The Chemistry Maths Book, Second Edition

604 Chapter 21Probability and statistics

Independent events

The possible outcomes of tossing a coin twice are {HH, HT, TH, TT}, each

with probability. The outcome of the second toss (event B) is independent of the

outcome of the first (event A), and the probability of event A andB is then the

product of the probabilities of A and B separately:

P(A and B) 1 = 1 P(A) 1 × 1 P(B) (independent events) (21.16)

EXAMPLE 21.4Find the probability of outcome 10 from two throws of a die.

Of the 36 possible outcomes, those equal to 10 are (5, 5), (4, 6), and (6, 4). The

second throw is independent of the first (or, for a single throw of two dice, the

outcome of each is independent of that of the other). Each outcome has probability

and because they are exclusive,

0 Exercises 9, 10

EXAMPLE 21.5Independent systems

In statistical thermodynamics the probabilityP(E)that a system is in a state with

energyEis a function of Eonly. Two noninteracting (independent) systems with

energiesE

1

andE

2

have combined energyE

1

1 + 1 E

2

and combined probability

P(E

1

1 + 1 E

2

) 1 = 1 P(E

1

) 1 × 1 P(E

2

) (21.17)

One function that satisfies this equation isP(E) 1 = 1 e

βE

, where βis a constant. We have

It can be shown that this function (multiplied by a constant) is the only one

that satisfies (21.17). In statistical thermodynamics the parameter βis inversely

proportional to the temperature,β 1 = 1 − 12 kT, where kis Boltzmann’s constant, and

the exponential probability function describes the Boltzmann distribution.

0 Exercise 11

21.5 The binomial distribution

The binomial distribution is the theoretical distribution that describes the results

of a given number of independent performances of an experiment that has only two

possible outcomes (see Section 21.3). We consider 4 tosses of a coin (or 4 observations

of the spin of an electron). Each toss (each ‘trial’) has two possible outcomes with

PE PE e e e PE E

EE EE

() () ( )

()

12 12

12 12

×= = =+

ββ β+

PP P P()()()()10 5 5 4 6 6 4 3

1

36

1

12

=+ + =×=and and and

1

6

1

6

1

36

×=

1

4