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8
PROBABILITY
DISTRIBUTIONS
AND STATISTICS
Copyright © Cengage Learning. All rights reserved.
8.4
The Binomial Distribution
Copyright © Cengage Learning. All rights reserved.
Probabilities in Bernoulli Trials
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Probabilities in Bernoulli Trials
Let’s reexamine the computations we performed in the last
example. There, it was found that the probability of
obtaining exactly one success in a binomial experiment
with four independent trials with probability of success in a
single trial p is given by
P(E) = 4pq3
(where q = 1 – p)
(11)
Observe that the coefficient 4 of pq3 appearing in Equation
(11) is precisely the number of outcomes of the experiment
with exactly one success and three failures, the outcomes
being
SFFF FSFF FFSF FFFS
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Probabilities in Bernoulli Trials
Another way of obtaining this coefficient is to think of the
outcomes as arrangements of the letters S and F.
Then the number of ways of selecting one position for S
from four possibilities is given by
C (4, 1) =
=4
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Probabilities in Bernoulli Trials
Next, observe that because the trials are independent,
each of the four outcomes of the experiment has the same
probability, given by
pq3
where the exponents 1 and 3 of p and q, respectively,
correspond to exactly one success and three failures in the
trials that make up each outcome.
As a result of the foregoing discussion, we may write
Equation (11) as
P(E) = C(4, 1)pq3
(12)
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Probabilities in Bernoulli Trials
We are also in a position to generalize this result.
Suppose that in a binomial experiment the probability of
success in any trial is p.
What is the probability of obtaining exactly x successes in n
independent trials? We start by counting the number of
outcomes of the experiment, each of which has exactly x
successes.
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Probabilities in Bernoulli Trials
Now, one such outcome involves x successive successes
followed by (n – x) failures—that is,
SS . . . S FF . . . F
(13)
The other outcomes, each of which has exactly x
successes, are obtained by rearranging the S’s (x of them)
and F’s (n – x of them).
There are C(n, x) ways of arranging these letters.
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Probabilities in Bernoulli Trials
Next, arguing as in Example 1, we see that each such
outcome has probability given by
pxqn – x
For example, for the outcome (13), we find
P(SS . . . S FF . . . F) = P(S)P(S) . . . P(S) P(F)P(F) . . .P(F)
= pp . . . p qq . . . q
= pxqn – x
Let’s now state this important result formally.
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Probabilities in Bernoulli Trials
If we let X be the random variable that gives the number of
successes in a binomial experiment, then the probability of
exactly x successes in n independent trials may be written
P(X = x) = C(n, x)pxqn – x
(x = 0, 1, 2, . . . , n)
(14)
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Probabilities in Bernoulli Trials
The random variable X is called a binomial random
variable, and the probability distribution of X is called a
binomial distribution.
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Example 2
A fair die is rolled five times. If a 1 or a 6 lands uppermost
in a trial, then the throw is considered a success.
Otherwise, the throw is considered a failure.
a. Find the probabilities of obtaining exactly 0, 1, 2, 3, 4,
and 5 successes in this experiment.
b. Using the results obtained in the solution to part (a),
construct the binomial distribution for this experiment,
and draw the histogram associated with it.
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Example 2(a) – Solution
This is a binomial experiment with X, the binomial random
variable, taking on each of the values 0, 1, 2, 3, 4, and 5
corresponding to exactly 0, 1, 2, 3, 4, and 5 successes,
respectively, in five trials.
Since the die is fair, the probability of a 1 or a 6 landing
uppermost in any trial is given by
p= = ,
from which it also follows that
q=1–p= .
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Example 2(a) – Solution
cont’d
Finally, n = 5, since there are five trials (throws of the die)
in this experiment. Using Equation (14), we find that the
required probabilities are
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Example 2(a) – Solution
cont’d
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Example 2(b) – Solution
cont’d
Using these results, we find the required binomial
distribution associated with this experiment given in
Table 12.
Table 12
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Example 2(b) – Solution
cont’d
Next, we use this table to construct the histogram
associated with the probability distribution (Figure 12).
The probability of the number of successes in five throws
Figure 12
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Probabilities in Bernoulli Trials
The following formulas (which we state without proof ) will
be useful in solving problems that involve binomial
experiments.
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Probabilities in Bernoulli Trials
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Probabilities in Bernoulli Trials
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Probabilities in Bernoulli Trials
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Example 4 – Solution
For the experiment in Example 2 compute the mean, the
variance, and the standard deviation of X by (a) using
Equations (15a), (15b), and (15c) and (b) using the
definition of each term.
Solution:
a. We use Equations (15a), (15b), and (15c),
with p = , q =
and n = 5, obtaining
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Example 4 – Solution
cont’d
We leave it to you to interpret the results.
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Example 4 – Solution
cont’d
b. Using the definition of expected value and the values of
the probability distribution shown in Table 12,
we find that
 = E(X)  (0)(.132) + (1)(.329) + (2)(.329)
+ (3)(.165) + (4)(.041) + (5)(.004)
 1.67
which agrees with the result obtained in part (a).
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Example 4 – Solution
cont’d
Next, using the definition of variance and  = 1.67, we find
that
Var(X) = (.132) (–1.67)2 + (.329) (–0.67)2 + (.329) (0.33)2
+ (.165) (1.33)2 + (.041) (2.33)2 + (.004) (3.33)2
 1.11
X =

 1.05
which again agrees with the preceding results.
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Practice
p. 472 Self-Check Exercises #2
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