# Inclusion/Exclusion Principle for 3 events

• guyvsdcsniper
In summary, the problem asks to prove the I/E formula for 3 events by first applying I/E to (A∪B)∪Z and then considering (A∪B) as a single event "Y" and applying I/E between "Y" and Z. The distributive law is used to simplify (A∪B)∪Z to (A∪Z)∩(B∪Z). However, this does not follow the instructions as I/E is not applied to Y and Z. Furthermore, in order to prove the identity, both the left and right side must be memorized and written correctly.
guyvsdcsniper
Homework Statement
Prove the I/E formula for 3 events by 1st applying
I/E to (A∪B) ∪Z. In other words, consider (A∪B) as a
single event "Y" and then apply I/E between "Y" and Z.
Relevant Equations
P(A∪B)=P(A)+P(B)−P(A∩B)

P(A∪B∪C)=P(A)+P(B)+P(C)−
−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)
Below is my attempt at the problem. I used the distributive law and applied to what was given, (A∪B)∪Z which equates to (A∪Z)∩(B∪Z). I then applied the 2 set I/E formula to each union. Since there is an intersection between these two sets I added them.
But I end up with 2 P(Z) which doesn't seem right.

I feel like I am got the distributive property part down but definitely getting something wrong past that.

I have a little trouble following it, but it looks like you've made a mistake in distribution e.g., ##(A \cup B) \cup Z \neq (A \cup Z) \cap (B \cup Z)## in general.

We have ##P(Y \cup Z) = P(Y) + P(Z) - P(Y \cap Z)##. We can then apply I/E to ##P(Y)## (i'll leave that to you). We can also use distribution (as you suggested) to get
$$Y \cap Z = (A \cup B) \cap Z = (A \cap Z) \cup (B \cap Z)$$

And now we can use I/E on ##P((A \cap Z) \cup (B \cap Z))## (what are the two events?).

Putting it all together should give us the identity!

PeroK
quittingthecult said:
Homework Statement:: Prove the I/E formula for 3 events by 1st applying
I/E to (A∪B) ∪Z. In other words, consider (A∪B) as a
single event "Y" and then apply I/E between "Y" and Z.
Relevant Equations:: P(A∪B)=P(A)+P(B)−P(A∩B)

P(A∪B∪C)=P(A)+P(B)+P(C)−
−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)

Below is my attempt at the problem. I used the distributive law and applied to what was given, (A∪B)∪Z which equates to (A∪Z)∩(B∪Z). I then applied the 2 set I/E formula to each union. Since there is an intersection between these two sets I added them.
But I end up with 2 P(Z) which doesn't seem right.

I feel like I am got the distributive property part down but definitely getting something wrong past that.

View attachment 301811
The instructions say to apply I/E to Y U Z. You didn't do that.

As for the rest, identities have a right side and a left side. You have to memorize both. You can't just memorize the right side and then write down whatever you feel like for the left side.

Regardless, there is no need to pick your favorite law from your list of identities. The instructions are really clear about what to do first.

## 1. What is the Inclusion/Exclusion Principle for 3 events?

The Inclusion/Exclusion Principle for 3 events is a mathematical principle that states the total number of elements in the union of three sets can be calculated by adding the individual sizes of the three sets, subtracting the size of the intersection of any two sets, and adding back the size of the intersection of all three sets.

## 2. How is the Inclusion/Exclusion Principle for 3 events used in probability?

In probability, the Inclusion/Exclusion Principle for 3 events is used to calculate the probability of the union of three events. This is done by adding the probabilities of the three events, subtracting the probabilities of the intersections of any two events, and adding back the probability of the intersection of all three events.

## 3. What is an example of using the Inclusion/Exclusion Principle for 3 events?

An example of using the Inclusion/Exclusion Principle for 3 events is when calculating the number of students who take either math, science, or history classes in a school. The total number of students can be calculated by adding the number of students in each class, subtracting the number of students who take both math and science, both math and history, and both science and history, and then adding back the number of students who take all three classes.

## 4. Can the Inclusion/Exclusion Principle for 3 events be extended to more than 3 events?

Yes, the Inclusion/Exclusion Principle can be extended to any number of events. The formula for calculating the union of n events involves adding the individual sizes of the sets, subtracting the sizes of the intersections of any two sets, adding back the sizes of the intersections of any three sets, and so on, until the final step of subtracting the size of the intersection of all n sets.

## 5. How does the Inclusion/Exclusion Principle for 3 events relate to the Principle of Inclusion and Exclusion?

The Inclusion/Exclusion Principle for 3 events is a special case of the Principle of Inclusion and Exclusion, which states that the total number of elements in the union of any number of sets can be calculated by alternating between adding and subtracting the sizes of the intersections of the sets. The Inclusion/Exclusion Principle for 3 events is simply the application of this principle to three sets.

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