Downloadable - About University

78 Introduction to probability
whether or not the river caused flooding. For example, there was light
rainfall and yet the river flooded in four out of the last 20 Aprils.
Suppose that in order to make a particular decision we need to calculate
the probability that next year there will either be heavy rain or the river
will flood. We decide to use the relative frequency approach based on
the records for the past 20 years and we then proceed as follows:
p(heavy rain or flood) = p(heavy rain) + p(flood)
= 11/20 + 13/20 = 24/20 which exceeds one!
The mistake we have made is to ignore the fact that heavy rain and
flooding are not mutually exclusive: they can and have occurred together.
This has meant that we have double-counted the nine years when both
events did occur, counting them both as heavy rain years and as
flood years.
If the events are not mutually exclusive we should apply the addition
rule as follows:
p(A orB) = p(A) + p(B) − p(A andB)
The last term has the effect of negating the double-counting. Thus the
correct answer to our problem is:
p(heavy rain or flood) = p(heavy rain) + p(flood)
Complementary events
− p(heavy rain and flood)
= 11/20 + 13/20 − 9/20 = 15/20 (or 0.75)
If A is an event then the event ‘A does not occur’ is said to be the
complement of A. For example, the complement of the event ‘project
completed on time’ is the event ‘project not completed on time’, while
the complement of the event ‘inflation exceeds 5% next year’ is the event
‘inflation is less than or equal to 5% next year’. The complement of
event A can be written as A (pronounced ‘A bar’).
Since it is certain that either the event or its complement must occur
their probabilities always sum to one. This leads to the useful expression:
p(A) = 1 − p(A)