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in which N DD is the total number of dry days in the data record under considera-
tion, and N DS is the number of dry periods in the s am e record. As can be seen from
Equation (13.17), this mean duration is also given by k
p ) 1 , which shows how
p can be determined immediately from the record by means of (13.32). The respective
conditional probabilities of dry day sequences obtained this way with the data from
Portland, Maine, on which Figure 13.6 is based, were found to be 0.708, 0.816 and 0.870
for the three thresholds; as was to be expected, these values are nearly the same as those
obtained from the slopes of the lines through the points mentioned above.
This example was presented for sequences of dry days, or rather days with precipita-
tion less than a certain finite threshold. The same reasoning also holds, mutatis mutandis ,
if p represents the probability of a rainy day (i.e. in excess of a certain threshold) being
succeeded by a rainy day. Finally, it should be kept in mind that p , as used here in this
example, is different from the probability of a dry day; the latter is defined as the ratio
of the number of dry days over the total number of days (dry and rainy) of the record.
However, Hershfield (1970b) has shown how the two may be related.
=
(1
13.3.2
The binomial distribution
Consider again an experiment with two possible outcomes, namely success and failure,
whose respective probabilities are p and q
p ). This distribution provides the
answer to the question what the probability is of k successes and ( n
=
(1
k ) failures in n
trials. This question is similar to the one leading to the geometric distribution. Indeed,
if the sequence of independent successes and failures were made to occur in a certain
specified order, the probability of the outcome would be given by p k q n k ; however,
there are n !
k )!] ways to select k items out of a total of n items. Therefore, if
K denotes again the random variable for the number of successes, the total probability
is the sum, or
/
[ k !( n
n !
k )! p k q n k
{
=
}=
P
K
k
(13.33)
k !( n
As before, it is assumed that the events are independent of one another.
Example 13.4. Probability of a year without freezing
An obvious application of the binomial distribution could be the following problem.
From a 45 y temperature record at a certain location it is known that in 26 y (out of 45)
the temperature did not go below 0 C. Thus the probability of a frost-free year can be
estimated to be p
45). Equation (13.33) then provides the probability of k frost-
free years during a period of n years; in Figure 13.7 the results are illustrated for the
different values of k during an 8 y period. A related problem is the determination of the
probability of at least a certain number of years without freezing temperatures during a
period of n years; for instance, what is the probability that during the next 5 y the site
will experience a year with freezing temperatures not more than twice? The answer is
=
(26
/
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