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Chapter 2: Probability
I. Introduction A. Terminology
- Sample space: A set of all possible elements of
an experiment (โPopulation).
- Sample point (elementary event): Any member of a given sample space (โscore, datum).
- Event: Any subgroup of a sample space (โsample).
- Probability of an event, event A.
of sample points in event A
P(A)= โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ Total # of sample points in the sample space
e.g. a) The probability of a randomly drawn card to be a King: P(King) = 4/ b) P(Spade) = 13/ c) More examples
B. Basic Facts about Probability
- 0 โค P(A) โค 1
- ฮฃP(A) = 1
- Small probability means "unusual."
- All hypothesis testing is a conditional probability.
C. Basic Concepts
- Independence: The occurrence of one event does not change the probability of the occurrence of the second event (related to multiplication rule).
- Mutually Exclusive (Mutual Exclusiveness): Two events have no sample points in common. Two events cannot happen simultaneously (related to addition rule).
- Exhaustive: A sample space should include all possible outcomes.
D. Basic Rules
- Addition Rule ("Or Rule") a) P(AUB) = P(A) + P(B) - P(AโฉB) e.g. What is the probability of a randomly drawn card to be a King "or" a Spade?
P(King U Spade) = P(King) + P(Spade) - P(KingโฉSpade) = 4/52 + 13/52 - 1/ = 16/ b) If A and B are mutually exclusive, P(AโฉB)= e.g. What is the probability of a randomly drawn card to be a King "or" a Queen?
P(King U Queen) = P(King) + P(Queen) -P(KingโฉQueen) = 4/52 +4/52 - 0 =8/
- Multiplication Rule ("And Rule") a) If A and B are independent, P(AโฉB) = P(A) x P(B) e.g. What is the probability of a randomly drawn card to be a King and Spade?
P(King โฉ Spade) = P(King) x P(Spade) = (4/52) x (13/52) = 1/ b) If A and B are dependent, P(AโฉB) = P(A) x P(B|A) P(B|A): The conditional probability of B given A If A and B are independent, P(B|A) = P(B). D. Joint, Marginal, and Conditional Probability
- Joint Probability a) The probability of the co-occurrence of two or more events. b) A series of multiplication rules. c) From a raw data set
of sample points in a case of A and B
P(AโฉB)= โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ Total # of sample points in the sample space
e) What is the probability that a driver has a child in the car given he/she wears seat belt? f) What is the probability that a driver has a child in the car given he/she does not wear seat belt? g) More questions.
II. Bayesian Theorem (expansion of conditional probability)
P(A|B)P(B)
P(B|A) = โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
_ _
P(A|B)P(B) + P(A|B)P(B)
III. Combinatorics A. Factorial
- The number of ways to order (arrange) a set of objects.
- n! = (n)(n-1)(n-2)... (2)(1)
- 0! = 1
- e.g. How many ways do we have to arrange 3 letters, A, B, C. ABC, ACB, BCA, BAC, CAB, CBA ๏ 6 ways. 3! = 3 X 2 X 1 = 6 5! = 120 B. Permutations
- The number of ways to select a certain number of objects(r) from a set of objects (N) and arrange the r (sometimes called ordered combination).
- P Nr = ( )!
N r
N
- Select 3 letters out of 5 letters and arrange the 3 letters.
P 53 = ( 5 3 )!
C. Combinations
- The number of ways to select a certain number of objects(r) from a set of objects(N).
- C Nr = ( )!!
N r r
N
- Compute the number of ways to select 3 out of 5 letters.
C 53 =
IV. Binomial Distribution A. Introduction
- Bernoulli trial: an independent trial which results in two mutually exclusive outcomes.
- Binomial distribution: a theoretical distribution built by a certain number of Bernoulli trials, discrete distribution.
- P(X=x 0 ) = ( ) ( ) ( )!!
N x Nx!^ x Nx x P Q N x x
N
C P Q โ โ
where P(X=x 0 ): the probability of the event of X=x 0 , Px, P: a theoretical (given) probability of the event x for one trial, Q(N-x): 1 - Px, X: the target event, x 0 : specific value for the target event, and N: sample size (# of total trials).
- Examples a)Having 3 children (N=3), what is the probability of having 2 boys (X=2)? P = ยฝ =.
P(X=2) = (. 50 ) (. 50 ). 375 1! 2!
All events 3 Boys 1 1 1 (.5)(.5)(.5)=. 2 Boys 1 1 0 =. 1 0 1 =. 0 1 1 =. 1 Boy 1 0 0 =. 0 1 0 =. 0 0 1 =. 0 Boy 0 0 0 =. Total =1. b) Tossing a fair coin ten times (N=10), what is the probability of having ten heads (X=10)? P = .50.
the coin is fair (P=.50). X=8 (8 heads). b) Procedure H 0 : P = .50 H 1 : P โ . ฮฑ =. TS: P(X โฅ 8|P=.50) =.
๏ P(X=8|P=.50) = (. 50 )^8 (. 50 )^2 2! 8!
P(X=9|P=.50) = (. 50 )^9 (. 50 )^1
P(X=10|P=.50) = (. 50 )^10 (. 50 )^0
Decision P(X โฅ 8|P=.50) = .0546875 / โค .025 = ฮฑ /2. โด Fail to reject H 0. C. Mean and variance of the binomial distribution
- Given N, P, and Q, a) Mean(X) = NP e.g., Tossing a fair coin 10 times (N=10) and build a binomial distribution, X= # of heads, P = .50, and Q = .50, Mean = NP = (10)(.50) = 5. b) Variance(X) = NPQ = (10)(.5)(.5) = 2.5. c) SD(X) = Var ( X ) = NPQ = 2. 5 โ 1. 58.
