Download MTH/STA 561 BERNOULLI AND BINOMIAL DISTRIBUTION A

MTH/STA 561
BERNOULLI AND BINOMIAL DISTRIBUTION
A simple experiment is one that may result in either of two possible outcomes. We can
think of many examples of such experiments: a toss of a coin (head or tail), performance of a
student in a course (pass or fail), the sex of a yet-to-be-born child (male or female), placing
a satellite in orbit around the earth (success or failure). We will call an experiment with two
possible outcomes a Bernoulli trial and will label the two outcomes success and failure.
De…nition 1. A Bernoulli trial is an experiment with two possible outcomes, success
or a failure. The sample space for a Bernoulli trial is S = fsuccess, failureg.
The probability distribution for a Bernoulli trial can be easily speci…ed and depends only
upon the single parameter p, the probability of success, and then the probability of failure
is q = 1 p. In order that it be possible for either outcome to occur, the parameter p lie
between 0 and 1, exclusively. Any experiment can be used to de…ne a Bernoulli trial simply
by labeling some event A as “success”and calling its compliment A a “failure”. In this case,
p = P (A) and q = P A . A Bernoulli random variable can be de…ned on a Bernoulli trial
or any more complicated sample space.
De…nition 2. Let S be the sample space for an experiment, let A
with p = P (A), 0 < p < 1, and de…ne
Y =
S be any event
if event A occurs
if event A occurs.
1
0
Then the random variable Y is called the Bernoulli random variable with parameter p. If
the experiment is actually a Bernoulli trial, we simply let A = fsuccessg.
The probability distribution for a Bernoulli random variable follows directly from the
probability distribution for S. Since Y = 1 if and only if event A occurs, we have P fY = 1g =
P (A) = p. Similarly, since Y = 0 if and only if event A occurs, we have P fY = 0g =
P A = 1 p = q. No other value can occur for Y .
Bernoulli Distribution. The probability distribution for a Bernoulli random variable
with parameter p is called Bernoulli distribution and given by
p (y) =
py q 1
0
y
if y = 0; 1
elsewhere.
The mean and variance of the Bernoulli distribution can be readily obtained as illustrated
in the following theorem.
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Theorem 1. Let Y be a Bernoulli random variable with success probability p. Then
and
= E (Y ) = p
2
= V ar (Y ) = pq
where q = 1 p.
Proof. By de…nition,
= E (Y ) = 1 p + 0 q = p
and
E Y 2 = 12 p + 02 q = p:
Thus,
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Example 1.
aircraft. Let
Y =
=E Y2
2
=p
p2 = p (1
p) = pq:
Suppose that a radar has a probability 0:98 of detecting an intruding
1
0
if the radar detects the intruding aircraft
if the radar fail to detect the intruding aircraft.
Clearly, Y is a Bernoulli random variable with parameter p = 0:98, and has probability
distribution
(0:98)y (0:02)1 y
if y = 0; 1
p (y) =
0
elsewhere.
That is, P fY = 1g = 0:98 and P fY = 0g = 0:02. We also …nd that
= 0:98
and
2
= (0:98) (0:02) = 0:0196:
If a Bernoulli trial is repeated n times independently, then the resulting experiment is
referred to as a Binomial experiment. For example, tossing a coin once is a Bernoulli trial
whereas tossing a coin 10 times is a binomial experiment.
De…nition 3. An experiment that consists of n (…xed) repeated independent Bernoulli
trials, each with probability of success p, is called a binomial experiment. The number successes observed during the n trials is called the binomial random variable with n trials and
parameter p.
More precisely, a binomial experiment is one that possesses the following properties:
1. The experiment consists of n repeated identical trials.
2. Each trial results in one of two outcomes. We shall call one outcome a success and
the other a failure.
3. The probability of success on a single trail is equal to p and remains the same from
trial to trial. The probability of a failure is equal to q = 1 p.
4. The repeated trials are independent.
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5. The random variable of interest is Y , the number of successes observed during the n
trials.
Example 2. The following are typical binomial experiments:
1. Toss a fair coin ten times and observe the number, Y , of heads. Then Y is a binomial
random variable with n = 10 and p = 1=2.
2. An early warning detection system for aircraft consists of four identical radar units
operating independently of one another. Suppose that each unit has a probability of 0:98
of detecting an intruding aircraft. When an intruding aircraft enters the scene the random
variable of interest is Y , the number of radar units that detect the aircraft. Clearly, Y is a
binomial random variable with n = 4 and p = 0:98.
3. Suppose that 40% of a large population of registered voters favor candidate Mr. Jones.
A random sample of …fteen voters will be selected and the number, Y , of voters favoring Mr.
Jones is to observed. Then Y is a binomial random variable with n = 15 and p = 0:40.
The natural sample space for a binomial experiment is the Cartesian product of Bernoulli
trial sample space with itself n times. That is, the binomial sample space is
S = S1
S2
Sn
where Si = fsuccess, failureg for i = 1; 2;
; n. Each element of S is an n-tuple (! 1 ; ! 2 ;
where ! i represents success or failure on the ith trial for i = 1; 2;
; n. That is,
S = f(! 1 ; ! 2 ;
; ! n ) j ! i is success or failure for i = 1; 2;
; ! n ),
; ng
Since, for each Bernoulli trial,
p = probability of success
q = 1 p = probability of failure
and the trials are independent, we compute the probability of occurrence of a single outcome (single n-tuple) of S by multiplying the probabilities of occurrence of the appropriate
outcomes for the given trials. For example, the probability of occurrence of outcome
(success, success,
is equal to p p
p = pn , whereas the probability of occurrence of outcome
(failure, failure,
is equal to q q
, success)
, failure)
q = qn.
Example 3. For a binomial experiment with n = 3 trials and the probability of success,
p, for each trial, the probabilities of occurrence of outcomes are given as follows.
3
Outcome
(success, success, success)
(success, success, failure)
(success, failure, success)
(success, failure, failure)
(failure, success, success)
(failure, success, failure)
(failure, failure, success)
(failure, failure, failure)
Probability
p p p = p3
p p q = p2 q
p q p = p2 q
p q q = pq 2
q p p = p2 q
q p q = pq 2
q q p = pq 2
q q q = q3
It can be easily veri…ed that the sum of the probabilities assigned to the outcomes does in
fact add to one, as it must.
Now let Y be the binomial random variable that counts the number of successes observed
in a binomial experiment with n trials. Each sample point in the sample space can characterized by an n-tuple involving the letters S and F corresponding to success and failure. A
typical sample point would thus appear as
SSF SF F F SF S
FS
where the letter in the ith position (proceeding from left to right) indicates the outcome
of the ith trial. Now consider a typical sample point with y successes and n y failures
which is hence contained in the event fY = yg. This sample point is the intersection of n
independent trials, y successes and n y failures, and hence its probability is
pp
p
q = py q n
qq
y
Every other sample point in the event fY = yg will appear as a rearrangement of the S 0 s
and F 0 s in the sample point described above and will therefore contain y S 0 s and n y F 0 s
and be assigned the same probability. Since the number of distinct arrangements of y S 0 s
and n y F 0 s is
n
n!
=
y
y! (n y)!
it follows that the probability distribution for Y is given by the following formula.
Binomial Distribution. Consider a binomial experiment with n trials and parameter
p, the probability distribution of the binomial random variable Y , the number of successes
in n independent trials, is
8
< n y n y
p q
for y = 0; 1; 2;
;n
p (y) =
y
:
0
elsewhere.
Example 4. Suppose that a student is given a test with 10 true-false questions. Also
assume that the student is totally unprepared for the test and guesses at the answer to every
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question. Then the probability that the student answers all questions correctly is
p (10) =
10
10
10
1
2
0
1
2
=
10
1
2
=
1
1024
whereas the probability that exactly four questions are answered correctly is
10
4
p (4) =
1
2
4
1
2
6
= 210
1
2
10
=
105
:
512
Example 5. Suppose that a huge lot of electrical components contains 5% defectives.
If a random sample of …ve components is tested, then the probability of …nding at least one
defective is
P (at least one defective) = 1
P (no defectives) = 1
5
(0:05)0 (0:95)5 0
0
0:774 = 0:226
= 1
= 1
p (0)
Note. The term “binomial distribution” derives from the fact that the probability
n y n y
p q , y = 0; 1; 2;
, n, are terms of the binomial expansion
y
(q + p)n =
n
P
y=0
=
n y n
p q
y
y
n n
n
q +
pq n
0
1
1
+
n 2 n
pq
2
2
+
+
n n
p
n
Observe that
p (0) =
n n
q ; p (1) =
0
n
pq n 1 ; p (2) =
1
n 2 n 2
pq ;
2
; p (n) =
n n
p
n
Since q + p = 1, we conclude that
n
P
p (y) =
y=0
n
P
y=0
n y n
p q
y
y
= (q + p)n = 1
A tabulation of binomial probability in the cumulative form
a
P
y=0
p (y) =
a
P
y=0
n y n
p q
y
y
presented in Table 1 of the Appendix, will greatly reduce the computations associated with
some of the exercises. We illustrate the use of this table in the following examples.
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Example 6. It is known from experience that 40% of all orders received by a shipping
company are placed by the Mallon Company. If ten orders are received by the shipping
company, then from Table 1 of the Appendix, we …nd the probability that
(1) no more than six are received from the Mallon Company is
P (Y
6) =
6
P
6
P
p (y) =
y=0
10
(0:4)y (0:6)10
y
y=0
y
= 0:9452
=1
0:6331
(2) at least …ve are received from the Mallon Company is
P (Y
5) = 1
P (Y
4
P
4) = 1
p (y)
y=0
= 1
4
P
y=0
10
(0:4)y (0:6)10
y
y
= 0:3669
(3) more than three but less than 8 are received from the Mallon Company is
P (4
Y
7) = P (Y
7)
P (Y
3) =
7
P
3
P
p (y)
y=0
7
P
p (y)
y=0
3
P
10
10
(0:4)y (0:6)10 y
(0:4)y (0:6)10
y
y
y=0
y=0
= 0:9877 0:3833 = 0:6054
=
y
(4) exactly six are received from the Mallon Company is
P (Y = 6) = P (Y
6)
P (Y
5) =
6
P
5
P
p (y)
y=0
5
P
10
10
(0:4)y (0:6)10 y
(0:4)y (0:6)10
y
y
y=0
y=0
= 0:9452 0:8338 = 0:1114:
=
6
P
p (y)
y=0
y
Example 7. Suppose that each electronic component from a large lot fails to survive a
shock test with probability 0:4. If …fteen components are selected at random from the lot,
then from Table 1 of the Appendix, we …nd the probability that
(1) no more than 8 components fail to survive is
P (Y
8) =
8
P
p (y) =
y=0
8
P
y=0
15
(0:4)y (0:6)15
y
y
= 0:9050
(2) at least 10 components fail to survive is
P (Y
10) = 1
P (Y
9) = 1
9
P
p (y)
y=0
=
9
P
y=0
15
(0:4)y (0:6)15
y
6
y
=1
0:9662 = 0:0338
(3) from 3 to 8 components fail to survive is
P (3
Y
8) = P (Y
8)
P (Y
2) =
8
P
p (y)
y=0
2
P
p (y)
y=0
2
P
15
15
(0:4)y (0:6)15
(0:4)y (0:6)15 y
y
y
y=0
y=0
= 0:9050 0:0271 = 0:8779
=
8
P
y
(4) exactly 5 components fail to survive is
P (Y = 5) = P (Y
5)
P (Y
4) =
5
P
p (y)
p (y)
y=0
y=0
4
P
15
15
(0:4)y (0:6)15 y
(0:4)y (0:6)15
y
y
y=0
y=0
= 0:4032 0:2173 = 0:1859
=
5
P
4
P
y
The mean and variance of the binomial distribution are presented in the following theorem.
Theorem 2. Let Y be a binomial random variable based on n trials and success
probability p. Then
= E (Y ) = np
2
and
= V ar (Y ) = npq
Proof. Recall that a binomial experiment with n trials and parameter p is in fact the
sum of n independent Bernoulli random variables. Suppose that n independent Bernoulli
random variables, each with probability, p, of success, are to be performed. De…ne
1
0
Xi =
for i = 1; 2;
if the ith trial result in a success with probability p
if the ith trial result in a failure with probability q
; n, where q = 1
Y =
p. Then
n
X
Xi = X1 + X2 +
+ Xn
i=1
gives the total number of success and thus it is a binomial random variable with parameters
n and p. In view of Theorem 1, E (Xi ) = p and V ar (Xi ) = pq for i = 1; 2;
; n. Therefore,
= E (Y ) = E (X1 ) + E (X2 ) +
+ E (Xn ) = np:
It will be shown in Chapter 5 that if X1 , X2 ,
, Xn are independent random variables,
then
V ar (X1 + X2 +
+ Xn ) = V ar (X1 ) + V ar (X2 ) +
+ V ar (Xn ) :
Consequently,
2
= V ar (Y ) = V ar (X1 ) + V ar (X2 ) +
7
+ V ar (Xn ) = npq
Example 8. Refer to Example 7, the mean and variance are
= E (Y ) = np = (15) (0:4) = 6
and
2
= V ar (Y ) = npq = (15) (0:4) (0:6) = 3:6:
8