Contents
<aside> <img src="/icons/book_orange.svg" alt="/icons/book_orange.svg" width="40px" /> A compendium for Mathematics NS1 written by Martin Deraas has been published on the Amazon store. It is available for purchase as a paperback for £9.95 (120 NOK), and as a PDF document on Google Drive/Adobe Acrobat for NOK 100.
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$$ A\cup B=\space \{ x: x \in A \text{ or }x\in B\} $$
For example, if A = {1,3,5,7} and B = {1,2,4,6}, then A∪B={1,2,3,4,5,6,7}. Notice that the element 1 is not listed twice in the union, even though it appears in both sets A and B.
This leads us to the general addition rule for the union of two events:
$$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $$
Where P(A∩B) is the intersection of the two sets. We must subtract this out to avoid double counting of the inclusion of an element.
If sets A and B are disjoint, however, the event A∩B has no outcomes in it, and is an empty set denoted as ∅, which has a probability of zero. So, the above rule can be shortened for disjoint sets only:
$$ P(A\cup B)=P(A)+P(B) $$
$$ \begin{alignedat}{4} P(&A\cap B&)&=&\space P(B)\space P(A\mid B)\\ &A\cap B&&=&\space \{x:x\in A{\text{ and }}x\in B\} \end{alignedat} $$
For example, if A = {1,3,5,7} and B = {1,2,4,6}, A∩B = {1} because 1 is the only element that appears in both sets A and B.
"The conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes *Pʙ*(*A*).
The complement of A, usually denoted as Ā or A', is the event in which A does not occur. From this we can conclude that:
$$ P(A \cup \text{\={A}})=1\\[5 pt]P(A)+P(\text{\={A}})=1\\[9 pt] $$
$$ P(B)=P(B\cap A)+(B\cap Ā)\\A\cap B = B\setminus \text{\=A}\\\text{\={A}}\cap B = B\setminus A $$
Where B∖A means the set that contains all those elements of B that are not in A.
The probability of an event X occurs at least k times is equal to 1 minus the probability that X occurs less than k times;
$$ P(X≥k)=1-P(k-1) $$
$$ P(A\mid B)={P(B\cap A)\over P(B)}={P(B\mid A)\cdot P(A)\over P(B)} $$
Events A and B are independent when P(A|B) = P(A) and P(B|A) = P(B).
$$ \text{events }A \text{ and }B \text{ are independent.}\Leftrightarrow P(A\cap B)=P(A)\cdot P(B) $$
With restoration
For a given probability n of every event and k draws:
$$ n_1\cdot n_2\cdot ...\cdot n_k=n^k $$
Without restoration
For n objects and k draws:
$$ n\cdot(n-1)\cdot...\cdot(n-k+1) $$
Nor n objects and k=n draws:
$$ n\cdot(n-1)\cdot ...\cdot 3\cdot 2\cdot 1=n! $$
$$ \text{Unordered selections}={\text{Ordered selections}\over \text{Ways in which drawn objects } k\text{ can be ordered}} $$
$$ \dbinom{n}{k}={n\cdot(n-1)\cdot...\cdot(n-k+1)\over k!} $$
Recursive formula
$$ \dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}\text{ for all integers }n,k:1<k<n-1, $$
Factorial formula
Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function:
$$ \dbinom{n}{k}={n!\over k!(n-k)!} $$
The probability of k to occur is the same as the probability of n\\k to occur;
$$ \dbinom{n}{k}=\dbinom{n}{n-k}\text{ for all integers }0≤k≤n, $$
From this we can conclude that:
$$ \dbinom{n}{0}=\dbinom{n}{n}=1\text{ for all integers }n≥0, $$
Identities involving binomial coefficients
The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then
$$ \dbinom{n}{k}={n\over k}\dbinom{n-1}{k-1} $$
and, with a little more work,
$$ \dbinom{n-1}{k}-\dbinom{n-1}{k-1}=\frac{n-2k}{n}\dbinom{n}{k}. $$
Moreover, the following may be useful:
$$ \dbinom{n}{h}\dbinom{n-h}{k}=\dbinom{n}{k}\dbinom{n-k}{h}. $$
For constant n, we have the following recurrence:
$$ \dbinom{n}{k}=\frac{n-k+1}{k}\dbinom{n}{k-1}. $$
(without restoration); completely accurate.
The probability of picking k₁ samples of n₁ and k₂ samples of n₂ when randomly picking k samples from n objects:
$$ \dbinom{n_1}{k_1}\dbinom{n_2}{k_2}\over \dbinom{n}{k} $$
(with restoration); easy for big selections.
The probability of an event A to occur exactly k times independently, where p is the constant probability of A:
$$ P(k)=\dbinom{n}{k}\cdot p^k \cdot (1-p)^{n-k} $$
A Venn diagram illustrating the intersection of two sets. Note that the solution/samples of A\B, A∩B, and B\A all belong to the union A∪B.
Photo: Cepheus - Own work, Public Domain, Wikimedia.