PS1Sol_f10 - Physics 112 Fall 2010 Professor William...

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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 πNN N e - N 2 π ( N - n )( N - n ) ( N - n ) e - ( N - n ) 2 πnn n e - n . a) The probability of getting 5000 heads and 5000 tails is thus P (5000) 1 2 10000 2 π 10000(10000) 10000 e - 10000 2 π 5000(5000) 5000 e - 5000 2 π 5000(5000) 5000 e - 5000 = 1 2 π 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
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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). It’s worthwhile to note that if each trial consisted of flipping a coin then we’d 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 λ = Np . More precisely, we are taking the limits p 0 and N → ∞ where λ = 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 πNN N e - N n ! 2 π ( N - n )( N - n ) N - n e - ( N - n ) p n (1 - p ) n = e - n ( Np ) n (1 - p ) ( N - n ) (1 - n/N ) n - 1 / 2 n !(1 - n/N ) N (3) 2
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after some algebra. Now, in the limits we’re taking we have 1 - n N N -→ e - n 1 - n N n - 1 / 2 -→ 1 (1 - p ) N - n = 1 - Np N N (1 - p ) - n -→ e - λ , and substituting these into (3) yields P ( n ) = λ n n !
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