This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 112 Fall 2010 Professor William Holzapfel Homework 1 Solution Problem 1. Sharpness of multiplicity function. The probability P ( n ) of getting n heads is the number of configurations that have n heads divided by the total number of configurations, or P ( n ) = 1 2 N N n = 1 2 N N ! ( N n )! n ! (1) 1 2 N 2 NN N e N p 2 ( N n )( N n ) ( N n ) e ( N n ) 2 nn n e n . a) The probability of getting 5000 heads and 5000 tails is thus P (5000) 1 2 10000 2 10000(10000) 10000 e 10000 2 5000(5000) 5000 e 5000 2 5000(5000) 5000 e 5000 = 1 2 50 = 0 . 00798 . b) The relative probability P (5100) /P (5000) is P (5100) P (5000) = ( 10000 5100 ) ( 10000 5000 ) = 5000!5000! 5100!4900! 5000 5100 * 4900 (5000) 10000 (5100) 5100 (4900) 4900 = 5000 5100 * 4900 50 51 5100 50 49 4900 = 0 . 135 . c) The relative probability P (6000) /P (5000) is P (6000) P (5000) = 5000!5000! 6000!4000! 5000 6000 * 4000 (5000) 10000 (6000) 6000 (4000) 4000 = 5 24 5 6 6000 5 4 4000 = 3 . 6 10 88 . 1 d) With 10 coins P (6) P (5) = ( 10 6 ) ( 10 5 ) = 10! 6!4! 5!5! 10! = 5 / 6 . Problem 2. Poisson Distribution. a) In this part, you are asked to justify the use of the binomial distribution P ( n ) = N n p n (1 p ) N n = N ! ( N n )! n ! p n (1 p ) N n (2) to describe an the probability that an event P characterized by probability p occurs n times in N trials. The probability that P occurs in a single trial is p . The probability that P does not occur (Not P) in a single trial is (1 p ). Now consider N trials. The probability that you get n outcomes of event P in N trials requires that you also get ( N n ) outcomes of Not P . Since the trials are independent, the probabilities multiply. For a particular microstate (this corresponds to running through N trials once), you get a factor of p for every time you get an event P , and you get a factor of (1 p ) every time you get Not P . This explains the factor of p n (1 p ) N n in P ( n )  it gives the probability of a single microstate in which P occurs n times. Now we just have to count the number of such microstates, but this amounts to choosing n trials out of the N total, so the number of such microstates is just ( N n ) . Putting this all together gives (2). Its worthwhile to note that if each trial consisted of flipping a coin then wed have p = 1 / 2, which when plugged into (2) just gives (1)! Thus, (2) generalizes the result of problem 1 to the case where p and Not p are not equally likely. b) Now we want to show that the binomial distribution becomes the Poisson distribution when n N , p 1, and = Np . More precisely, we are taking the limits p 0 and N where = Np stays constant and n is fixed. Using the Stirling approximation for N ! and ( N n )! (but not for n !) we get P ( n ) 2 NN N e N n !...
View
Full
Document
This note was uploaded on 11/18/2010 for the course PHYSICS 112 taught by Professor Steveng.louie during the Fall '06 term at Berkeley.
 Fall '06
 StevenG.Louie
 Physics, Work

Click to edit the document details