A.1 Some Basic Probability Concepts
591
A natural question to ask is therefore: ‘What is the probability that n of the N
marbles are found in one of the two boxes?’. Our figures tell us that
for N = 2 : p 0 =
1
4
; p 1 =
2
4
; p 2 =
1
4
for N = 4 : p 0 =
1
16
; p 1 =
4
16
; p 2 =
6
16
; p 3 =
4
16
; p 4 =
1
16
.
Thus, more generally then, if we let C(n) be the number of ways of distributing
the marbles between the boxes so that n of them are found in one (knowing
which), then the probability of finding n marbles in that particular box is
p n =
C(n)
2 N .
(A.1.3)
We can also infer from our diagrams that if N is large, as long as n is either close
to N or close to zero, then C(n) 2 N , so that p n 1. Now, if p n is small, then
it is very unlikely that all of the marbles (or none, as the case may be) are to be
found in one particular box. Such situations are very rarely realized: we refer to
such situations as non-random, or orderly. We also can see that C(n) is maximal
when n n ≈
N
2 and, in this case, p n is larger, i.e., such configurations are
more probable: these are called random, or disorderly situations.
We have just examined probability concepts based upon two possible outcomes (i.e.,
RHS or LHS of the box for a marble). An event in general need not have just two
outcomes: for example a ‘fair’ die will have 6 possible outcomes, corresponding to
the six faces of the die, and each outcome will have a probability
1
6 (rather than
1
2 ).
The theory of probability is a tool that allows us to make better guesses in a consistent
manner.
Let us work with what we shall refer to as a system A, which we can take for
the purposes of illustration to be a coin. We shall then perform an experiment, or
observation, on it (for the present example, toss the coin), and record the outcome
(for the present example, H or T). We must assume that the situation is timeindependent, i.e., the conditions at the end of the experiment are the same as those
at the beginning, and we then repeat the experiment many times, e.g., N = 500,
and record all the outcomes. Having carried out the set of observations, we then
ask the question ‘If I toss the coin just once, can you predict whether the outcome
will be H or T?’. In order to attempt to answer this question we might, for example,
examine the outcomes for the first 100 experiments, and perhaps find that N H = 42,
N T = 58 (N = 100), so that p H = 0.42, p T = 0.58. This answer we would, of
course, not find particularly satisfactory, since we would intuitively know (from our
lifelong experience) that the real answer should be closer to p H = p T = 0.50.
What do we do? Well, we analyse yet more of our data. If we examined 200
experiments, we would perhaps find N H = 106, N T = 94 (N = 200), which
implies that p H =
166
200 = 0.53 and p T =
94
200 = 0.47. We see immediately that the
591
A natural question to ask is therefore: ‘What is the probability that n of the N
marbles are found in one of the two boxes?’. Our figures tell us that
for N = 2 : p 0 =
1
4
; p 1 =
2
4
; p 2 =
1
4
for N = 4 : p 0 =
1
16
; p 1 =
4
16
; p 2 =
6
16
; p 3 =
4
16
; p 4 =
1
16
.
Thus, more generally then, if we let C(n) be the number of ways of distributing
the marbles between the boxes so that n of them are found in one (knowing
which), then the probability of finding n marbles in that particular box is
p n =
C(n)
2 N .
(A.1.3)
We can also infer from our diagrams that if N is large, as long as n is either close
to N or close to zero, then C(n) 2 N , so that p n 1. Now, if p n is small, then
it is very unlikely that all of the marbles (or none, as the case may be) are to be
found in one particular box. Such situations are very rarely realized: we refer to
such situations as non-random, or orderly. We also can see that C(n) is maximal
when n n ≈
N
2 and, in this case, p n is larger, i.e., such configurations are
more probable: these are called random, or disorderly situations.
We have just examined probability concepts based upon two possible outcomes (i.e.,
RHS or LHS of the box for a marble). An event in general need not have just two
outcomes: for example a ‘fair’ die will have 6 possible outcomes, corresponding to
the six faces of the die, and each outcome will have a probability
1
6 (rather than
1
2 ).
The theory of probability is a tool that allows us to make better guesses in a consistent
manner.
Let us work with what we shall refer to as a system A, which we can take for
the purposes of illustration to be a coin. We shall then perform an experiment, or
observation, on it (for the present example, toss the coin), and record the outcome
(for the present example, H or T). We must assume that the situation is timeindependent, i.e., the conditions at the end of the experiment are the same as those
at the beginning, and we then repeat the experiment many times, e.g., N = 500,
and record all the outcomes. Having carried out the set of observations, we then
ask the question ‘If I toss the coin just once, can you predict whether the outcome
will be H or T?’. In order to attempt to answer this question we might, for example,
examine the outcomes for the first 100 experiments, and perhaps find that N H = 42,
N T = 58 (N = 100), so that p H = 0.42, p T = 0.58. This answer we would, of
course, not find particularly satisfactory, since we would intuitively know (from our
lifelong experience) that the real answer should be closer to p H = p T = 0.50.
What do we do? Well, we analyse yet more of our data. If we examined 200
experiments, we would perhaps find N H = 106, N T = 94 (N = 200), which
implies that p H =
166
200 = 0.53 and p T =
94
200 = 0.47. We see immediately that the