Chapter 4. Discrete Probability
Distributions
Section 4.6: Negative Binomial Distribution
Jiaping Wang
Department of Mathematical Science
02/25/2013, Monday
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Outline
Probability Function
Mean and Variance
Examples
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Part 1. Probability Function
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In the last section, the geometric distribution models the
probability of the number of failures prior to the first success in
a sequence of independent Bernoulli trials. What if we were
interested in the number of failures prior to the second success,
or the third success or (in general) the r-th success?
Let X denote the number of failures prior to the r-th success, p denotes the
common probability.
P(X=x) = p(x) =P(The 1st (x+r-1) trials contain (r-1) successes
and (x+r)th trial is a success)
= P(The 1st (x+r-1) trials contain (r-1)
successes)P((x+r)th trial is a success)
𝑟 − 1 1 − 𝑝 𝑥𝑝 = 𝑥+𝑟−1 𝑝𝑟 1 − 𝑝 𝑥
= 𝑥+𝑟−1
𝑝
𝑟−1
𝑟−1
For x=0, 1, …
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Probability Function
The negative binomial distribution function:
𝑟𝑞𝑥 , x= 0, 1, 2, …., q=1-p
P(X=x)=p(x)= 𝑥+𝑟−1
𝑝
𝑟−1
If r=1, then the negative binomial distribution becomes the geometric distribution.
Example 4.19: As in Example 4.15, 20% of the applicants for a certain sales position
Are fluent in English and Spanish. Suppose that four jobs requiring fluency in English
And Spanish are open. Find the probability that two unqualified applicants are
Interviewed before finding the fourth qualified applicant, if the applicants are interviewed
sequentially and at random.
Answer: so r=4, x=2, p=0.2, p(X=2)=
2+4−1
4−1
0.24(1-0.2)2=10*(0.2)4(0.8)2=0.01
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Part 2. Mean and Variance
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Mean and Variance
Let W1 be the number of failures prior to the 1st success,
W2 be the number of failures between 1st and 2nd
success, so for W3, …, then X=∑Wi, and Wi follows the
geometric distribution and independently,
So E(X)=E(∑Wi)= ∑q/p=rq/p and V(X)= ∑V(Wi)=rq/p2.
In summary, 𝐸 𝑋 =
𝑟𝑞
,𝑉
𝑝
𝑋 =
𝑟𝑞
𝑝2
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Example
Suppose we are at a rifle range with an old gun that misfires
5 out of 6 times. Dene “success" as the event the gun fires
and let X be the number of failures before the third success.
Then X is a negative binomial random variable with
parameters (3, 1/6 ). Find E(X) and Var(X).
Answer: E(X)=rq/p=3*5/6*6=15, V(X)=rq/p2=3*5/6*36=90.
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An alternative way, if let n=x+r, then the probability function
becomes
𝑛−1 𝑟
𝑃 𝑋=𝑛 =𝑝 𝑛 =
𝑝 1 − 𝑝 𝑛 − 𝑟, 𝑛 = 𝑟, 𝑟 + 1, 𝑟 + 2, …
𝑟−1
Which can be used to model the count data, such as the
number of accidents in a year, the number of trees in a plot,
or the number of insects on a plant.
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Example 4.21
Barnacle often attach to hulls of ships. Their presence
speeds corrosion and increases drag resistance,
leading to reduced speed and maneuverability. Let X
denote the number of barnacles on a randomly
selected square meter of a ship hull. For a particular
shipyard, the mean and variance of X are 0.5 and 0.625,
respectively. Find the probability that at least one
barnacle will be on a randomly selected meter of a ship
hull.
Answer: E(X)=r(1-p)/p=0.5, V(X)=r(1-p)/p2=0.625, so we have
V(X)=E(X)/p=0.5/p=0.625 p=0.5/0.625=0.8, so P(X ≥ 1)=1-P(X=0)=10.8*0.8=0.36.
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Example
A research scientist is inoculating rabbits, one at a time, with
a disease until he finds two rabbits which develop the disease.
If the probability of contracting the disease 1/6. What is the
probability that eight rabbits are needed?
Let X be the number of rabbits needed until the first rabbit to contract the
disease. Then X follows a negative binomial distribution with r = 2; x = 6;
and p = 1/6. Thus,
P(X=6)=C(2+6-1,6)(1/6)2(5/6)6
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Example
Suppose that 3% of computer chips produced by a certain
machine are defective. The chips are put into packages of 20
chips for distribution to retailers.
(a) What is the probability that a randomly selected package of
chips will contain at least 2 defective chips?
(b) What is the probability that the ten-th pack selected is the
third to contain at least two defective chips?
(a) n=20, p=3%, so P(X≥2)=1-P(X=0)-P(X=1)=120
3% 1 97% 20
1
(b) r=3, x=10, p=P(X ≥2), so P(X=10)=
3+10−1
10
20
0
3%
𝑃 𝑋≥2
0
97%
3[1
20
−
− 𝑃(𝑋 ≥ 2)]10
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HW#6
Page 150: 4.66, 4.67
Page 152: 4.84.
Additional Problem:
1. Find the expected value and the variance of the number of times one must
throw a die until the outcome 1 has occurred 4 times.
2. If the probability is 0.40 that a child exposed to a certain contagious disease
will catch it, what is the probability that the tenth child exposed to the
disease will be the third to catch it?
Due Wed., 03/06/2013
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