If A is an event, and A={E1,E2,....En}, Then P(A)=P(E1)+P(E2)+- - - +P(En)
If S is a Sample space containing N equally likely outcomes, and if A is an event containing k outcomes, then
P(A)=k/N
1.1 Fundamental Principle of Counting
Annotations:
Assume that k operations are to be performed. If there are n1 ways to preform the first operation, and if for each of these ways there are n2 ways to preform the second operation, and if for each choice of ways to perform the first two operations there are n3 ways of preforming third operation, and so on, then the total number of ways to preform the sequence of k operations is n1n2----nk
1.2 1) Calculate Sample Space
1.2.1 Sample Space (S)
Annotations:
The set of all possible outcomes of an experiment is called the sample space for the experiment
1.2.1.1 Let S be a sample space. Then P(S)=1
1.2.1.2 The number of permutations of n objects is n!
1.2.2 2) Calculate Events
1.2.2.1 Event
Annotations:
A subset of a sample space is called an event
1.2.2.1.1 Mutually Exclusive
Annotations:
The events A and B are sid to be mutually exclusive if they have no outcomes in comment
More generally, a collection of events A1,A2,...An is said to be mutually exclusive if no two of them have any outcomes in common.
1.2.2.1.1.1 If A and B are mutually exclusive events then, Then P(A U B) = P(A)+P(B)
Annotations:
More generally, if A1,A2,.... are mutually exclusive events, then P(A1 U A2 U- -)= P(A1)+P(A2)+- - -
1.2.2.1.2 For any event A, 0 ≤ P(A) ≤ 1
1.2.2.1.3 Complament
Annotations:
For any A, P(A^c)=1-P(A)
1.2.2.1.4 Let A and B be any events. Then P(A U B)=P(A)+P(B)-P(A ∩ B)
1.2.2.1.5 nPr
Annotations:
Example, in choosing between A,B,C,D & E , how many ways can you choose any 3 arranged letters. (This means ABC is different from ACB or BCA or...) Then 5P3.
1.2.2.1.5.1 When order is importent
1.2.2.1.6 nCr
Annotations:
Example , how many ways can you get three letters from A,B,C,D & E. (here ABC is no different from BAC since you just got A&B&C.) Then 5C3.