Homework 4

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
nE
PE 
nS 
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 e100 200
2100
.





2
2
50
 50   50!
5050 e50 100

N

As the sample space has n(S) = 2 equally likely outcomes, the probability of this event is
nE
2100
2100
1
1
PE 

 2100 
 2100 

 0.0798 .
nS 
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.