilities

##P(A) = \frac{P(A)P(B)}{P(B)} = \frac{P(A\cap B)}{P(B)} = P(A | B)## and similarly for ##P(B)##.

So, you can see in a more intuitive manner that the occurrence of one event does not affect the probability of occurrence of the other.
 
  • #3
Addez123 said:
If A and B are independent then the probability of both happening at once should be 0.
This is "mutually exclusive", not "independent". Think about it in terms of wagers. Suppose you are betting that it will rain tomorrow. Depending on your location you might accept 10:1 odds on such a wager.

Now, suppose that you also know that tomorrow is Tuesday, would that change the odds you would accept? If not, then they are independent.

If two events are mutually exclusive then knowing one definitely changes the wager you would accept on the other. So they are not independent.
 
  • Like
Likes QuantumQuest
  • #4
Addez123 said:
If A and B are independent then the probability of both happening at once should be 0.
If we drew a ven diagram it'd be just two circles who don't intersect.

I'm guessing I got wrong definition of independent, can someone explain please?

If event A is Chelsea beating Spurs in the FA cup and event B is Arsenal beating Man City tomorrow, then those are independent, but they could both happen.

But, if event B was Spurs beating Chelsea, then events A and B are not independent, but mutually exclusive.

An example of two events that are dependent are event A that Arsenal win and event B that Sanchez scores a hat trick.
 
  • Like
Likes QuantumQuest
  • #5
PeroK said:
An example of two events that are dependent are event A that Arsenal win and event B that Sanchez scores a hat trick.

Although, as it turned out, one goal from Sanchez was enough!
 
  • #6
The most confusing thing here is the symbol ##\cap##. The meaning of this symbol in set theory is very different from the meaning of the same symbol in probability theory. In set theory it can be interpreted in terms of Venn diagrams, but such an interpretation is not very useful in probability theory.
 
  • Like
Likes QuantumQuest
  • #7
Demystifier said:
n set theory it can be interpreted in terms of Venn diagrams, but such an interpretation is not very useful in probability theory.
Why not? I usually consider it in terms of Venn diagrams also where the area of the region is the probability.
 
  • #8
Dale said:
Why not?
Because it can mislead you to a wrong conclusion as in post #1.
 
  • #9
Demystifier said:
Because it can mislead you to a wrong conclusion as in post #1.
But that is because he drew the Venn diagram wrong, not because a correctly drawn Venn diagram is not useful in probability.
 
  • #10
Dale said:
But that is because he drew the Venn diagram wrong, not because a correctly drawn Venn diagram is not useful in probability.
So how to correctly draw the Venn diagram in this case?
 
  • #11
Demystifier said:
So how to correctly draw the Venn diagram in this case?
Draw a square of area 1, a circle of area P(A) and a circle of area P(B). Position them such that both circles are inside the square and their overlap has area P(A∩B). The shape of the circles can be distorted if needed.
 
  • #12
Dale said:
Draw a square of area 1, a circle of area P(A) and a circle of area P(B). Position them such that both circles are inside the square and their overlap has area P(A∩B). The shape of the circles can be distorted if needed.
How such a diagram would tell us that A and B are independent?
 
  • #13
Demystifier said:
How such a diagram would tell us that A and B are independent?
You have it backwards. We were given that A and B were independent, and drew the diagram accordingly.

If instead we were given a Venn diagram where the area represents probability then you can simply check if P(A∩B) = P(A) P(B) by looking at the corresponding areas.
 
  • #14
Dale said:
You have it backwards. We were given that A and B were independent, and drew the diagram accordingly.

If instead we were given a Venn diagram where the area represents probability then you can simply check if P(A∩B) = P(A) P(B) by looking at the corresponding areas.
OK, you can do it, but in my opinion it's not very useful. Do you know any reference where such diagrams are really used in practice?
 
  • #15
Demystifier said:
Do you know any reference where such diagrams are really used in practice?
Such diagrams were used to explain conditional probability and Bayes theorem to me when I was a student. Here is a lecture that takes the same approach: http://math.arizona.edu/~sreyes/math115as08/S08Proj1-BayesThm.ppt

I have not done a survey, but I have the impression that it is a common pedagogical technique. Certainly I would guess that the OP's teacher took that approach.
 
  • #16
Demystifier said:
OK, you can do it, but in my opinion it's not very useful. Do you know any reference where such diagrams are really used in practice?

It's the way I remember Bayes' theorem.

##P(A|B) = \frac{P(B|A)P(A)}{P(B)}##

For some reason this is something I always find hard to remember. So, I draw a Venn diagram of two overlapping sets ##A, B## and note that:

##P(A|B) = \frac{P(A \cap B)}{P(B)}##

That's just the area of ##A \cap B## divided by the area of ##B##.

Likewise:

##P(B|A) = \frac{P(A \cap B)}{P(A)}##

And then I put the two together.

One day, perhaps, I will memorise Bayes' Theorem directly!
 
  • #17
I still find the Venn diagrams in probability calculus more confusing than useful. But perhaps that's just me.
 
  • #18
Here are two example Venn diagrams of independent and dependent events. But I only use it to visualize the concept and don't know if it is a practical method for determining independence.
independentEvents.png
 

Related to Understanding the Concept of Independent Events: Venn Diagrams Explained

1. What does P(A∩B) mean in probability?

P(A∩B) represents the probability of the intersection of two events, A and B, occurring simultaneously. This means that both events must occur for P(A∩B) to be true.

2. Why is P(A∩B) sometimes equal to P(A)P(B)?

This is because when two events, A and B, are independent, the probability of both events occurring is equal to the product of their individual probabilities. In other words, the occurrence of one event does not affect the likelihood of the other event happening.

3. Can P(A∩B) be greater than P(A) or P(B)?

Yes, P(A∩B) can be greater than P(A) or P(B) if the events A and B are not independent. In this case, the occurrence of one event may increase or decrease the probability of the other event happening.

4. How can P(A∩B) be calculated?

P(A∩B) can be calculated by multiplying the probabilities of the two events, A and B, occurring. This is represented by the formula P(A∩B) = P(A)P(B).

5. What is the difference between P(A∩B) and P(A∪B)?

P(A∩B) represents the probability of both events A and B occurring simultaneously, while P(A∪B) represents the probability of either event A or B occurring. In other words, P(A∪B) includes the possibility of both events happening, while P(A∩B) does not.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
12
Views
5K
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Special and General Relativity
Replies
15
Views
511
Replies
1
Views
790
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
Back
Top