Chapter 1. Probability

Transcription

1 Chapter. Probablty Mcroscopc propertes of matter: quantum mechancs, atomc and molecular propertes Macroscopc propertes of matter: thermodynamcs, E, H, C V, C p, S, A, G How do we relate these two propertes? Statstcal thermodynamcs (mechancs)

2 Basc Probablty Theory Varables: quanttes that can change n value throughout the course of an eperment or seres of events e.g., the sde of con observed after tossng the con. Dscrete varables: assume only a lmted number of specfc values e.g., the outcome of toss = two values (head or tal) sample space of the varables = the possble values a varable can assume e.g., the outcome of toss {+, } Contnuous varables: assume any value n the certan range e.g., temperature, 0 < T <.

3 Imagne a lottery where balls numbered to 50 are randomly med. The probablty of selectng s /50. Ths requres an nfnte number of eperments. Consder a varable for whch the sample space conssts of n values denoted as {,, n }. The probablty that a varable wll assume one of these values (P ) s: 0 < P <, =,,, M The sum of the probabltes for selectng each ndvdual ball must be equal to. M P+ P+ + P = n P= =

4 Consder the probablty assocated wth a gven outcome for a seres of eperments,.e, the event probablty. Imagne tossng a con four tmes. What s the probablty that at least two heads are observed after four tosses? Fgure Potental outcomes after tossng a con four tmes. Red sgnfes heads and blue sgnfes tals. The probablty = /6 The probablty (P E ) that the outcome or event of nterest, E, occurs n N values n sample space E PE = N

5 The Fundamental Countng Prncple For a seres of manpulatons {M, M,, M j } havng n j ways to accomplsh each manpulaton {n, n,, n j }, the total number of ways to perform the entre seres of manpulatons (Total M ) s the product of the number of ways to perform each manpulaton under the assumpton that the ways are ndependent: Total = ( n )( n )( n ) ( n ) M 3 j - Assemble 30 students n a lne. - How many arrangements of students are possble? The total number of ways 3 W = = 30! =.65 0 Eample: How many fve-card arrangements are possble from a standard deck of 5 cards? Soluton: Total M = (n ) (n ) (n 3 ) (n 4 ) (n 5 ) = (5) (5) (50) (49) (48) = 3,875,00

6 Permutatons How many permutatons are possble f only a subset of objects s employed n constructng the permutaton? P(n, j): the number of permutatons possble usng a subset of j objects from the total group of n Pn (, j) = nn ( )( n ) ( n j+ ) nn ( )( n ) ()() n! = = ( n j)( n j ) ()() ( n j)! Eample: The coach of a basketball team has plays on the roster, but can only play 5 plays on one tme. How many 5-player arrangements are possble usng the -player roster? Soluton:! Pn (, j) = P(,5) = = 95,040 ( 5)!

7 Confguratons permutatons = the number of ordered arrangements confguratons = the number of unordered arrangements Fgure Illustraton of confguratons and permutatons usng four colored balls. The left-hand column presents the four possble three-color confguratons, and the rght-hand column presents the s permutatons correspondng to each confguraton.

8 C(n, j) = the number of confguratons that are possble usng a subset of j objects from a total number of n objects. Pn (, j) n! Cnj (, ) = = j! j!( n j)! Eample: How many possble 5 card combnatons or hands from a standard 5 card deck are there? Soluton: 5! C (5,5) = =,598,960 5!(5 5)!

9 Bnomal Probabltes Defne the complement of P E as the probablty of an outcome other than that assocated wth the event of nterest, as denoted by P EC P + P = E EC Bernoull tral: the outcome of a gven eperment wll be a success (.e., the outcome of nterest) or a falure (.e., not the outcome of nterest). Bnomal eperment: a collecton of Bernoull trals The probablty of observng heads every tme when a con s tossed four tmes P E = = 6

10 The probablty of obtanng j successes n a tral consstng of n trals for a seres of Bernoull trals n whch the probablty of success for a sngle tral s P E : P( j) = Cnj (, )( P) ( P) E j n j E n! = ( PE) ( PE) j!( n j)! j n j C(n, j): the number of confguratons that are possble usng a subset of j successes n n trals

11 Eample: Toss a con 50 tmes. What are the probabltes of havng the con land heads up 0 tmes and 5 tmes? Soluton: P = Cnj (, )( P) ( P) P 0 5 j n j E 0 40 = C(50,0) E ! = = 9. 0 (0!)(40!) 5 5 = C(50, 5) ! = = 0. (5!)(5!) 6

12 Strlng s Appromaton Calculatons of factoral quanttes becomes etremely large: 57 00! = Need an appromaton: Strlng s appromaton ln N! = Nln N N Dervaton ln N! = ln[( N)( N )( N ) ()()] = ln N+ ln( N ) + ln( N ) + + ln + ln N = ln n n= N ln ndn = Nln N N (ln ) Nln N N

13 Probablty Dstrbuton Functons Number of Heads Probablty Number of Heads Probablty

14 Ths nformaton can be presented graphcally by plottng the probablty as a functon of outcome. Fgure 3. Plot of the probablty of the number of heads beng observed after flppng a con 50 tmes. The red curve represents the dstrbuton of probabltes for P E =0.5, the blue curve for P E =0.3, and the yellow curve for P E =0.7 The probablty of observng j successful trals followng n total trals n! j n j P( j) = ( PE) ( PE) j = 0,,,, n j!( n j)!

15 A probablty dstrbuton functon ( f ) represents the probablty of a varable () havng a gven value, wth the probablty descrbed by a functon P ( ) f. P( ) = Cf M = P ( P ( ) = M = Cf = Cf + Cf + + Cf = C f C = M = = M = f ) = M f = f M

16 Probablty Dstrbutons Involvng Dscrete and Contnuous Varables If the varable s contnuous, P() s the probablty that the varable has a value n the range of d P( ) d = Cf ( ) d, where f () s a functon not yet defned. P( ) d = C f ( ) d = C = f ( ) d P( ) d = Cf ( ) d = f ( ) d f ( ) d

17 Characterzng Dstrbuton Functons Average Values Consder a functon, g(), whose value s dependent on. M g ( ) = g ( ) P ( ) = = Dstrbuton moments g ( ) n M = g ( ) f M = : the frst moment of the dstrbuton functon : the second moment of the dstrbuton f : the root-mean-squared (rms) value Eample a P( ) = C e 0 Are the mean and rms values for ths dstrbuton the same?

18

19 Varance The varance ( σ ) : a measure of the wdth of a dstrbuton defned as the average devaton squared from the mean of dstrbuton ( ) σ = = + Note: b ( ) + d ( ) = b ( ) + d ( ) cb ( ) = cb ( ) σ = + = + =

20 - Gaussan dstrbuton: the bell-shaped curve ( ) ( δ ) σ P d e d = σ : the wdth of the dstrbuton ( πσ ) Fgure 4. The nfluence of varance on Gaussan probablty dstrbuton functons. Notce that an ncrease n the varance corresponds to an ncrease n the wdth of the dstrbuton.

21 Eample: a ( ) σ? P = C e =