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L. Discrete Distributions.
1. Binomial Distribution.
L1. Text 5.18-5.21, 5.23, 5.27[5.14a-d, 5.15, 5.16, 5.19, 5.20, 5.21] (5.15, 5.16, 5.18*, 5.19, 5.20).
2. Geometric Distribution.
L2-L5.
3. Poisson Distribution.
L6-L8. Text 5.30, 5.31, 5.33, 5.34, 5.35, 5.38[5.29 – 5.30, 5.34 - 5.37] (5.22 - 5.23, 5.26-5.29).
4. Hypergeometric Distribution.
Text 5.44-5.45, 5.49-5.50 [5.22 – 5.23, 5.27 – 5.28] (5.30, 5.31, 5.35, 5.36). L9, L10.
5. Review.
Note that problem L11 is a review of the continuous uniform distribution.
This document includes Problems with the Poisson and Hypergeometric distribution.
----------------------------------------------------------------------------------------------------------------------------- ----
Poisson Distribution Problems
Note that in my notes, m replaces the more common  .
Exercise 5.30 [5.29 in 9th] (5.22in 8th edition): find the following.
a) Px  2 when m  2.5. Solution: Px  
e m m x
is the formula for the Poisson distribution.
x!
e 2.5 2.5 x
. According to my calculator
x!
1
1
 .0820849 . If you can’t manage that, try e 2.5  2.5 
 .0820849 . If you can’t
12
.
182494
e
With a parameter of m  2.5 it becomes Px  
e 2.5
e 2.5 2.52 .0820849 6.25 

2!
2 1
 .25652 . Of course this can be found on Table 14, Poisson Distribution, in the first column of
the Poisson 2.5 part of the table in the k  2 line as .256516.
find e, use e  2.7182818 or get a new calculator. So P2 
e 8.0 8.08 .00035463 16777216 

8!
8  7  6  5  4  3  2 1
 .13959 . But you should find this on the table instead of doing all this work.
c) Px  1 when m  0.5. Solution: m  0.5, x  1. Look in the first column of the Poisson (0.5)
table at the k  8 line. It says P1  303265 .
b) Px  8 when m  8. Solution: m  8.0, x  8. P8 
d) Px  0 when m  3.7. Solution: m  3.7, x  0. Since Px  
e m m x
,
x!
e  m m 0 e  m 1

 e  m . For m  3.7, e 3.7  .02472
0!
1
e) Px  7 when m  4.4. (This is not in the 10th edition). Solution: m  4.4, x  7.
P0 
P7  
e 4.4 4.47 .0122773 31927 .781 

 .077754
7!
7  6  5  4  3  2 1
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Exercise 5.31[5.30 in 9th] (5.23in 8th edition): All of these solutions use the Poisson table. The right-hand
column in each section is the Cumulative distribution. Find the following.
a) Px  2 when m  2. Solution: If m  2.0 , Px  1  Px  2  1  Px  1
 1  .40601  .59399
b) Px  3 when m  8. Solution: If m  8.0 , Px  3  1  Px  2  1  .01375  .98625
c) Px  1 when m  0.5. Solution: If m  0.5 , Px  1  .90980 . This is the probability of at
most one!
d) Px  1 when m  4. Solution: If m  4.0 , Px  0  Px  1  1  Px  0
 1  P0  1  .01832  .98168 . This is the probability of at least one!
e) Px  3 when m  5. Solution: If m  5.0 , Px  4  Px  3  .26503 .
Exercise 5.32 (not in 8th or 9th edition): For a Poisson distribution with a parameter of 5, find (a) P(X =
1), (b) P(X < 1), (c) P(X > 1) and (d) P(X  1).
Solution: All answers are copied from the Instructor’s Solutions Manual and are unchecked. You should be
able to get them using the Poisson table. If you cannot get them, please tell me.
(a)
P(X = 1) = 0.0337
(b)
P(X < 1) = 0.0067
(c)
P(X > 1) = 0.9596
(d)
P(X  1) = 0.0404
Exercise 5.33 (not in 8th or 9th edition): A LAN experiences an average of 2.4 errors per day. For a given
day, find the probability of a) no errors, b) exactly one error, c) at least two errors and d) fewer than 3
errors.
Solution: All answers are copied from the Instructor’s Solutions Manual and are unchecked. You should be
able to get them using the Poisson table. If you cannot get them, please tell me.
(a)
P(X = 0) = 0.0907
(b)
P(X = 1) = 0.2177
(c)
P(X  2) = 0.6916
(d)
P(X < 3) = 0.5697
Exercise 5.34 (5.26 in 8th edition): Chocolate chip cookies contain an average of 6 chip parts per cookie.
What is the probability that a given cookie will have a) fewer than 5 chip parts, b) exactly five chip parts,
c) at least five chip parts or d) five or six chip parts. In the 9 th edition, you were asked to do the problem
again e) assuming an average of 5 chip parts per cookie.
Solution: The distribution is Poisson with a parameter of 6.0.
a) Px  5  Px  4  .28506
b) Px  5  .160623
c) Px  5  1  Px  4  1  .28506  .71494
d) P4  x  5  Px  5  Px  3  .44568  .15120  .29448
Instead of answering the above, the Instructor’s Solutions Manual provides the following Excel
table.
251solnL2 3/29/06 (Open this document in 'Page Layout' view!)
e) Now the parameter is 5.0.
(a) Px  5  Px  4  .44049
(b) Px  5  .175467
(c) Px  5  1  Px  4  1  .44049  .55951
(d) P4  x  5  Px  5  Px  3  .61596  .26503  .35093
Exercise 5.35 (5.27 in 8th edition): This is a variation on the previous problem. Assume an average of 6
chips per cookie. Now assume that an inspector will discard any cookies with fewer than 4 chips. How
many will be discarded from a batch of 100.
Solution: The parameter remains m  6.0. We discard cookies with fewer than 4 chips.
Px  4  Px  3  .15120 . This represents the proportion that have fewer than 4 chips, so the quantity is
100  0.15120   15.120 . Seems like it’s 15 or 16.
[Exercise 5.36 in 9th] (5.28 in 8th edition) This is not in the 10th edition but see 5.36 and 5.37, which are
lots harder: An airline averages 9 claims per day for missing baggage. What is the probability that it will
get a) fewer than 3 claims, b) exactly 3 claims, c) at least 3 claims and d) more than three claims?
Solution: The distribution is Poisson9
a) Px  3  Px  2  .00623
b) Px  3  .014994
c) Px  3  1  Px  2  1  .00623  .99377
d) Px  3  1  Px  3  1  .02123  .97877
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Exercise 5.38 [5.37 in 9th] (5.29 in 8th edition): The number of flaws in a roll of paper follows a Poisson
distribution with a mean of 1 flaw in 5 feet of paper. Find a) the probability of at least two flaws in a 1 foot
roll, b) at least one flaw in a 12 foot roll and between 5 and 15 flaws in a 50 foot roll.
Solution: One flaw in 5 feet can be changed to 1 5  0.2 flaws per foot. The mean of 0.2 flaws per foot
must be customized for each part of this problem.
a) For 1 foot, m  0.2 . Px  2  1  Px  1  1  .98248  .01752
b) For 12 feet, m  0.2 12   2.4 . Since Px  
e m m x
, P0  e 2.4  .0907179.
x!
Px  1  1  Px  0  1  .0907179  .90928
c) For 50 feet, m  0.2 50   10 .0. P5  x  15   Px  15   Px  4
 .95126  .02925  .92201
Problem L6: If average sales for a six week period are 20 units, and it takes six weeks to get a delivery,
what is my reorder point if I want a 95% certainty of avoiding a stockout? What about a 90% certainty?
Solution: In order to do this problem we must have the correct mean for the relevant period or space. In
this case it is 6 weeks and the mean is 20. In the Poisson table with a mean of 20, we find that
Px  27   .94752 , Px  28   .96567 , etc. Since the smallest value of x with a cumulative probability
above 95% is 28, our reorder point is 28. If we lower the probability to 90%, we find the smallest value of
x with a cumulative probability above 90% is 26, so our reorder point is 26.
Problem L7: In Frunze there are, on the average, 36 earthquakes a year.
a. If I take a one month vacation in Frunze, what is my chance of at least one earthquake?
b. What about between five and eight earthquakes?
Solution: a) the relevant unit of time and space this time is one month and the mean for one month is about
36  3 . If we use the Poisson table with m  3 , we find that Px  1  1  P0  1  .04979  .95021 .
12
b) P5  x  8  Px  8  Px  4  .99620  .81526  .18094 .
Problem L8: After showing that the Poisson Distribution can be used for this problem, find the solution to
the Binomial problem Px  3 when p  .01 and n  100 using the Poisson Distribution. How does this
compare with the solution from a binomial table?
n
n 100
 500 . In this case

 10000  500 , so use the Poisson distribution.
Solution: The usual test is if
p
p .01
To find the mean, use m  np  100 .01  1 . So from the Poisson table with parameter of one,
Px  3  .98101 . From the Binomial table, Px  3  .98163 . The error is about 6 in 10000, which is
very small for most purposes.
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 5
Hypergeometric Distribution Problems The text uses A where I use M and C 35    .
 3
Exercise 5.44 [5.22 in 9th] (5.30 in 8th edition):
Find the following: (a) P3 if n  4, N  10, M  5 ; (b) P 1 if n  4, N  6, M  3 ; (c) P0 if
n  5, N  12, M  3 and (d) P3 if n  3, N  10, M  3 .
Solution: For the Hypergeometric distribution we have Px  
C nNxM C xM
C nN
or Px  
C xM C nNxM
C nN
or
 M  N  M 
 

 x  n  x 
, which gives the probability of x successes in a sample of n taken from a
Px  
N
 
 n
M 
M!
population of N in which there are M successes. Remember that C xM    
. The
x

M
 x ! x!
 
Instructor’s Solutions Manual provides the following solutions.
(a) n  4, N  10, M  5
 5  10 – 5  5  4  3! 5  4!
   


5
 3  4 – 3 
P3  P( x  3) 
 3!2 1 4!1! 
 0.2381
10  9  8  7  6! 3  7
10 
 
6!4  3  2 1
4
(b) n  4, N  6, M  3
 3   6 – 3  3  2! 3!
   


 1   4 – 1  2!1 3!0! 1
P( x  1) 

  0.2
6  5  4!
5
 6
 
4
!

2

1
 4
(c) n  5, N  12, M  3
 3  12 – 3 
3! 9  8  7  6  5!
   


7
 0   5 – 0  3!0! 5!4  3  2 1
P( x  0) 


 0.1591
12

11

10

9

8

7
!
44
12 
 
7!5  4  3  2 1
5
(d) n  3, N  10, M  3
 3  7 – 0 
3! 7!
   


1
 3   3 – 3  3!0! 7!0!
P( x  3) 


 0.0083
10  9  8  7! 120
10 
 
7!3  2 1
3
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Exercise 5.45 [5.23 in 9th] (5.31in 8th edition): Find the mean and standard deviation for the following
Hypergeometric distributions:: (a) n  4, N  10, M  5 ; (b) n  4, N  6, M  3 , (c)
n  5, N  12, M  3 and (d) n  3, N  10, M  3
Solution:
M
N n
npq , so
If p 
, the mean is   np and the variance is  2 
N
N 1
N n
 N n
 npq
 . q  1 p .
N 1
 N 1 
The Instructor’s Solutions Manual provides the following solutions (heavily edited):
  npq
p
(a) n  4, N  10, M  5
5
 .5, q  1  .5  .5
10
10 – 4
 0.8165
10 – 1
3
p   .5, q  1  .5  .5
6
 = np = 4•(0.5) = 2
  4(0.5)(0.5) 
(b) n  4, N  6, M  3
 = np = 4•(0.5) = 2
6–4
 0.6325
6 –1
3
p
 .25, q  1  .25  .75
12
  4(0.5)( 0.5) 
(c) n  5, N  12, M  3
12 – 5
 0.7724
12 – 1
3
p
 .3, q  1  .3  .7
10
 = np = 5•(0.25) = 1.25
  5(0.25)(0.75) 
(d) n  3, N  10, M  3
  3(0.3)( 0.7) 
 = np = 3•(0.3) = 0.9
10 – 3
 0.70
10 – 1
Exercise 5.49 [5.27 in 9th] (5.35 in 8th edition): A state lottery is conducted in which the 6 winning
numbers are selected from a total of 54 numbers. This seems to mean that the 54 numbers consist of 6
winning numbers and 48 non-winning numbers. A sample of six is selected form the population of 54
numbers. In the 9th edition we are asked the probability that the six randomly chosen numbers include a) all
six winning numbers, b) exactly five winning numbers, c)( not in 10th) exactly four winning numbers, d)
(not in 10th) exactly 3 winning numbers, and e) ( c) in 10 th) no winning numbers. f) ( d) in 10th). Revise
your probabilities assuming that the 6 are selected from 40 numbers instead of 54.
Solution: The Instructor’s Solutions Manual provides the following solutions.
(a)
If n = 6, M = 6, and N = 54, P(X = 6) = 3.8719 x 10–8  0.00000004
(b)
If n = 6, M = 6, and N = 54, P(X = 5) = 1.1151 x 10–5  0.00001
(c)
If n = 6, M = 6, and N = 54, P(X = 4) = 0.0007
(d)
If n = 6, M = 6, and N = 54, P(X = 3) = 0.0134
(e)
If n = 6, M = 6, and N = 54, P(X = 0) = 0.4751
(f)
(a)
If n = 6, M = 6, and N = 40, P(X = 6) = 2.6053 x 10–7  0.0000003
(b)
If n = 6, M = 6, and N = 40, P(X = 5) = 5.3147 x 10–5  0.00005
(c)
If n = 6, M = 6, and N = 40, P(X = 4) = 0.0022
(d)
If n = 6, M = 6, and N = 40, P(X = 3) = 0.0312
(e)
If n = 6, M = 6, and N = 40, P(X = 0) = 0.3504
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Exercise 5.50 [5.28 in 9th] (5.36 in 8th edition):
1 of a shipment of 15 hard disks are defective. 4 disks are inspected. What is the chance of finding a)
3
exactly one defective, b) at least one defective and c) no more than two defective. d) What is the mean
number of defective disks in a sample of 4?
Solution: p  13 , n  4, N  15, M  np  15 13   5
a) P1 
C15 C 310
C 415
10!
10  9  8
5
7!3!
3  2 1  5 10  9  8  4  .4395604


15!
15 14 13 12 15 14 13 12
11! 4!
4  3  2 1
5
 10! 
 1


  1   6! 4!   1  10  9  8  7  1  .1538461  .84615
 15! 

15 14 13 12



11
!
4
!


c) Px  2  P0  P1  P2  .153846  .43956  .32967  .9231
Instead of answering the above, the Instructor’s Solutions Manual provides the following Excel
table.
 C 5 C 10
b) Px  1  1  P0  1   0 154
 C
4

(d)
 = np = 4•(0.333) = 1.3
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Note that the next 2 problems involve common exam glitches. In problem L9, remember that if a
sample is less than 1/20 of a population, the Binomial distribution substitutes for the
Hypergeometric. In Problem L10, a majority of 10 is not exactly 6, it is 6 or more.
Problem L9: We have fifteen units of equipment in a bin of which five are defective. Pull three out at
random. What is the probability that exactly one will be defective if we:
a. sample without replacement?
b. sample with replacement?
C N M C M
Solution: For the Hypergeometric distribution we have Px   n  x N x , which gives the probability of
Cn
x successes in a sample of n taken from a population of N in which there are M successes. If p 
the mean is   np and the variance is  2 
p
M
,
N
N n
npq . In this case N  15 , M  5 , n  3 and
N 1
N n
15  3 1  2 
5 1
1
npq 
3    0.57143 .   0.75593
 , so that   np 3  1 and  2 
N 1
15  1 3  3 
15 3
3
a) P1 
C 31515 C15
C 315

C 210 C15
C 315
10! 5!
8! 2! 4!1!

 .4945
15!
12!3!
b) If we sample with replacement or take our sample of 3 from a population that is larger than 60 (20 times
1
3), but is still 13 defective, we can use the Binomial distribution with p  , and n  3 .
3
1
2
1  2
 1  4  4
Px  C xn p x q n x , so P1  C13      3     .4444
 3  3
 3  9  9
Problem L10: In a population of 10 students 60% prefer Coke, the remainder Pepsi. If a sample of four is
taken, (a) what is the chance that exactly three prefer Coke? (b) What is the chance that a majority prefer
Coke? (Hint: The answer to (a) is not the same as the answer to (b) .) (c, d) Redo (a) and (b) assuming that
there are many students in the population (but the sample is still 3 individuals).
Solution: a) Hypergeometric: N  10 , n  4 , M  Np  10.6  6 . Px  
P3 
C36C14
C410
C xM CnNxM
C Nn
 6! 
654

4
4
3!3! 


 3  2  1  .3810 .
10  9  8  7
 10! 


4  3  2 1
 6!4! 
b) Px  3  P3  P4
P4 
C46C04
C410
 6! 
65

1
1
2!4! 

2 1


 .0714
10  9  8  7
 10! 


4  3  2 1
 6!4! 
Px  3  .3810  .0714  .4524 . A majority is 3 or 4!
c) Binomial: p  .6 , n  4 . In 4 tries, 3 successes = 1 failure, 4 successes = 0 failures.
P3; p  .6  P1; p  .4  Px  1  Px  0  .4752  .1296  .3456 .
d) Px  3; p  .6  Px  1; p  .4  .4752 .
.