2.3 KiB
Lecture Topic: Probability
Experiment: An act where the outcome is subject to uncertainty
Sample space: The set of all possible outcomes of an experiment Example:
- Flip a coin:
S = \{H, T\}
- Throw a dice:
S = \{1, 2, 3, 4, 5, 6\}
Events: An event is a collection (subset) of outcomes contained in the sample space S
S =
\begin{Bmatrix}
1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\
2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\
3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\
4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\
5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\
6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6
\end{Bmatrix}
A = First and second elements are the same
A = \{(1,1) \ (2,2) \ (3,3) \ (4,4) \ (5,5) \ (6,6)\}
Union: Union of two events, A and B, (A \cup B
)
Compliment: The compliment of an event A, is the set of all outcomes in S that are not in A
Mutually Exclusive Events: When and B have no common outcomes, they are said to be mutually exclusive or disjoint events, i.e. P=(A \cap B) = 0
Axiom:
- For event A, P(A) >= 0
- P(S) = 1
- (Need to look at slides to correct)
(a) If
A_1, A_2 ... A_k
is a finite collection of mutually exclusive events then: $$P(A_1 \cup A_2 \cup ... \cup A_k) = \sum^{\infty}{i=1} P(A_i) $$ (b) IfA_1, A_2 ... A_k
is a finite collection of mutually exclusive events then: $$P(A_1 \cup A_2 \cup ... \cup A_k) = \sum^{\infty}{i=1} P(A_i) $$ Proposition: - For event A, P(A) = 1 - P(A')
- If and and B are mutually exclusive then
P(A \cap B) = 0
Permutation: Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from n objects is denoted by P_{k,n}
and calculated by:
P_{k,n} = \frac{n!}{(n-k)!}
Combination: Given a set of n, any unordered sub-set of k of the objects is called a combination. The set of combinations of size k that can be formed from n distinct objected will be denoted by C_{k,n}
and calculated by:
C_{k,n} = \frac{n!}{k!(n-k)!} = \begin{pmatrix}n \\ k \end{pmatrix}
(NEED TO REVIEW THIS)
Example: S = {1, 2, 3, 4, 5}
A_1
= {1,2,3}, A_2
= {2,3,5}, A_3
= {2,3,1}
Permutation: 5! = 120 (5-3)! = 2! = 2 120 / 2 = 60
Combination: 5!/2!3! = 60/6 = 10 0! = 1 5 out of 5 5!/(5-5)! = 120/1 = 120