Events are frequently one-offs: Pretty Lady will run in the 4.20 at Sandown
Park only once this year, and if she enters the race next year, her form and
the field will be different. The probability that we want is for this year’s
race. Sometimes events can be repeated, however. For example, there is
no obvious difference between one throw of a die and the next throw, so
it makes sense to assume that the probability of throwing a 5 is the same
on each throw. When events can be repeated in this way we seek to assign
probabilities in such a way that when we make a very large number N of
trials, the number nA of trials in which event A occurs (for example 5 comes
up) satisfies
nA ≃ pAN. (1.1)
In any realistic sequence of throws, the ratio nA/N will vary with N, while
the probability pA does not. So the relation (1.1) is rarely an equality. The
idea is that we should choose pA so that nA/N fluctuates in a smaller and
smaller interval around pA as N is increased.
Events can be logically combined to form composite events: if A is the
event that a certain red die falls with 1 up, and B is the event that a white
die falls with 5 up, AB is the event that when both dice are thrown, the red
die shows 1 and the white one shows 5. If the probability of A is pA and the
probability of B is pB, then in a fraction ∼ pA of throws of the two dice the
red die will show 1, and in a fraction ∼ pB of these throws, the white die
will have 5 up. Hence the fraction of throws in which the event AB occurs is
∼ pApB so we should take the probability of AB to be pAB = pApB. In this
example A and B are independent events because we see no reason why
the number shown by the white die could be influenced by the number that
happens to come up on the red one, and vice versa. The rule for combining
the probabilities of independent events to get the probability of both events
happening, is to multiply them:
p(A and B) = p(A)p(B) (independent events). (1.2)
Since only one number can come up on a die in a given throw, the
event A above excludes the event C that the red die shows 2; A and C are
exclusive events. The probability that either a 1 or a 2 will show is obtained
by adding pA and pC . Thus
p(A or C) = p(A) + p(C) (exclusive events). (1.3)
In the case of reproducible events, this rule is clearly consistent with the
principle that the fraction of trials in which either A or C occurs should be
4 Chapter 1: Probability and probability amplitudes
the sum of the fractions of the trials in which one or the other occurs. If
we throw our die, the number that will come up is certainly one of 1, 2, 3,
4, 5 or 6. So by the rule just given, the sum of the probabilities associated
with each of these numbers coming up has to be unity. Unless we know that
the die is loaded, we assume that no number is more likely to come up than
another, so all six probabilities must be equal. Hence, they must all equal
1
6
. Generalising this example we have the rules
With just N mutually exclusive outcomes, X
N
i=1
pi = 1.
If all outcomes are equally likely, pi = 1/N.