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