| Multiplication symbol | · (this should appear as a middle dot; however, some browsers may not properly render it) |
| Complement symbol | ' (read as prime) |
| Union/Or symbol | U |
| Intersection/And symbol | ∩ (this should appear as an inverted U; however, some browsers may not properly render it) |
| Conditional probability | P(A|B) means the probability of A given B has occurred |
| Mutually exclusive | Events A and B are mutually exclusive if the occurence of A precludes the occurence of B, i.e., if A ∩ B = null. |
| Independence | Events A and B are independent if and only if P(A|B) = P(A) or P(B|A) = P(B) |
| Complement rule | P(A) + P(A') = 1 |
| Complement rule for conditional probabilities | P(A|B) + P(A'|B) = 1 |
| Addition rule | P(A U B) = P(B U A) = P(A) + P(B) - P(A ∩ B) |
| Addition rule if A and B are mutually exclusive | P(A U B) = P(B U A) = P(A) + P(B) |
| Addition rule for three events | P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C) |
| Addition rule for four events | P(A U B U C U D) = P(A) + P(B) + P(C) + P(D) - P(A ∩ B) - P(A ∩ C) - P(A ∩ D) - P(B ∩ C) - P(B ∩ D) - P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ C ∩ D) + P(A ∩ B ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D) |
| Multiplication rule | P(A ∩ B) = P(B ∩ A) = P(A)·P(B|A) = P(B)·P(A|B) |
| Multiplication rule for m events | P(A1 ∩ A2 ∩ A3 ∩ ... ∩ Am) = P(A1)·P(A2|A1)·P(A3|(A1 ∩A2))· ... ·P(Am|(A1 ∩ A2 ∩ ... ∩ Am-1)) ... |
| Multiplication rule if A and B are independent events | P(A ∩ B) = P(B ∩ A) = P(A)·P(B) |
| Intersection of complements | P(A' ∩ B') = 1 - P(A U B) |
| Rule A1 (see appendix below for proof) | P(A ∩ B) + P(A' ∩ B) = P(B) |
Appendix
Proof that P(A ∩ B) + P(A' ∩ B) = P(B)
Given the multiplication rule:
P(A ∩ B) = P(B)·P(A|B)
P(A' ∩ B) = P(B)·P(A'|B)
Therefore, P(A ∩ B) + P(A' ∩ B)
= P(B)·P(A|B) + P(B)·P(A'|B)
= P(B)·[P(A|B) + P(A'|B)]
From the complement rule
P(A|B) + P(A'|B) = 1
Therefore,
P(A ∩ B) + P(A' ∩ B) = P(B)
Diagramatically, the proof can be shown through a Venn diagram. Note that A' is everything outside of A, so that A' includes part of B, specifically, B less (A ∩ B).
As can be seen the union of the two shaded areas--A ∩ B and A' ∩ B--is = B. Therefore, P(A ∩ B) + P(A' ∩ B) = P(B)
Print References
- Robert Johnson and Patricia Kuby. 2000. Elementary Statistics, 8th ed. Pacific Grove, CA: Duxbury.
- Ronald E. Walpole. 1982. Introduction to Statistics, 3rd ed. New York: Macmillan.