Chapter 29
Probability
29.1 Why Probability
29.2 Basic Probability Theory
29.3 Sterling’s Approximation
29.4 Probability Distribution Functions
29.5 Probability Distributions Involving Discrete
and Continuous Variables
29.6 Characterizing Distribution Functions
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.1 Why Probability
** Quantum mechanics
-- A set of principles underlying the most fundamental known description of all
physical systems at the microscopic scale (at the atomic level).
-- One of the greatest human accomplishments in 20th century
Example: Explain the observation of discrete emission from the H atom.
** Thermodynamics
-- The study of the conversion of heat energy into different forms of energy, and its
relation to macroscopic variables such as temperature, pressure, and volume.
Example: Predict the reaction spontaneity.
1023 HCl
?? Why the reaction occurs in the first place?
Internal energy
?? What are the molecular details that give rise to Gibbs energy?
Heat capacity
?? Why should ΔG vary from one species to the next?
Entropy
** Statistical mechanics: An application of probability theory.
A framework for relating the microscopic properties of individual atoms
and molecules to the macroscopic or bulk properties of materials.
Quantum mechanics, molecular details
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
HCl molecule
29.2 Basic Probability Theory
** Variable – Quantities that can change in value throughout the
course of an experiment or series of event.
Example: The outcome of a coin toss; two values: head or tail.
-- Discrete variables: A limited number of specific numbers.
Sample space: The collection of possible values a variable can
assume, e. g., {1, 2, 3, …., 100}.
-- Continuous variables: Any values within a set of limits.
The sample space is defined by the limiting values of the variables,
e. g., 1 ≦ X ≦ 100.
** To what extent will the variable assume any individual value from
the sample space?
Probability.
sample space + corresponding probability
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
probability model
29.2 Basic Probability Theory
sample space + corresponding probability
probability model
Lottery:
Sample space
Probability
= {1, 2, 3,
4, …..…….., 50} ball #
= {1/50, 1/50, 1/50, 1/50, ………, 1/50}
Tossing a coin:
Sample space
Probability
= {head, tail}
= {1/2, 1/2}
A deck of cards: Sample space
Probability
= { ♥A - K; ♦A – K; ♣A – K; ♠A – K}
= {1/52, …………………………, 1/52}
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.2 Basic Probability Theory
** Sample space, S = {s1, s2, s3, …., sN}
(1) Probability of observing an individual value is equivalent = 1/N
(2) there are E values in S corresponding to the outcome of interest.
1 1 1
1 E
PE = + + + ..... + =
N N N
N N
E times
There are N values in sample space, and E of these values correspond to the
event of interest, then
NE
PE =
N
Example 29.2:
The probability of selecting a heart from a standard deck of 52 cards?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.2.1 The Fundamental Counting Principle
-- ! (exclamation point symbol): factorial.
你應該可以馬上回答下列問題:
** A class of 30 students to form a line. How many arrangements of students are
possible?
The total number of ways (W) = 30x29x28…..x2x1 = 30! = 2.65 x 1032
** Example 29.3: How many 5-card arrangements are possible for a standard deck
of 52 cards?
** Example 29.4: How many possible spin states for He 1s12s1?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.2.1 The Fundamental Counting Principle
你即將要會回答:
** If you tossed a coin four times, what is the probability that at least
two heads are observed after four tosses?
** If we toss a coin 50 times, what is the probability of the coin
landing heads 20 times out of the 50 tosses?
** How to describe the rotational level populations of I2 at 300 K?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.2.2 Permutations (排列
排列)
排列
** Permutation - An ordered sequence of elements selected from a given finite
set, without repetitions, and not necessarily using all elements of the given set.
The number of permutations – The total number of unique ways to be arranged.
-- There are n! total permutations for a set of n objects.
?? If only a subset of j objects is employed in constructing the permutations,
what is the number of permutations?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Ex. 29.5, basketball team
5 player arrangement
out of 12 players
29.2.3 Configurations (combination,組合
組合)
組合
** Configurations – The number of
unordered arrangements of objects.
** The number of permutations:
P(4, 3) = 4!/1! = 4 x 3 x 2 = 24
** The number of unordered
arrangements, the number of
configurations:
C(4,3) = 4!/3!1! = 4
組合
configuration
combination
排列
permutation
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Ex. 29.6
5 card combinations with a
standard 52-card deck.
29.2.4 A Counting Example: Bosons and Fermions
** Application of the probability theory to chemical problems:
-- Bosons (photons, particles of integer spin such as 4He) described by
Bose-Einstein (BE) statistics.
?? How many ways can a set of n indistinguishable particles be arranged into x
equally accessible states, with each state capable of holding any number of particles?
Simple example: Four particles and three states,
4 particles (n particles) +
2 movable walls
(x states – 1)
Step1: Counting the number of permutations with a
collection of six distinguishable objects = (n+x-1)!
Step 2: The particles and walls are indistinguishable.
P
= (n + x -1)!/n!(x-1)!
Grace H. Ho, Department
BE of Applied Chemistry, National University of Kaohsiung
不在乎牆1、牆2;
球 1、2 、3 、4的排列
29.2.4 A Counting Example: Bosons and Fermions
Example 29.7 How many quantum states are possible for 1s22s22p2 (C)?
PFD = 6! / 4!2! = 6 x 5 / 2! = 15
How about d5s1?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.2.5 Binomial Probabilities
** Bernoulli trials: The probability of the event of interest (success, PE) combined
with the event not of interest (fail, complementary, PEC) must be equal to unity.
PE + PEC = 1
-- The outcome of each trail is either a success or a failure.
-- The outcome of each trial is independent of the outcome of any other trial in the
experiment.
Example: Binomial experiment, coin toss.
?? A coin is tossed four times.
(1) What is the probability of observing “head” every time?
Example 29.8
 1  1  1  1  1
PE =      =
 2  2  2  2  16
(2) What is the probability of observing [H, T, T, H]?
 1  1  1  1  1
PE =      =
 2  2  2  2  16
10H40T, 25H25T?
P10 = 9.1x10-6
P25 = 0.11
(3) What is the probability of obtaining j successes in a trial consisting of n trials?
n
j
n− j
j
E
E
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
P(j) = C (P ) (1 − P )
29.3 Stirling’s Approximation
** Calculation P(n, j) or C(n, j) involving a real-world system deals with an
extreme large number!
100! = 9.3 x 10157
1023! = ?
** Sterling’s approximation:
ln (N!) = N ln N - N
∫ ln ax dx
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
= x ln ax − x
29.4 Probability Distribution Functions
n
j
j
P(j) = C (PE ) (1 − PE )
n− j
Probability Distribution function
PE = 0.5
PE = 0.3
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
PE = 0.7
P(X i ) =
fi
M
∑f
i =1
i
29.5 Probability Distribution Involving Discrete and
Continuous Variables
** Discrete variables
Summation
P(X i ) =
fi
M
∑f
i
Continuous variables
Integration
-- Probability density, P(x)
-- P(x)dx
The probability that the variable x has in the
range of dx.
i =1
=1
P( x)dx =
f ( x )dx
∫
x2
x1
f ( x )dx
-- Example: The probability density for the
spatial location of the particle.
∫
∞
−∞
*
ψ ( x, t ) ψ ( x, t ) dx = 1
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
P(x)
Probability distributions for the translational kinetic energy of an ideal gas.
∫
∞
∫
∞
0
0
x e − x dx =
1
π
2
 1 
P(E)dE = 2π 

 πRT 
3/ 2
3/2
RT RT ∫
1
1
= 2π   ×
π =1
2
π 
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
∞
0
E / RT e − E / RT d (E / RT )
29.6 Continuous Representation of Discrete Variables
** Treating a discrete variable as continuous simplifies an probability evaluation
involving the summation of a large number of terms.
100
Ptotal =
∑e
-0.3X
= 3.86
Ptotal =
X=0
100
∫
0
e -0.3X dX = 3.3333
14% difference
100
Ptotal =
∑e
-0.05X
= 20.5
Ptotal =
X=0
100
∫
0
e -0.05X dX = 19.86
3% difference
-- The difference between the discrete and continuous treatments is smaller for a
more “fine-grain” distribution.
= If the difference between two consecutive values of a function is small relative to
the domain of interest, treating a discrete variable as continuous is appropriate,
Examples: the distribution function of translational and rotational states.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.6 Characterizing Distribution Functions
29.6.1 Average Value
** Average value = Expectation value, < g(X) >.
g(X): A function, which is dependent on variable X.
P(X): The probability distribution of g(X) associated with the variable X.
Ex. 29.12
A3
A2
A1
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.6.2 Distribution Moments
** Distribution moments involve functions of the form xn.
-- n = 1, the first moment = the average value = <x1> = <x>
-- n = 2, the second moment = <x2>
Root-mean-squared, rms = square root of <x2> = <x2>1/2
Example 29.13 (P. 738)
1.
2 − ax 2
P( x) = Cx e
Normalization of a distribution function.
∫
∞
2 n − ax 2
x e
(
2n − 1)!! π
dx =
2
n +1
n
(n :1,2,3,...; a > 0)
a
a
∞
n!
2 n +1 − ax 2
∫0 x e dx = 2a n+1 (n : 0,1,2,3,...; a > 0)
0
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
4a3 / 2 2 − ax 2
P(x) =
x e
π
2. Mean value = ?
Example 29.13
n!
∫0
2a n +1
(n : 0,1,2,3,...; a > 0)
∞
x
2 n +1 − ax 2
e
dx =
4a 3 2 1
2
n = 1, < x >=
=
2
π 2a
πa
3. <x2> = ?
∫
∞
0
2
x 2 n e − ax dx =
(2n − 1)!! π
2
n +1
a
n
a
4a 3 2 3 ⋅1
n = 2, < x >=
3 2
2
a
π
2
(n :1,2,3,...; a > 0)
π
3
=
a 2a
4. Root-mean-squared value, <x2>1/2 = ?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
29.6.3 Variance
** Variance (σ2), a measure for the width of a distribution.
σ2 = < (x - <x>)2 > = <x2 – 2x<x> + <x>2>
-- The variance of a distribution is equal to the difference between the second
moment and the square of the first moment.
-- The standard deviation σ is the square root of the variance of x.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
Example 29.14
< x>=
2
P( x ) =
4a 3/2
π
2 − ax 2
xe
4 
 3
1σ =  − 
 2a πa 
πa
3
< x >=
2a
2
** Gaussian distribution
1
−( x -δ ) 2
P( x) =
e
σ 2π
-- “Bell-shaped curve”.
-- Applications: social science, chemistry, physics,
grades, error distribution.
-- Maximum at x = δ.
-- The width of the distribution is determined
by the variance σ2.
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
2σ 2
1
−( x -u ) 2
P( x) =
e
σ 2π
2σ 2
− ln 2 = − x 2 / 2σ 2
FWHM?
2(ln 2)σ 2 = x 2
(
)
x = ± 2 ln 2 σ
(
)
FWHM = 2 2 ln 2 σ
FWHM? for a simple Gaussian form:
P( x ) =
1
π
e
− x2
∫
∞
−∞
P( x)dx = 2
1
π
− ln 2 = − x 2
x = ± ln 2
FWHM = 2 ln 2
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
∫
∞
0
e
− x2
1 1
dx = 2
π =1
π 2
1
− x2
P( x) =
e
σ 2π
2σ 2
n!
∫0 x e dx = 2a n+1
(n : 0,1,2,3,...; a > 0)
∞
In the equation,
a=
1
2σ
2
, σ=
1
2a
∫
∞
0
2 n − ax 2
x e
dx =
2 n +1 − ax 2
(2n − 1)!! π
2
n +1
a
n
a
(n :1,2,3,...; a > 0)
Homework:
(1) Calculate <x> .
(2) Calculate the value of <x2> .
(3) What is <x2> - <x>2?
(4) What is one standard deviation?
Grace H. Ho, Department of Applied Chemistry, National University of Kaohsiung
0
σ2
σ2
σ