PHYS 224 Homework 4 Sections 2.1 – 2.4: Multiplicity, Two-State Systems, Einstein Oscillators, 2nd Law. Problem 3.5 (From Statistical and Thermal Physics. Fundamentals and Applications. By M. D. Sturge, A, K. Peters, Ltd., 2003) a) In N successive tosses of a coin, what is the probability that a particular sequence of heads and tails, chosen in advance, will occur? If an event E has n(E) equally likely outcomes and its sample space S has n(S) equally likely outcomes, the probability of event E is nE PE nS Coins are 2-state objects. Therefore, in N successive tosses of a coin, the sample space has n(S) = 2N equally likely outcomes. The event E, which in this case is a particular sequence of heads and tails, chosen in advance, can occur only once. Then n(E) = 1. Therefore, its probability is 1 P N 2 b) What is the probability that the number of heads and number of tails will be equal after 100 tosses? The number of combinations of 100 elements taken 50 at a time is n(50) = 100 100! 100! . To approximately find n(50), apply Stirling’s 2 50 50!100 50 ! 50! approximation. Then 100! 100100e100√(200) and 50! 5050e50√(100). With these approximate values, we have 100 100! 100100 e100 200 2100 . 2 2 50 50 50! 5050 e50 100 N As the sample space has n(S) = 2 equally likely outcomes, the probability of this event is nE 2100 2100 1 1 PE 2100 2100 0.0798 . nS 50 50 50 12.5 c) Suppose that you win 1 cent for a head and lose 1 cent for a tail, and that at some point in the game you have won 50 cents. What is the probability that after another 50 tosses your opponent will have got exactly his or her money back? Why is this probability different from the result in (b)? Probability of getting a head in one tossing is ½ . To get all 50 tosses a head each time is 1 (½)50 = 50 . In the case (b), you have the total of 100 tosses and the outcome of each toss is 2 uncertain. All we know is the total of 50 heads and 50 tails.
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