Each such trial is called a Bernoulli trial. For convenience, we let Xi be a Bernoulli random variable for trial i. Such a random variable is assigned the value 1 if the trial is a success and the value 0 if the trial is a failure.

**THE BINOMIAL DISTRIBUTION**

As introduced in the previous section, the binomial
random variable is the count of the number of successes in *n* independent trials when the probability of success on any given
trial is *p*. The binomial distribution
applies in situations where there are only two possible outcomes, denoted as *S* for success and *F* for failure.

Each such trial is called a Bernoulli trial. For
convenience, we let *X _{i}* be
a Bernoulli random variable for trial

For *Z*
(the number of successes in *n* trials)
to be *Bi*(*n*, *p*), we must have *n* inde-pendent Bernoulli trials with
each trial having the same probability of success *p*. *Z* then can be
represented as the sum of the *n*
independent Bernoulli random variables *X _{i}
*for

The binomial distribution arises naturally in many
problems. It may represent appropriately the distribution of the number of boys
in families of size 3, 4, or 5, for example, or the number of heads when a coin
is flipped *n* times. It could
represent the number of successful ablation procedures in a clinical trial. It
might represent the number of wins that your favorite baseball team achieves
this season or the number of hits your favorite batter gets in his first 100 at
bats.

Now we will derive the general binomial
distribution, *Bi*(*n*, *p*). We simply
gener-alize the combinatorial arguments we used in the previous section for *Bi*(3, 0.50). We consider *P*(*Z*
= *r*) where 0 ≤ *r* ≤ *n*. The
number of elementary events that lead to *r*
successes out of *n* trials (i.e.,
getting exactly *r* successes and *n – r* failures) is *C*(*n*,* r*) =* n*!/[(*n *–*
r*)!* r*!].

Recall our earlier example of filling slots.
Applying that example to the present situation, we note that one such outcome
that leads to *r* successes and *n – r* failures would be to have the *r* successes in the first *r* slots and the *n – r* failures in the re-maining *n – r* slots. For each slot, the probability of a success is *p*, and the probabil-ity of a failure is
1 – *p*. Given that the events are
independent from trial to trial, the multiplication rule for independent events
applies, i.e., products of terms which are either *p* or 1 – *p*. We see that
for this particular arrangement, *p* is
multiplied *r* times and 1 – *p* is multiplied *n – r* times.

The probability for a success on
each of the first *r* trials and a
failure on each of the remaining trials is *p ^{r}*(1
–

**TABLE 5.2. Binomial Distributions for n = 8 and p ranging from
0.05 to 0.95**

The number of such arrangements is just the number
of ways to select exactly *r *out of
the* n *slots for success. This number
denotes combinations for selecting* r *objects
out of *n*, namely, *C*(*n*,
*r*). Therefore, *P*(*Z* = *r*) = *C*(*n*, *r*)(1
– *p*)* ^{n–r}* = {

Table 5.2 shows for *n* = 8 how the binomial distribution changes as *p* ranges from small values such as 0.05 to large values such as 0.95.
From the table, we can see the relationship between the probability
distribution for *Bi*(*n*, *p*)
and the one for *Bi*(*n*, 1 – *p*). We will derive this relationship algebraically using the
formula for* P*(*Z *=* r*).

Suppose *Z*
has the distribution *Bi*(*n*, *p*);
then *P*(*Z* = *r*) = *n*!/[(*n*
– *r*)!*r*!]*p ^{r}*(1 –

Earlier in this chapter, we noted that *Bi*(*n*,
*p*) has a mean of *μ* = *np* and a variance of *σ*^{2} = *npq*,
where *q* = 1 – *p*. Now that you know the probability mass function for the *Bi*(*n*,
*p*), you should be able to verify
these results in Exercise 5.21.

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