A.1 Some Basic Probability Concepts
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• N s systems in the ensemble exhibit outcome s
• N r + N s systems in the ensemble exhibit either r or s
Therefore p(r or s) = (N r +N s )/N = p r +p s . This result can be generalized.
Note that either · · · or excludes the possibility of r and s together.
2. What is the probability of the joint occurrence of both outcomes r and s? If the
probability for s is unaffected by the occurrence of r (we say that outcomes r and
s are statistically independent or uncorrelated), then if N r → r and N rs of
these systems also exhibit s, then
p s =
N rs
N r
⇒ N rs = N r p s
(A.1.6)
and hence
p rs ≡
N rs
N
=
N r
N
p s = p r p s .
(A.1.7)
Thus, if outcomes r and s are statistically independent, then p rs = p r p s .
Example Let us consider an ideal gas consisting of N atoms enclosed in a container
of volume V T . Let us now divide the box into two parts, of volumes V and V , such
that V + V = V T . We treat the atoms as statistically independent, and construct
an ensemble of such containers of gas. Let p be the probability that a given atom is
found in volume V , and q = 1 − p be the probability that a given atom is found in
the remaining volume V . At equilibrium the atoms tend to be uniformly distributed
throughout the volume V 0 , so that
p =
V
V T
;
q =
V
V T
.
We now ask the question: ‘What is the probability p(n) that n of the N atoms are
found in the volume V , while the remaining N − n atoms are found in the volume
V ?’.
ppp · · · p
n factors
qqq · · · q
N −n factors
= p
n q
N −n .
Each set of n atoms out of a total N atoms can be chosen in
N
n
= C(N, n) =
N!
n!(N − n)!
(A.1.8)
ways. Hence we see that
p(N, n) =
N
n
p
n q
N −n .
(A.1.9)