Download 5.5 Geometric Distributions and Negative Binomial Distributions

5.5 Geometric Distribution and
Negative Binomial Distribution
Example: Roll a die repeatedly until number 6 is observed.
Let X be the number of times the die has to be rolled to observe
a first 6.
The possible values for X are 1, 2, 3,……What is P(X = 5)?
Solution:
{X=5}={First 4 rolls are not 6 and 5th one is a 6.}
The probability that a roll does not result in a 6 is 5/6.
The probability that a roll results in a 6 is 1/6.
P(X = 5)=(5/6)4(1/6)
P(X = k)= (5/6)k-1(1/6) , k = 1, 2, …
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
Geometric Distribution If repeated independent
trials (Bernoulli trial) can result in a success with
probability p and a failure with probability q = 1 – p,
then the probability distribution of random variable X,
the number of the trial on which the first success
occurs is
g(x; p) =P(X = x) = qx-1p,

x = 1, 2,3, …
Theorem 5.4 The mean and variance of a random
variable following the geometric distribution g(x; p)
are
 = 1/p
,
1 p
  2
p
2
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Example
It is known that the probability of being able to log on to a
computer from a remote terminal at a given time is 0.90. Let
X denote the number of attempts that must be made to gain
access to the computer.
(a) Find a close form for f(x), the probability distribution of X.
f(x) =(0.1) x 1 (0.9)
 = 1/p
, x=1,2,···
=1/0.9=10/9
x
(b) Find a close form for F(x), 
the
(0cumulative
.1) k 1 (0.9) distribution of X.
F(x) = P(X  x) =
x
 (0.1)
= 0.9
k 1
k 1
k 1
1  0 .1 x
= 0.9 1  0.1
1 0.1x
=
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Example
I am selling my house, and have decided to accept the
first offer exceeding $ K. Assume that offers (money
offered to buy the house) are independent random
variables with common distribution F, find the
expected number of offers received before I sell the
house.
Solution: N = the number of offers before I sell the
house. Find E(N)
N has a geometric distribution.
N ~g(x, p)
X = a price offered by a potential buyer.
X has probability distribution F(x).
q = P(X ≤ K) = F(K)
E(N) = 1/p =1/[1 – F(K)].
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Negative Binomial Distribution
Consider repeated independent trials that can result
in “success” with probability p and a “failure” with
probability q = 1 – p. Let X be the number of the trial on
which the kth success occurs. X is said to be a negative
binomial random variable, and its probability distribution
is called the negative binomial distribution with
parameters k and p.
b*(x; k, p) =P(X = x)
E(X) = k/p
 x  1 k x  k
=
p q
 k  1
x=k,k+1, …
and D(X) = kq/p2
When k=1, negative binomial distribution becomes
geometric distribution .
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Example
Toss a coin over and over again until the
third head is observed. What is probability
that the third head occurs at 5th toss?
Solution:
(k = 3, x = 5)
P(X = 5) =
=
 5  1

(1 / 2) 3 (1 / 2) 53
 3  1
 4
 (1 / 2)3 (1 / 2) 2
 2
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