Introduction to Probability 1. What is the “chance” that sales will decrease if the price of the product is increase? 2. How likely that the Thai GDP will increase? 3. What is the “chance” of Thai-rak-Thai wining the parliament again?

Sample Space Roll a die S={1,2,3,4,5} Sample Point Toss a coin S={1,2} Sample Point Sample Space Head tail Head Tail First CoinSecond Coin Head Tail Head Sample Point (H,T) (H,H) (T,H) (T,T) Sample Space = {(H,H), (H,T), (T,H), (T,T)} Tree Diagram

A Counting Rule for Multiple-step Experiments If an experiment can be described as a sequence of k steps in which there are n 1 possible outcomes on the first step, n 2 possible outcomes on the second step, and so on, then the total number of experimental outcomes is given by (n 1 )(n 2 )…(n k ). That is, the number of outcomes for the overall experiment is found by multiplying the number of outcomes on each step. Counting Rule for Combinations The number of combinations of N objects taken n at a time is so follows: N = N!/(n!(N-n)!) Where N! = N(N-1)(N-2)…(2)(1) n! = n(n-1)(n-2)…(2)(1) And 0! = 1 n

Ex: A quality control procedure where and inspector randomly selects two of five parts to test for defects. In a group of five parts, How many combinations of two parts may be selected? 5!/(2!(5-2)!) = (5)(4)(3)(2)(1)/((2)(1)(3)(2)(1))=120/12=10 Assigning Probabilities to Experimental outcomes Requirement: 1. 0  P(Ei)  1 2.  P(Ei) =1 The three method assigning probability 1.Classical Method: assign equal probability to outcomes 2.Relative Frequency Method: Refer to past event 3.Subjective Method: Use human judgement

An event is a collection of sample points The Event of tossing a die with the number less than 4 Probability of an Event The probability of any event is equal to the sum of the probabilities of the sample points in the event /6 P(E<4) = = 0.5

P(A) = 1 – P(A c ) Ex: P(A c ) = 0.2 P(A) = = 0.8 AAcAc AB Union All sample points belonging to A or B or both A  B Intersection Sample points belonging to A both A  B AB Additional Law P(A  B) = P(A)+P(B)- P(A  B) AB Additional Law for Mutually Exclusive P(A  B) = P(A)+P(B

Conditional Probability: A chance of outcome A occurring is influenced by the occurrence of outcome B. P(A|B) = P(A  B)/P(B) or P(B|A) = P(A  B)/P(A) Promotion status over 2 years Joint Probability Table P(M  A) P(W  A) P(W  A C ) P(M  A C ) P(M)P(W) P(A) P(A C ) P(A|M) = P(A  M)/P(M) = 288/960 or.24/.80 =.3 P(A|W) = P(A  W)/P(W) = 36/240 or.03/.2 =.15 Are there any discrimination against female officers?

Conditional Probability P(A|B) = P(A  B)/P(B) or P(B|A) = P(A  B)/P(A) Independent Events P(A|B) = P(A) or P(B|A) = P(B) P(A  B) = P(B) P(A|B) or P(A  B) = P(A) P(B|A) P(A  B) = P(B) P(A) The probability of occurrence of an outcome A is the same regardless of whether or not an outcome B occurs

If Independence  P(A|M) = P(A) .27 P(A|W) = P(A) .27 That means there shouldn’t be discrimination and that the chance of male and female officer to be promoted should be equal to the over all promoting opportunity.

Bayes’s Theorem P(Ai|B) = P(Ai)P(B|Ai) Prior Probabilities P(Ai) Information P(B|Ai) Apply Bayes’ Theorem to get Post Probabilities (P(A 1 )P(B|A 1 )+ P(A 2 )P(B|A 2 )+…+ P(A i )P(B|A i ) P(A 1 ) P(B|A 1 ) P(A 2 ) P(A 1 ) P(B|A 2 ) P(A 2 ) P(A 1 |B)P(A 2 |B) P(A 1 |B) = P(B) P(A 1  B) P(A 1 )P(B|A 1 ) P(A 1  B)+P(A 2  B)= P(A 1 )P(B|A 1 )+P(A 1 )P(B|A 1 )

Supplier1 P(A1) = 0.65 Supplier2 P(A2) = 0.65 P(G|A 1 )=.98 P(B|A 1 )=.02 P(G|A 2 )=.95 P(B|A 2 )=.05 Parts are used in manufacturing process: Find the probability of machine breaks down because bad part from Supplier1