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 /4/44/4 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,TT},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.(Non-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 8!=40320 The number of target arrangements is 4！×4!×2=1152 The probability #{target arrangements} = 1152 1 #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
Question If 4 boys and 4 girls randomly sit in a row,what is the probability of the following seating arrangement (b)only the boys must sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4！×5！=2880 The probability #{target arrangements} 2880 1 #all arrangements} 40320 14
Question If 4 boys and 4 girls randomly sit in a row, what is the probability of the following seating arrangement (b) only the boys must sit together The total number of all arrangements is 8!=40320 The number of target arrangements is 4! ×5! =2880 The probability #{𝑡𝑎𝑟𝑔𝑒𝑡 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠} #{𝑎𝑙𝑙 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠} = 2880 40320 = 1 14
Question If 4 boys and 4 girls randomly sit in a row,what is the probability of the following seating arrangement (c)no two people of the same sex are allowed 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} 1152 1 #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 (c) no two people of the same sex are allowed 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
Question Eight balls are randomly withdrawn from an urn that contains 8 red,8 blue,and 8 green balls.Find the probability that (a)4 red,2 blue,and 2 green balls are withdrawn The total number of all combinationsis (2)=735471 The number of target combinations is (8)×()×(8)=54880 The probability #(target combinations} 54880 #all combinations} 735471
Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (a) 4 red, 2 blue, and 2 green balls are withdrawn The total number of all combinations is 24 8 =735471 The number of target combinations is 8 4 × 8 2 × 8 2 =54880 The probability #{𝑡𝑎𝑟𝑔𝑒𝑡 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} #{𝑎𝑙𝑙 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} = 54880 735471
Question Eight balls are randomly withdrawn from an urn that contains 8 red,8 blue,and 8 green balls.Find the probability that (b)at least 3 red balls are withdrawn The total number of all combinationsis (2)=735471 The number of target combinations is ()-()x()-()×(,)-()=406857 The probability #(target combinations} 12329 #all combinations} 22287
Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (b) at least 3 red balls are withdrawn The total number of all combinations is 24 8 =735471 The number of target combinations is 24 8 − 8 2 × 16 6 − 8 1 × 16 7 − 16 8 =406857 The probability #{𝑡𝑎𝑟𝑔𝑒𝑡 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} #{𝑎𝑙𝑙 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} = 12329 22287
Question Eight balls are randomly withdrawn from an urn that contains 8 red,8 blue,and 8 green balls.Find the probability that (c)all withdrawn balls are of the same color The total number of all combinationsis (2)=735471 The number of target combinations is 3 The probability #{target combinations} 1 #all combinations} 245247
Question Eight balls are randomly withdrawn from an urn that contains 8 red, 8 blue, and 8 green balls. Find the probability that (c) all withdrawn balls are of the same color The total number of all combinations is 24 8 =735471 The number of target combinations is 3 The probability #{𝑡𝑎𝑟𝑔𝑒𝑡 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} #{𝑎𝑙𝑙 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠} = 1 245247