We all use the word, or at least the concept of,
probability in everyday speech. For example, "it will probably rain tomorrow",
"the Swans will probably win", "he (she) will
probably get drunk again". If we make a bet we are making a judgment about
relative probabilities. When used in a statistical context the word has a more
formal description. A probability provides a quantitative description of the
likely occurrence of a particular event. It is usually expressed on a scale from
0 to 1 (proportion rather than percentage). An improbable event has a
probability close to 0, a very common event has a probability close to 1.
A subjective probability describes an
individual's opinion about how likely a particular event is to occur. It is not
based on any precise computation but is often a reasonable assessment by a
knowledgeable person. Indeed this kind of information, especially when derived
from experts, is used in many expert systems (a computer based artificial
intelligence technique). Most statistical techniques require a more robust
estimate of a probability which is often based on a large sample.
David Grimmett asked the following
question on one of the statistics mailing lists….
If a weatherman has historically been 80% accurate in
his forecasts, and now forecasts an 80% chance of rain for tomorrow, what
is the true chance of rain?
See below for an answer

Statisticians make use of a particular set of terms
to avoid confusion!
 An outcome is what we observe or measure in any
situation where the outcome is subject to some uncertainty, i.e. a particular
outcome has a certain probability of occurring. In some situations it may be
possible to list all of the possible outcomes.
 This exhaustive list is called the sample space.
For example the sample space for a coin toss is (Heads, Tails), for the roll
of a dice it is (1,2,3,4,5,6), heart rate is any real number within the range
(0 to maximum possible heart rate).
 Strictly an event is a collection of outcomes.
Thus, it is a subset of the sample space.
The probability of more than one
event
Two events are said to be independent if the
occurrence of one of them provides no information about whether or not the other
event will occur. For example, if we know a person's gender we gain no extra
information about whether they will be left or right handed. The same is not
true for colour blindness since colour blindness is more common in males. In
probability theory we say that two events, A and B, are independent if the
probability that they both occur is equal to the product of the probabilities of
the two individual events.
If two events are independent then they cannot be
mutually exclusive (disjoint) and vice versa. In statistical notation
this is written as P(A Ç B) = P(A).P(B). The
symbol Ç
means 'and'.
The addition rule is used to determine the
probability that event A or event B occurs or both occur. However,
some events are mutually exclusive (or disjoint) and they cannot occur together.
For example, a person cannot be both colour sighted and colour blind. If two
events are mutually exclusive, they cannot be independent and vice
versa.
The formal notation for calculating probabilities is
shown below.
P(A È B) = P(A) +
P(B)  P(A Ç B). where:
 P(A) = probability that A occurs
 P(B) = probability that B occurs
 P(A È B) = probability that A or B occurs
 P(A Ç B) = probability that A and B
occur
For mutually exclusive events, P(A Ç B) must equal 0 so the addition rule simplifies to P(A
È B) = P(A) + P(B) or the Probability that A OR B will occur is P(A) +
P(B)
For independent events, P(A Ç B) = P(A).P(B) so the addition rule
reduces to P(A È
B) = P(A) + P(B)  P(A).P(B).
Example
Suppose we wish to find the probability of
picking either a king or an ace from a pack of 52 playing cards. These are
mutually exclusive events. A card cannot be both a king and an ace. We define
the events A = 'pick a king' and B = 'pick an ace'. Since a card cannot be both
a king and an ace P(A Ç B) = 0. So we ignore this and find.
P(A È
B) = P(A) + P(B)
Hence, the probability of picking either a
king or an ace is 1/13 + 1/13 = 0.154.
Suppose now we wish to find the probability of
picking either a king or a diamond from a pack of 52 playing cards. This time
one of the kings is a diamond so these are not mutually exclusive events. If we
define the events as:
A = 'pick a king' and
B = 'pick a diamond'.
then
P(A È B)
= P(A) + P(B) + P(A Ç
B)
There are 4 kings and 13 diamonds in the pack, but 1
card is both a king and a diamond. So, the probability of drawing either a king
or a diamond is 1/13 + 1/4  1/52 = 0.308.
The multiplication rule is used to determine the
probability that two events, A and B, both occur simultaneously. For independent
events, that is events which have no influence on one another, there is a simple
rule: P(A Ç B) = P(A).P(B).
Example
What is the probability that a person selected at
random will be female and lefthanded? For the purposes of this illustration we
will assume that 20% of the population is lefthanded and that gender and
handedness are independent of each other.
Let p1 = probability of being female = 0.5
Let p2 = probability of being lefthanded =
0.2
Therefore the probability that a person selected at
random will be female and lefthanded is p1p2 = 0.5 * 0.2 = 0.10 ( 10% or
approximately 1 in 10).
Conditional Probabilities
(Bayes Theorem)
sometimes the probability of an event is
conditional on some other event, e.g. the probability of developing
lung cancer is influenced by the number of cigrattes smoked. This type of
conditional probability is best dealt with using Bayes's theorem. Bayes was an
English cleric who invetigated the way in which probabilties changed as more
more information becomes available. If A and B are two related (dependent)
events the fact that A has occured will alter the probability that B occurs. The
basic form of the the theorem is.
P(A) is the probability that A will occur.
P(B) (not used so far) is the probability
that B occurs.
P(A,B) is the joint probability that both A and B
occur.
P(BA) is the conditional probability that B will
occur if A occurs.
For example, if P(A) is the probability that someboby
smokes and P(B) is the probability of developing lung cancer, then P(A,B) is the
probability that a person smokes and develops lung cancer. However, P(BA) is
more useful since it gives us the probability that lung cancer will develop
given the fact that someone smokes. How do we find P(BA)?
Because A and B are related events we can also write
Bayes's theorem as:
P(BA). P(A)= P(AB) .
P(B) 
which means that we can rearrange it to
give:
P(BA) = [P(AB) . P(B)] /
P(A) 
This turns out to be calculation that has great
practical significance. For example, using the lung cancer example:
P(AB) is the probability that a person with lung
cancer is also a smoker. We can find this information from patient
records.
P(B) and P(A) can be found from general population
surveys and hospital records.
Hence, we can now estimate P(BA), the probability
that a smoker will develop lung cancer.
One of the big advantages of Bayes's theorem is that
we can use additional information to refine our estimate of P(BA). Thus, as
more data become available we can be more confident about the
relationship.
The following simple example is taken from Gelman et al
(1995).
Haemophilia is a sexlinked disorder. Thus
most sufferers are male, having inherited the abnormal gene from their
mother. Consider a mother whose brother is haemophilic, what is the
probability that she is a carrier? In order for the brother to be a
sufferer their mother must have been a carrier, thus the unconditional
probability that his sister is a carrier is 0.5, i.e. P(carrier) =
0.5.
If the mother has sons how can we use this
information about the sons to refine our estimate that the mother is a
carrier?
Even if only one son has haemophilia the mother
must be a carrier, and P(carrier) = 1 (ignoring a very small probability
of a mutation).
Suppose she has two sons, both normal, what
does this tell us? Presumably we begin to think that prehaps she isn't a
carrier, or P(carrier) < 0.5. Bayes's theorem allows us to refine our
estimate of P(carrier) a little more precisely. Let P(C+) = P(carrier) and
P(C) = P(not carrier).
P(C+ ,2 normal sons)
= 
P(2 normal
sonsC+) . P(C+) 

P(2 normal
sonsC+).P(C+) + P(2 normal sonsC) .
P(C) 
P(C+) and P(C) are both 0.5 if the mother was
a carrier.
P(2 normal sonsC+) is the probability of
having 2 normal sons if the mother is a carrier. The probabilty that a
carrier will have a nomal son is 0.5, there the probability of having two
normal sons (normal son AND normal son) is 0.5 . 0.5 = 0.25.
However, if the mother is not a carrier then
P(2 normal sonsC) must be 1.0, i.e. a noncarrier mother must have
normal sons.
Substituting the above P values
gives
P(C+ ,2 normal sons)
= 
(0.25) .
(0.5) (0.25) . (0.5) + (1.0) .
(0.5) 


P(C+ ,2
normal sons) = 
0.125 /
0.625 
= 
0.20 
Hence, we now estimate the probability that the
mother is a carrier to be 0.2, given the information that she has had
two normal sons.
One of the biggest advantages of Bayes's
theorem is that we can use new evidence or data to refine our
estimates.

Suppose the mother has another normal son. In
the previous calculation we began with the assumption that P(C+) = P(C) =
0.5, but finished with estimates of P(C+) = 0.2 and P(C) = 0.8.
Therefore, we use these new estimates, P(C+) = 0.2 and P(C) = 0.8, to
estimate P(C+3 normal sons). For example, P(3 normal sons  C+) is (0.8 .
0.8 . 0.8) . (0.2) = (0.512) .(0.2).
P(C+3 normal sons) =

(0.512) .
(0.2) (0.512) . (0.2) + (1) .
(0.8) 


= 
0.113 
As we might expect the birth of a 3rd normal
son increases our confidence that she is not a
carrier. 
Possible answers to the weather
man problem…
Milo Schield's reply
I'd say 68% (assuming base rate = actual rate = 50%.
Say the past history was as follows:
Suppose the forecaster says there is an 80% chance of rain. There
is an 80% chance the forecaster should forecast "RAIN". There is a 20%
chance the forecaster should forecast "NO RAIN" If rain is forecast, there
is an 80% chance of actual rain. If norain is forecast, there is a 20%
chance of actual rain.
Assuming independence, there is a 68% chance of rain: 64% + 4%.
Note that 64% = .8 x .8 and 4% = .2 x .2.
Note that the 20% forecast of norain actually implies a 32% real
chance of no rain….. Obviously a different table could give different
values.
Don Esslemont said..
I don't think the question as formulated can be answered. What can
it mean to say the the forecasts have been "80% accurate" when they took
the form of statements about the probability of rain?
Derek Christie had a different view!
I agree that the 80% weather problem is not well defined but here
is a possible interpretation.
The god of weather has an adjustable spinner with two sections 
rain and fine. Each evening she sets the size of the rain sector to
whatever she fancies (say a uniform distribution from 0% to 100%). The
forecaster has to try and guess what she has set it to. He observes her
from afar through binoculars. If he has a record of 80% accuracy, this
means that 80% of the time he gets a clear view of the spinner and knows
where it is set. The remaining 20% of the time his view is obscured and he
just has to guess. The weather god now spins the pointer and presses the
rain or fine button on the weather machine, whichever is appropriate.
Take 100 occasions on which the forecaster has predicted 80%. On 80
of those occasions he got a good view and the spinner was indeed set at
80%. 64 of those 80 days will have rain. On the other 20 occasions he has
no idea of the spinners setting. Assuming a random setting, it will rain
on 10 of those 20 days. Total  74 days rain out of 100 or a probability
of rain of 74%. ( If the actual distribution of settings is known this
figure can be adjusted accordingly.)

This page(slightly modified for an Australian audience), with acknowledgement, from a web site on univariant statistics by Dr Alan Fielding BSc MSc PhD FLS FHEA, Senior Learning and Teaching Fellow, School of Biology, Chemistry and Health Science, Manchester Metropolitan University. Alan has a new site with information on monitoring and statistics. He may be contacted at alan@alanfielding.co.uk or via his web page.
