Page 110 - Applied statistics and probability for engineers

88 Chapter 3/Discrete Random Variables and Probability Distributions

Example 3-21 Wafer Contamination The probability that a wafer contains a large particle of contamination
is 0.01. If it is assumed that the wafers are independent, what is the probability that exactly 125
wafers need to be analyzed before a large particle is detected?
Let X denote the number of samples analyzed until a large particle is detected. Then X is a geometric random variable
with p = 0 01. The requested probability is
.
(
.
.
P X = 125 ) = (0 99. ) 124 0 01 = 0 0029
The mean of a geometric random variable is
∞ k −1 ∞
μ = ∑ kp (1 − ) p = ∑ kq k −1
p
k =1 k =1
where q = p − 1. The right-hand side of the previous equation is recognized to be the partial
derivative with respect to q of
∞ pq
p∑ q =
k
k=1 1 − q
where the last equality is obtained from the known sum of a geometric series. Therefore,

μ = ∂ ⎢ pq ⎥ ⎥ = p = p = 1
⎢
⎣
∂q 1 − q ⎦ (1 − ) q 2 p 2 p
and the mean is derived. To obtain the variance of a geometric random variable, we can i rst
2
derive E X ( ) by a similar approach. This can be obtained from partial second derivatives with
respect to q. Then the formula V X( ) = E X ) ( EX) is applied. The details are a bit more
−
(
2
2
work, and this is left as a mind-expanding exercise.
Mean and Variance
If X is a geometric random variable with parameter p,
X
E
V X
μ = ( ) = 1 p and È 2 = ( ) = (1 − ) p p 2 (3-10)
Example 3-22 Mean and Standard Deviation Consider the transmission of bits in Example 3-20. Here p = 0 1.
.
=
.
The mean number of transmissions until the irst error is 1 0 1 10. The standard deviation of the

number of transmissions before the irst error is
− )
.
σ = ( ⎡ 1 0 1 0 1. / . 2 ⎤ ⎦ 1 2 / = 9 49
⎣
Practical Interpretation: The standard deviation here is approximately equal to the mean, and this occurs when p is

small. The actual number of trials until the irst success may be much different from the mean when p is small.
Lack of Memory Property
A geometric random variable has been deined as the number of trials until the i rst success.

However, because the trials are independent, the count of the number of trials until the next
success can be started at any trial without changing the probability distribution of the random

variable. For example, if 100 bits are transmitted, the probability that the irst error, after bit 100,
occurs on bit 106 is the probability that the next six outcomes are OOOOOE. This probability
is ( . ) ( . )0 9 5 0 1 = . 0 059, which is identical to the probability that the initial error occurs on bit 6.
The implication of using a geometric model is that the system presumably will not wear
out. The probability of an error remains constant for all transmissions. In this sense, the geo-
metric distribution is said to lack any memory. The lack of memory property will be dis-
cussed again in the context of an exponential random variable in Chapter 4.