Understanding Set Theory: Explanation and Examples for Beginners

[Philip Wong]

Hi guys,
I someone help me to set my logic straight and help me understand the following situation.

For three I was given the following definition:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

I don't understand why we need to + P(A ∩ B ∩ C), instead of - P(A ∩ B ∩ C).
My reasoning are as follows:

1) I understand from other definitions that we − P(A ∩ B), − P(A ∩ C), − P(B ∩ C) from the union equation, because we would added the probability of A, B, and C twice if we don't do so.

2) Thus for P(A ∩ B ∩ C), it should follows the reason from 1), if we +P(A ∩ B ∩ C) instead of -P(A ∩ B ∩ C). Then wouldn't we added the probability of A,B, and C twice?

I tried to draw the Venn Diagram, to see if I can make some sense out of it. But that didn't help, as I don't get it at all and see any relationship at all. So please help.

 

[Stephen Tashi]

Write out each set X as the union of mutually exclusive sets. Then P(X) becomes the sum of the probabilities of these sets. If you write the right hand side of the rule this way, you can see what cancels out what.

For example, $A = (A \cap B \cap C) \cup (A \cap B \cap C^c) \cup A (A \cap B^c \cap C) \cup (A \cap B^c \cap C^c)$

This might be easier to write if you use the product notation for intersections. $ABC = A \cap B \cap C$

$P(A) = P(ABC) + P(ABC^c) + P(AB^cC) + P(AB^cC^c)$

$AB = ABC^c \cup ABC$

$P(AB) = P(ABC^c) + P(ABC)$

 

[Philip Wong]

Write out each set X as the union of mutually exclusive sets. Then P(X) becomes the sum of the probabilities of these sets. If you write the right hand side of the rule this way, you can see what cancels out what.

For example, $A = (A \cap B \cap C) \cup (A \cap B \cap C^c) \cup A (A \cap B^c \cap C) \cup (A \cap B^c \cap C^c)$

This might be easier to write if you use the product notation for intersections. $ABC = A \cap B \cap C$

$P(A) = P(ABC) + P(ABC^c) + P(AB^cC) + P(AB^cC^c)$

$AB = ABC^c \cup ABC$

$P(AB) = P(ABC^c) + P(ABC)$

Hi Stephen,
I don't quite follow you, mind to explain it in a more layman terms? thanks!

p.s. just to make sure, P stands for probability in the formula I type in the beginning
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

 

[Stephen Tashi]

I don't know what the explanation would be in layman's terms. Do the algebra and then explain the algebra to yourself verbally. (Laymen can't understand probability anyway!)

If you can't do the algebra, analyze a specific example involving the cardinality of sets instead of probability. A = {1,2,3,4,5,6}, B = {4,5,6,7}, C= {3,4,5,8}. What's the cardinality of $A \cup B \cup C$? Is it 6 + 4 + 4 -3 -3 -2? Or is it 6 +4 + 4 -3 -3 -2 + 2 ?

 

[D H]

Hi guys,
I someone help me to set my logic straight and help me understand the following situation.

For three I was given the following definition:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

I don't understand why we need to + P(A ∩ B ∩ C), instead of - P(A ∩ B ∩ C).

Here's an intuitive way to see understand this. Calculating P(A) + P(B) + P(C) accounts for that intersection three times. Calculating P(A ∩ B) + P(A ∩ C) + P(B ∩ C) also accounts for that intersection three times. Thus P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) doesn't account for it at all (3-3=0). You need to add that joint probability back in.

Or you could look at it from the perspective of disjoint sets. Split A into A_d, the part of A that is neither a member of B nor C; (A∩B)_d the part of A that also is in B but not in C; and A∩B∩C. Do the same for sets B and C. The union A∪B∪C can be split into seven disjoint parts A_d, B_d, C_d, (A∩B)_d, (A∩C)_d, (B∩C)_d, and A∩B∩C. Since these subsets are disjoint, the probability of the union is the sum of the probabilities of these subsets:

P(A∪B∪C) = P(A_d) + P(B_d) + P(C_d) + P((A∩B)_d) + P((A∩C)_d) + P((B∩C)_d) + P(A∩B∩C)

You can do the same for P(A), P(B), P(C), P(A∩B), P(A∩C), and P(B∩C). For example,

P(A) = P(A_d) + P((A∩B)_d) + P((A∩C)_d) + P(A∩B∩C)
P(A∩B) = P((A∩B)_d) + P(A∩B∩C)

With these, you will find that

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

is an identity.