Transcript The UNIVERSITY of NORTH CAROLINA at CHAPEL

Chapter 4. Discrete Probability
Distributions
Section 4.5: Geometric Distribution
Jiaping Wang
Department of Mathematical Science
02/13/2013, Monday
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Outline
Probability Function
Mean and Variance
An Alternative Parameterization
Homework #5
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Part 1. Probability Function
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Suppose that a series of test firing of a rocket engine can be
represented by a sequence of independent Bernoulli random
variables with Yi=1 if ith trial is a success and Yi=0, otherwise.
Assume the probability p of each trial is constant, denote X as
the number of failures before the first success, then what is
P(X=x)?
P(X=x) = p(x) = P(Y1=0, Y2=0, …, Yx=0, Yx+1=1)
= P(Y1=0)P(Y2=0)…P(Yx=0)P(Yx+1=1)
= (1-p)(1-p)…(1-p)p
= (1-p)xp=qxp, x= 0, 1, 2, …., q=1-p
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Probability Function
The geometric distribution function:
P(X=x)=p(x)=(1-p)xp=qxp, x= 0, 1, 2, …., q=1-p
P(X=x) = qxp = p[qx-1p]
= qP(X=x-1)
<P(X=x-1)
as q ≤ 1, for x=1, 2, …
A Geometric Distribution Function
with p=0.5
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Example 4.15
A recruiting firm finds that 20% of the applicants for a
particular sales position are fluent in both English and
Spanish. Applicants are selected at random from the
pool and interviewed sequentially. Find the probability
that five applicants are interviewed before finding the
first applicant who is fluent in both English and
Spanish.
Solution: X=5, p=0.2, using the geometric distribution function, we have
P(X=5)=(0.8)5(0.2)=0.066
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Part 2. Mean and Variance
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Geometric Series and CDF
The geometric series: {tx: x=0, 1, 2, …}
𝟏
∞
𝒙
𝒙=𝟎 𝒕 =𝟏−𝒕
Sum of Geometric series: For |t|<1, we have
+𝟏
Sum of partial series:
Then we can verify
∞
𝑥=0 𝑝
𝑥 =
𝒏
𝒙=𝟏−𝒕𝒏
𝒕
𝒙=𝟎
𝟏−𝒕
∞
𝑥=0
1 − 𝑝 𝑥𝑝 = 𝑝
∞
𝑥=0
1−𝑝
𝑥
1
= 𝑝 1−(1−𝑝) = 1
The cumulative distribution function:
+𝟏
F(x)=P(X≤x)=
𝟏−𝒒𝒙
𝒙
𝒕
𝒕=𝟎 𝒒 𝒑=𝐩 𝟏−𝒒
=1-qx+1
And P(X≥x)=1-F(x-1)=qx
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Mean and Variance
The Expected Value
The Variance V(X)=
𝒒
E(X)=
𝒑
𝒒
𝒑𝟐
∞
𝑥
2
3
3
E(X)= ∞
𝑥=0 𝑥𝑝 𝑥 = 𝑝 𝑥=0 𝑥𝑞 = 𝑝 0 + 𝑞 + 2𝑞 + 3𝑞 + ⋯ = 𝑝𝑞[1 + 2𝑞 + 3𝑞 + ⋯ ]
So E(X)/(pq) =[1 + 2𝑞 + 3𝑞3 + ⋯ ]
And E(X)/p = [0 + q + 2q2 + … ]
Thus, E(X)/(pq)-E(X)/p = 1+q+q2+q3+ • • • = 1/(1-q)
1 𝑝𝑞
 E(X)=1−𝑞 1−𝑞 = 𝑞/𝑝
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Example 4.16
Referring to Example 4.15, let X be the number of
unqualified applicants interviewed before the first
qualified one. Suppose that the first applicant who is
fluent in both English and Spanish is offered the
position, and the applicant accepts. Suppose each
interview costs $125.
1. Find the expected value and the variance of the total
cost of interviewing until the job is filled.
2. Within what interval should this cost be expected to
fall?
As X+1 is the number of the trial on which the interviewing ends, so the total
cost C=125(X+1)=125X + 125, thus E(C) = 125E(X)+125 =
125(0.8/0.2)+125=625, V(C)=1252V(X) =1252(0.8/0.04)=312500, the standard
deviation is 559.
By Chebysheff’s Inequality, the cost C will lie within two standard deviations of
its mean at least 75% of the time, so the cost will be between 625-2(559) and
625+2(599).
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Memoryless Property
Memoryless Property means if we have observed j
straight failures, then the probability of observing at
least k more failures (at least j+k failures in total)
before a success is the same as if we were just
beginning and wanted to determine the probability of
observing at least k failures before the first success,
that is:
P(X ≥ j+k | X ≥ j) = P(X ≥ k).
𝑃( 𝑋≥ 𝑗+𝑘 ∩ 𝑋≥ 𝑗 )
P(X≥ j+k|X ≥j)=
𝑃(𝑋≥ 𝑗)
=
𝑃(𝑋≥ 𝑗+𝑘)
𝑃(𝑋≥ 𝑗)
=
𝑞𝑗+𝑘
𝑞𝑗
= 𝑞𝑗 = 𝑃(𝑋≥k)
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Example 4.17
Referring to Example 4.15, suppose that 10 applicants have
been interviewed and no person fluent in both English and
Spanish has been identified. What is the probability that
15 unqualified applicants will be interviewed before
finding the first applicant who is fluent in English and
Spanish?
P(X=15|X≥10)=
𝑃( 𝑋=15 ∩ 𝑋≥ 10 )
𝑃(𝑋≥ 10)
=
𝑃(𝑋=15)
𝑃(𝑋≥ 10)
=
𝑝𝑞15
𝑞10
= 𝑝𝑞5 = 𝑃(𝑋=5)
Which means the probability that 15 unqualified applicants will be interviewed
before finding an applicant who is fluent in English and Spanish, given that the
first 10 are not qualified, is equal to the probability of finding the first qualified
candidate after interviewing 5 unqualified applicants.
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Part 3. An Alternative
Parameterization
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The geometric distribution can serve as a model for a
number of applications that are not associated with
Bernoulli trials. Counting data, such as the number of
insects on a plant or the number of weeds within a
square foot area, may be well modeled by the
geometric distribution.
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Example 4.18
The number of weeds within a randomly selected
square meter of a pasture has been found to be well
modeled using the geometric distribution. For a
given pasture, the number of weeds per square
meter averages 0.5. What is the probability that no
weeds will be found in a randomly selected square
meter of this pasture?
In this example, it doesn’t make sense to talk about Bernoulli trials and the
probability of success. Instead, we have counts 0, 1, 2, … So let X denote the
number of weeds in a randomly selected square meter of the pasture. We know
E(X)=0.5 = (1-p)/p  p=2/3, then we can find P(X=0)=p=2/3, so we can say
there is a probability of (1-2/3) of seeing one or more weeds in a randomly
selected square meter.
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Homework #5
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