Tutorial 1:Sample Space and Probability 1 Baoxiang WANG bxwang@cse Spring 2017
Tutorial 1: Sample Space and Probability 1 Baoxiang WANG bxwang@cse Spring 2017
Probability models A probability model is an assignment of probabilities to every element of the sample space. Probabilities are nonnegative and add up to one. Examples S=HH,HT,TH,TT /44/4/4 107 models a pair of coins with equally likely outcomes
Probability models A probability model is an assignment of probabilities to every element of the sample space. Probabilities are nonnegative and add up to one. S = { HH, HT, TH, TT } ¼ ¼ ¼ ¼ Examples models a pair of coins with equally likely outcomes
Elements of a Probabilistic Model Event:a subset of sample space. -A is a set of possible outcomes -Example.A ={HH,T'T},the event that the two coins give the same side. The probability law assigns our knowledge or belief to an event Aa number P(A)0. It specifies the likelihood of any outcome
Elements of a Probabilistic Model • Event: a subset of sample space. – 𝐴 ⊆ Ω is a set of possible outcomes – Example. 𝐴 = 𝐻𝐻, 𝑇𝑇 , the event that the two coins give the same side. • The probability law assigns our knowledge or belief to an event 𝐴 a number 𝑃 𝐴 ≥ 0. – It specifies the likelihood of any outcome
Probability Axioms 1.(Won-negativity)P(A)≥0，for every event A. 2.(Additivity)For any two disjoint events A and B,P(AUB)=P(A)+P(B) In general,if A1,A2,..are disjoint events, then P(A1UA2U…)=P(A1)+P(A2)+… 3.(Normalization)P()=1
Probability Axioms 1. (Non-negativity) 𝑃(𝐴) ≥ 0, for every event 𝐴. 2. (Additivity) For any two disjoint events 𝐴 and 𝐵, 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 In general, if 𝐴1, 𝐴2, … are disjoint events, then 𝑃 𝐴1 ∪ 𝐴2 ∪ ⋯ = 𝑃 𝐴1 + 𝑃 𝐴2 + ⋯ 3. (Normalization) 𝑃(Ω) = 1
Question If 4 boys and 4 girls randomly sit in a row,what is the probability of the following seating arrangement (a)the boys and the girls are each to sit together The total number of all arrangements is 81=40320 The number of target arrangements is 4！×4！×2=1152 The probability #target arrangements] 11521 #all arrangements) 40320 35
Question If 4 boys and 4 girls randomly sit in a row, what is the probability of the following seating arrangement (a) the boys and the girls are each to sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4! ×4! ×2=1152 The probability #{𝑡𝑎𝑟𝑔𝑒𝑡 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠} #{𝑎𝑙𝑙 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠} = 1152 40320 = 1 35