Review for the previous lecture

Transcription

1 Review fo the pevious letue Definition: sample spae, event, opeations (union, intesetion, omplementay), disjoint, paiwise disjoint Theoem: ommutatitivity, assoiativity, distibution law, DeMogan s law Pobability theoy: sample spae, sigma algeba, pobability funtion Theoem: define a pobability funtion on finite and ountable infinite sample spaes Chapte 1 Pobability Theoy Chapte 1.2 Basi of Pobability Theoy The Calulus of Pobability Theoem 1.2.8: If P is a pobability funtion and A is any set in B, then 1. P( ) = PA ( ) 1 3. PA ( ) = 1 PA ( ) Theoem 1.2.9: If P is a pobability funtion and A is any set in B, then 1

2 1. PB ( A) = PB ( ) PA ( B). 2. PA ( B) = PA ( _) + PB ( ) PA ( B) 3. If A B, then PA ( ) PB ( ) Example (Bonfeoni s Inequality): Let A and B be two events suh that P(A) = P(B) = What an we say about the pobability that both events will ou? Solution: PA ( B) = PA ( ) + PB ( ) PA ( B) PA ( ) + PB ( ) 1= = 0.90 Theoem : If P is a pobability funtion, then PA ( ) = PA ( C ) i fo any patition C1, C2, of the sample spae S ; i= 1 i= 1 i i= 1 i fo any sets 1 2 P( A) P( A) A, A, (Boole s Inequality) Example: A total of 28% of Ameian males smoke igaettes, 7% smoke igas, and 5% smoke both. Questions: (1) define the sample spae, S, the event that desibes Ameian males who smoke igaettes, the event that desibes Ameian males who smoke igas, the sigma algeba B on S, and the pobability funtion P on B. (2) alulate the peent of males who smoke neithe igaettes no igas. (3) alulate the peent of males who smoke igas but not igaettes. Solution: 2

3 (1) S ={All Ameian males}, A={Ameian Males who smoke igaettes}, B ={Ameian Males who smoke igas}, we know that PA ( ) = 28%, PB ( ) = 7%, PA ( B) = 5%. B ={All subsets of S }. P (Eah Ameian Male)=1/#Ameian Males. Then we know PA ( ) = 28%, PB ( ) = 7%, PA ( B) = 5% (2) C ={Ameian Males who smoke neithe igaettes no igas}, then PC ( ) = PA ( B) = 1 P(( A B) ) = 1 PA ( B), and PA ( B) = PA ( ) + PB ( ) PA ( B) = 30%, thus PC ( ) = 70%. C A B =, so (3) D ={Ameian Males who smoke igas but not igaettes}, then PD ( ) = PB ( A) = PB ( ) PB ( A) = 2%. D= B A, so Counting Main Pupose: assign pobabilities on finite sample spaes. Theoem : If a job onsists of k sepaate tasks, the i th of whih an be done in n i ways, i= 1, 2,, k, then the entie job an be done in n1 n2 nk ways. Example : If you an pik up 6 numbes fom 1 to 44 in a lottey game, how many ways an you hoose in the fist two steps? If the ode does not matte, how many ombinations of two numbes an you get in the fist two steps? Solution: If you an not hoose the same numbe twie, then we have = 1892 ways. But if you ae allowed to hoose the same numbe twie, then we have = 1936 ways. If you an not hoose the same numbe twie 3

4 and the ode does not matte, then we have 44 43/ 2 = 946 ombinations. If you ae allowed to hoose the same numbe twie and the ode does not matte, the we have 44 43/ = 990. Atually, the afoementioned example highlights two impotant fatos on methods of ounting: eplaement and ode, we an have fou situations: Possible methods fo ounting Odeed Unodeed Without Replaement With Replaement Definition : Fo a positive intege n, n! (ead as n fatoial) is the podut of all the positive integes less than o equal to n, i.e. n! = n ( n 1) 2 1. Futhemoe, we define 0! = 1. Examples (ontinue): If you an pik up 6 numbes fom 1 to 44 in a lottey game, how many ways do you have? odeed, without eplaement: = 44! 38! odeed, with eplaement: 4

5 = 44 unodeed, without eplaement ! = !38! Definition : Fo nonnegative integes n and, whee n, we define the symbol n, ead n hoose, as n n! = (binomial oeffiients).!( n )! unodeed, with eplaement: 49! 6!43! In summay, we have the following table: Numbe of possible aangements of size fom n subjets Odeed Without Replaement ( n) n! ( n )! With Replaement n 5

6 Unodeed n n Enumeating Outomes Equally likely outomes: Conside the sample spae S = { s1, s2,, s N } whee ( ) Ps i 1 # of elements in A PA ( ) = P({ s)) = s i i A si AN = # of elements in S 1 =. Then fo any event A: N Example : In a poke game (daw 5 ads without eplaement), find the pobability of getting (i) fou aes (ii) fou of a kind (iii) exatly one pai Example: Suppose thee ae nn ( 365) students in ou lass, then what is the pobability that at least two students have the same bithday (ignoing the yea of thei bith and onsideing thee ae only 365 days in a yea). 365 Solution: P= 1 ( n!)/365 n n. Define the event A={at least two students have the same bithday} then A ={All students have the diffeent bithday} 6

7 Table: # of elements in the sample spae S = 365 n # of elements in the event A = 365 n! n n PA ( ) Example (Sampling with eplaement): Conside sampling = 2 items fom n = 3 objets, with eplaement. The outomes in the odeed and unodeed sample spaes ae these: Unodeed {1,1} {2,2} {3,3} {1,2} {1,3} {2,3} Odeed (1,1) (2,2) (3,3) (1,2),(2,1) (1,3),(3,1) (2,3),(3,2) Pobability 1/9 1/9 1/9 2/9 2/9 2/9 Example (Sampling without eplaement): Conside sampling = 2 items fom n = 3 objets, without eplaement. The outomes in the odeed and unodeed sample spaes ae these: Unodeed {1,2} {1,3} {2,3} Odeed (1,2),(2,1) (1,3),(3,1) (2,3),(3,2) Pobability 1/3 1/3 1/3 Note: 7

8 1. To alulate the ight pobability, you must use the ight sample spae to ount the numbe of elements in the event and the sample spae. The ight sample spae means that all elements in the sample spae have the same pobabilities. 2. Fo sampling without eplaement, if we want to alulate the pobability of an event that does not depend on the ode, we an use eithe the odeed spae o the unodeed spae. 3. Fo the sampling with eplaement, if we want to alulate the pobability of an event that does not depend on the ode, we must use the odeed spae. This is oesponds to the ommon intepetation of sampling with eplaement. Fo example , that means one of thee objets is hosen, eah with pobability of 1/3, the objet is noted and eplaed; thee objets ae mixed and again one of them is hosen, eah with pobability of 1/3. Thus, six unodeed outomes do not have the equal pobability, but nine odeed outomes have the equal pobability unde this sampling sheme. 8