1. Discrete Random Variables and Probability Distributions3

2. 3.1
Random Variables

3. Random Variables Definition
A random variable (rv) is any rule that associates a number with each outcome in sample space S.
Mathematically, a random variable is a function
whose domain is the sample space S, and
whose value range is a set of real numbers.

4. Random Variables Random variables are denoted by X, Y, etc.
Use x, y to represent specific values of rvs X, Y.
→ X (s) = x means that x is the value of rv X associated with the outcome s.

5. Example 3.1
Calling help desk for computer support will have two possible outcomes: (i) Get someone (S, for success); and
(ii) put on hold (F, for failure). With sample space {S, F}, define rv X by X (S) = X (F) = 0 The rv X indicates whether the student can immediately speak to someone (1), or not (0).

6. Random Variables Definition
A random variable with only two possible values of 0 and 1 is called a Bernoulli random variable.
Example
Let X denote the outcomes of tossing a quarter, with
X(W) = 1, and X(E) = 0.
Then this X is a Bernoulli rv.

7. Example 3.3
Observe the # of pumps in use at 2 6-pump gas stations. Define rv’s X, Y, and U by X = the total # of pumps in use at the two stations Y = the difference between the # of pumps in use at station 1 and the # in use at station 2 U = the maximum of the numbers of pumps in use at the two stations
X, Y, and U are rvs.

8. Example 3.3 For observation (2, 3), determine the values of X, Y, & U.
cont’d
For observation (2, 3), determine the values of X, Y, & U.
X((2, 3)) = = 5, so we say that the observed value of X is x = 5.
Y((2, 3)) = 2 – 3 = – 1 → The observed value of Y is y = –1.
U((2, 3)) = max(2, 3) = 3 → Observed value of U is u = 3.

9. Example 3.3 X, Y, & U all have the same domain: The sample space
cont’d
X, Y, & U all have the same domain: The sample space
S = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6)
(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
There are 49 possible outcomes in S.

10. Example 3.3 They have different value ranges.
cont’d
They have different value ranges.
Value range of rv X: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
Range of rv Y: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6}.
Range of rv U: {0, 1, 2, 3, 4, 5, 6}.

11. Two Types of Random Variables
Definitions
A discrete random variable: One whose range is a finite set or countable infinite set of numbers.
X, Y, U in Example 3.3 are all discrete random variables.
A continuous random variable: One
(1) whose range includes all numbers in a single interval or the union of several disjoint intervals (e.g., [0, 10][20, 30]).
(2) P(X = c) = 0 for any c in the range of rv X.
Remark: P(X = c) = 0 for any c NOT in the range of rv X.

12. Example 3.5
Randomly select a spot in continental US and find its height above sea level, denoted by Y.
The Y is a continuous rv.
The domain of Y is sample space S = continental US.
The highest spot in US is 14,494 ft at Mt. Whitney
The lowest spot is –282 ft at Death Valley
Therefore, the value range of Y is [–282 ft, 14,494 ft].

13. Example 3.6
Married couples are selected at random, and a blood test is done on both until we find a couple with the same Rh factor.
Define rv X = # of blood tests to be performed, the value range of X is D = {2, 4, 6, 8, …}.
Since the value range is a countable infinite set of numbers, X is a discrete rv.
What is the domain of X?
It depends on what is of interest!
All married couples in Florida State
All married couples of age in Hillsborough County

14. More Examples Randomly select a student at USF. Define
X = 0, if the selected student is male; and 1, otherwise.
Range: {0, 1}
Y = the weight of the selected student.
Range: [70, 400] lbs
Z = the height in meters of the selected student.
Range: (4, 7) ft
U = 0, 1, and 2, if the selected student is from Florida, out-of-state, and a foreign country, respectively.
Range: {0, 1, 2}
X, Y, Z, U have the same domain: USF student body.

15. Exercise Problem 4
X = # of non-zero digits in a randomly selected zip code. Domain = Sample space S = {all US zip codes}.
A zip code has 5 digits
→ In theory, the value range can be {0, 1, 2, 3, 4, 5}.
But is not a valid zip code, & there are no zip codes with 4 zeros.
→ Value range of X is {2, 3, 4, 5}.
3 zip code examples: , 33620, 90210
X(33647) = 5, X(33620) = 4, X(90210) = 3

16. Section Summary Concepts discussed Random Variables
Bernoulli Random Variables
Discrete Random Variables
Continuous Random Variables

17. Probability Distributions
for Discrete Random Variables
3.2

18. Probability Distributions for Discrete Random Variables
Probabilities assigned to various outcomes in in turn determine probabilities associated with the values of any particular rv X.
The probability distribution of X says how the total probability of 1 is distributed among (allocated to) the various possible X values.
Suppose, for example, that a business has just purchased four laser printers, and let X be the number among these that require service during the warranty period.

19. Probability Distributions for Discrete Random Variables
P(X = x) = the probability of rv X = x, denoted by p(x).
p (0) = P(X = 0) = the probability of X equal to 0.
p (1) = P(X = 1) = the probability of X value to 1.

20. Example 3.7
The Statistics Dept. at Cal Poly has a lab with 6 computers.
Let X = the # of computers in use at a particular time of day.
Suppose the probability distribution of X is as given below.
p(0) = → 5% of time no one uses computers.
p(3) = → 25% of time 3 computers are in use.

21. Example 3.7 P(X  2) = P(X = 0 or X = 1 or X = 2)
cont’d
P(X  2) = P(X = 0 or X = 1 or X = 2)
= P(X=0) + P(X=1) + P(X=2)
= p(0) + p(1) + p(2) =
= 0.30

22. Example 3.7 A = {X  3}: the event at least 3 computers are in use.
cont’d
A = {X  3}: the event at least 3 computers are in use.
B = {X  2}: the event at most 2 computers are in use.
Then B = A'.
→ P(X  3) = 1 – P(X  2)
= 1 – 0.30
= 0.70

23. Example 3.7 P(2  X  5) = P(X = 2, X = 3, X = 4, or X = 5)
cont’d
P(2  X  5) = P(X = 2, X = 3, X = 4, or X = 5) =
= 0.75
P(2 < X < 5) = P(X = 3 or X = 4) =
= 0.45

24. Probability Distributions for Discrete Random Variables
Definition
The probability mass function (pmf) f of a discrete rv is defined for every number x by f (x) = P(X = x) such that
(i) f (x)  0, and
(ii) all possible x f(x) = 1.

25. Example 3.8 6 lots of components ready for shipment lot 1 2 3 4 5 6
# defects
X = # of defects in the lot selected for shipment.
The value range of X is {0, 1, 2}. Its pmf f is
f (0) = P(X=0) = P(lot 1, 3 or 6 is shipped) = 3/6 = 0.500
f (1) = P(X=1) = P(lot 4 is shipped) = 1/6 = 0.167
f (2) = P(X=2) = P(lot 2 or 5 is shipped) = 2/6 = 0.333

26. A Parameter of a Probability Distribution
The pmf p of the Bernoulli rv X:
where α is a parameter (0 < α < 1) of the above pmf p.
Each different value of  between 0 and 1 determines a different member of the Bernoulli family of distributions.
(3.1)

27. Example 3.12
Observe the gender of the next newborn at a certain hospital, and the observation stops when a boy (B) is born. Let B be event a girl is borm.
Let p = P (B). Then P(B) = 1 – p.
Assume that successive births are independent.
Define rv X = # of births observed.
Then the range of X is {1, 2, 3, 4, 5, 6, …}.
p(1) = P(X = 1) = P(B) = p.

28. Example 3.12 p(2) = P(X = 2) = P(GB) = P(G)  P(B) = (1 – p)p and
cont’d
p(2) = P(X = 2)
= P(GB)
= P(G)  P(B)
= (1 – p)p
and
p(3) = P(X = 3)
= P(GGB)
= P(G)  P(G)  P(B)
= (1 – p)2p

29. Example 3.12
cont’d
→ The pmf p is
Where the parameter p is in open set (0, 1).
pmf given by (3.2) defines the geometric distributions.
In the gender example, p = 0.51 might be appropriate.
But if we were looking for the first child with Rh-positive blood, then we might have p = 0.85.
(3.2)

30. The Cumulative Distribution Function
We often wish to compute the following probability
P(the observed value of X will be at most x) = P(X ≤ x)
P(X is at most 1)
= P(X  1) = p(0) + p(1) = = 0.667

31. The Cumulative Distribution Function
Similarly,
P(X  1.5) = P(X  1) = .667
P(X  0) = P(X = 0) = 0.5
P(X  .75) = P(X = 0) = 0.5
In fact for any x satisfying 0  x < 1, P(X  x) = 0.5.

32. The Cumulative Distribution Function
The largest possible X value is 2,
→ P(X  2) = 1 P(X  3.7) = 1 P(X  20.5) = 1
and so on.

33. The Cumulative Distribution Function
Definition
The cumulative distribution function (cdf) F(x) of a discrete rv X with pmf p(x) is definedby
F (x) = P(X  x) = ∑y: y ≤ x p(y) (3.3)
F(x) = P(the observed value of X is at most x).

34. Example 3.13
A store has flash drives of 1, 2, 4, 8, or 16 GB of memory.
Let Y = the memory size of a purchased drive.
The pmf of Y is given below.

35. Example 3.13 Then (Assume that all purchases are independent)
cont’d
Then (Assume that all purchases are independent)
F (1) = P (Y  1)
= P (Y = 1)
= p (1)
= 0.05
F (2) = P (Y  2)
= P (Y = 1 or Y = 2)
= p (1) + p (2)
= 0.15

36. Example 3.13 F(4) = P(Y  4) = P(Y = 1 or Y = 2 or Y = 4)
cont’d
F(4) = P(Y  4)
= P(Y = 1 or Y = 2 or Y = 4)
= p(1) + p(2) + p(4)
= 0.50
F(8) = P(Y  8)
= p(1) + p(2) + p(4) + p(8)
= 0.90
F(16) = P(Y  16)
= 1.00

37. Example 3.13 For any non-integer y in (1, 16), F (y) = F(int(y)).
cont’d
For any non-integer y in (1, 16), F (y) = F(int(y)).
For example,
F(2.7) = P(Y  2.7)
= P(Y  2)
= F(2)
= 0.15
F(7.999) = P(Y  7.999)
= P(Y  4)
= F(4)
= 0.50

38. Example 3.13 If y < 1, F (y) = 0 [e.g. F(.58) = 0], and
cont’d
If y < 1, F (y) = 0 [e.g. F(.58) = 0], and
if y > 16, F (y) = 1[e.g. F(25) = 1].
→ The cdf is

39. Example 3.13
cont’d
This cdf is depicted below.

40. The Cumulative Distribution Function
cdf F (x) of a discrete rv X is a step function, and the step size at each possible value x of X is equal to p(x).
Proposition
For any a and b with a  b,
P (a  X  b)
= P(X ≤ b) – P(X < a)
= F (b) – F (a–)
where “a–” represents the largest possible X value that is strictly less than a.

41. The Cumulative Distribution Function
If all possible values, and a and b are integers, then
P (a  X  b) = P(X = a or X = a + 1 or or X = b)
= F (b) – F (a – 1).
Taking a = b yields
P(X = a) = F (a) – F (a – 1)
in this case.

42. Example 3.15
X = # of days of sick leave taken by a randomly selected employee of a large company per year.
If the maximum # of allowable sick days per year is 14,
value range of X is {0, 1, 3, 4, , 14}.

43. Example 3.15 Given F(0) = 0.58, F(1) = 0.72, F(2) = 0.76,
cont’d
Given F(0) = 0.58, F(1) = 0.72, F(2) = 0.76,
F(3) = 0.81, F(4) = 0.88, F(5) = 0.94,
P(2  X  5) = P(X = 2, 3, 4, or 5)
= F(5) – F(1)
= 0.22
and
P(X = 3) = F(3) – F(2)
= 0.05

44. Exercise Problem 12 55 tickets sold for a 50 seat flight.
Y = # of ticketed passengers actually showed up. y
p(y)
a. P(accommodate all ticketed passengers showed up) = ?
b. P(cannot take all ticketed passengers showed up) = ?
c. P(1sf standby person can take the flight) = ?
P(3rd standby person can take the flight) = ?

45. Exercise Problem 12
a. P(accommodate all ticketed passengers showed up) = ?
= P(Y≤50)
= P(Y=45 or 46 or 47 or 48 or 49 or 50)
= P(Y=45)+P(Y=46)+ … + P(Y=50)
= = 0.83
b. P(cannot take all ticketed passengers showed up)
= P(Y ≥ 51)
= 1 – P(Y ≤ 50)
= 0.17

46. Exercise Problem 12 c. P(1sf standby person can take the flight)
= P(Y≤49)
= 0.66
P(3rd standby person can take the flight)
= P(Y ≤ 47)
= 0.27

47. Exercise Problem 23 The cdf of rv X is given. a. p(2) = P(X=2)
= F(2) – F(1)
= 0.39 – 0.19
= 0.20
b. P(X > 3) = 1 – P(X ≤ 3)
= 1 – F(3)
= 1 – 0.67
= 0.33

48. Exercise Problem 23 c. P(2 ≤ X ≤ 5) = P(X≤5) – P(X≤1) = F(5) – F(1)
= 0.92 – 0.19
= 0.73
d. P(2 < X < 5) = P(3 ≤ X ≤ 4)
= P(X≤4) – P(X≤2)
= F(4) – F(2)
= 0.92 – 0.39
= 0.53

49. Section Summary Concepts discussed
Probability Mass (Distribution) Function (pmf)
Parameters of a Probability Distribution
Cumulative Distribution Function (cdf)

50. 3.3
Expected Values

51. The Expected Value of X Definition
rv X has value range D and pmf p (x). The expected value or mean of X, denoted by E(X) or X or just , is

52. Example 3.16 A university has 15,000 students.
X = # of courses a student is registered. The pmf of X follows.
 = 1p(1) + 2p(2) + 3p(3) + 4p(4) + 5p(5) + 6p(6) + 7p(7)
= 1(.01) + 2(.03) + 3(.13) + 4(.25) + 5(.39) + 6(.17) + 7(.02) =
= 4.57

53. The Expected Value of a Function
Given a function h (X) of rv X, determine E [h(X)].
Remark: h (X) is also a random variable. Proposition
For rv X with value range D and pmf p (x), the expected value of any function h (X), denoted by E [h (X)] or h(X), is computed by

54. Example 3.23
A store has purchased 3 computers at $500 apiece, and will sell them for $1000 apiece.
The manufacturer has agreed to repurchase any computers unsold after a specified period at $200 apiece.
Let X denote the number of computers sold.
Suppose that
p(0) = 0.1, p(1) = 0.2, p(2) = 0.3 and p(3) = 0.4.
Then E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 2.0

55. Example 3.23 h (X) = profit from selling X units. Then
cont’d
h (X) = profit from selling X units. Then
h (X) = revenue – cost
= 1000X + 200(3 – X) – 1500
= 800X – 900
Value range of rv h(X) is {–900, –100, 700, 1500}
The expected profit is then
E [h (X)] = h(0)  p(0) + h(1)  p(1) + h(2)  p(2) + h(3)  p(3)
= (–900)(.1) + (– 100)(.2) + (700)(.3) + (1500)(.4)
= $700

56. Rules of Expected Value
Proposition E (aX + b) = a  E(X) + b.
In Example 23,
h (X) = 800X – 900, and E(X) = 2.
E[h(x)] = 800E(X) – 900
= 800(2) – 900
= $700.

57. The Variance of X
Definition Let X have pmf p (x) and expected value . Then the variance of X, denoted by V(X) or 2X or just 2, is The standard deviation (SD) of X is

58. Example 3.24
A library has an upper limit of 6 on # of videos that can be checked out by an individual.
Let X = # of videos checked out by an individual.
The pmf of X is as follows:
The expected value of X is easily computed as  = 2.85.

59. Example 3.24 The variance of X is then
cont’d
The variance of X is then
= (1 – 2.85)2(0.30) + (2 – 2.85)2(0.25) (6 – 2.85)2(0.15) =
The standard deviation of X is  =

60. A Shortcut Formula for 2
Proposition
V(X) = 2 = – 2 = E(X2) – [E(X)]2

61. Rules of Variance
The variance of h (X) is the expected value of the squared difference between h (X) and its expected value: V [h (X)] = 2h(X) =
For linear function h (X) = aX + b, h (x) – E [h (X)] = ax + b – (a + b) = a (x – ). → V (a X+b) = a2V(X)
(3.13)

62. Example 3.26 In Example 23, E(X) = 2, and
→ V(X) = 5 – (2)2 = 1.
The profit function h(X) = 800X – 900
→ V[h(X)] = (800)2 V(X) = (640,000)(1) = 640,000
→ Standard deviation of profit function h(X) is 800.

63. Another Example a. What value of c will make the following a pmf?
f(x) = c(1/2)x, x = 1, 2, 3.
b. Find E(X).
c. Find V(X).

64. Another Example a. c(1/2)1 + c(1/2)2 + c(1/2)3 = 1 → c = 8/7
cont’d
a. c(1/2)1 + c(1/2)2 + c(1/2)3 = 1 → c = 8/7
b. E(X) = 1[(8/7)(1/2)] + 2[(8/7)(1/2)2] + 3[(8/7)(1/2)3]
= 11/7
c. V(X)
= 12[(8/7)(1/2)] + 22[(8/7)(1/2)2] + 32[(8/7)(1/2)3] – (11/7)2
= 26/49 .

65. Section Summary Concepts discussed Expected Value of a rv
Expected Value of Function of a rv
Expected Value of a Linear Function of a rv
Variance of a rv
Variance of Function of a rv
Variance of a Linear Function of a rv

66. 3.4 The Binomial Probability Distribution
Copyright © Cengage Learning. All rights reserved.

67. The Binomial Probability Distribution
Many experiments conform to the following:
The experiment consists of a sequence of n smaller experiments called trials, with n fixed in advance.
Each trial results in 1 of the same 2 possible outcomes, denoted by success (S) and failure (F).
The trials are independent.
The probability of success P(S), denoted by p, is constant from trial to trial. And P(F) = 1 – p

68. The Binomial Probability Distribution
Definition
An experiment satisfying Conditions 1~4 is called a binomial experiment.

69. Example 3.27 A quarter is tossed successively & independently n times.
Use S to denote the outcome W (Washington) and F the outcome E (Eagle), for each toss.
Then this results in a binomial experiment with
p = P(S) = 0.5, and
P(F) = 1 – p = 0.5.

70. Another Example A die is tossed n times.
Use S to denote outcome 1, 2, 3, or 4, and F outcome 5 or 6, for each toss.
Then this also results in a binomial experiment with
p = P(S) = 4/6 = 2/3, and
P(F) = 1 – p = 1/3.

71. The Binomial Random Variable and Distribution
Definition
Let X = # of successes (S) resulted from a binomial experiment (i.e., # of times S is observed in the n trials).
Then X is a rv, called a binomial random variable, and
its probability distribution is called a binomial distribution, denoted as Bin(n, p).
Any binomial random variable is discrete!

72. The Binomial Random Variable and Distribution
Consider n = 3. There are 8 possible outcomes:
SSS SSF SFS SFF FSS FSF FFS FFF
By definition, X(SSF) = 2, X(SFS) = 2, X(SFF) = 1, & so on.
The value range of X for n = 3 is {0, 1, 2, 3}.
Range of X for an n-trial experiment is {0, 1, 2, , n}.

73. The Binomial Random Variable and Distribution
Write X ~ Bin(n, p) to indicate that
X is a binomial rv with n trials and success probability of p.
Use b(x; n, p) to denote the pmf of a binomial rv X.
Use B(x; n, p) to denote the cdf of a binomial rv X.

74. The Binomial Random Variable and Distribution
Formulas to compute b(x; n, p) and B(x; n, p)
b(x; n, p) = Cx,n px (1 – p)n – x, x = 0, 1, 2, 3, …, n; and
b(x; n, p) = 0, for all other x values.
B(x; n, p) = P(X  x) = b(y; n, p) x = 0, 1, , n

75. Example 3.31
Randomly select 6 cola drinkers. Each is given a glass of cola S and a glass of cola F. Glasses are all identical except for a code on the bottom to identify the cola. A cola drinker has no tendency to prefer 1 cola to the other.
Then p = P(a cola drinker prefers S) = 0.5.
Let X = the # of cola drinkers (of the six) who prefer S. Then X ~ Bin(6, 0.5).
P(X = 3) = b(3; 6, 0.5) = (0.5)3(0.5)3 = 20(0.5)6 = 0.313

76. Example 3.31 The probability that at least three prefer S is
cont’d
The probability that at least three prefer S is
P(3  X) = b(x; 6, 0.5)
= (0.5)x(1–0.5)6 – x
= 0.656
and the probability that at most one prefers S is
P(X  1) = b(x; 6, 0.5)
= 0.109

77. Using Binomial Tables (Table A.1)
Consider n = 20, and p = 0.25.
P(X ≤ 10) = B(10; 20, 0.25)
= 0.996
P(5 ≤ X ≤ 10) = P(X ≤ 10) – P(X ≤ 4)
= B(10; 20, 0.25) – B(4; 20, 0.25)
= – 0.415
= 0.581

78. The Mean and Variance of X
If X ~ Bin(n, p), then
mean E(X) = np,
variance V(X) = np(1 – p) = npq, and
sd X = ,
where q = 1 – p.

79. Example 3.34
75% purchases at a store are made with a credit card. 10 purchases are observed, and let X = # of purchases made with a credit card. Then X ~ Bin(10, .75).
Thus E(X) = np = (10)(0.75) = 7.5,
V(X) = npq = 10(0.75)(0.25)
= 1.875,
and  =
= 1.37.

80. Example 3.34
cont’d
The probability that X is within 1 standard deviation of its mean value is
P(7.5 – 1.37  X  ) = P(6.13  X  8.87)
= P(X = 7 or X = 8)
= P(X = 7) P(X = 8)
= b(7; 10, 0.75)b(8; 10, 0.75) =
P(7.5–1.37  X  ) = P(7  X  8)
= B(8; 10, 0.75) – B(6; 10, 0.75)
= 0.532

81. Exercise Problem 54
cont’d
P(A customer chooses an oversize tennis racket) = 0.6.
X = # of customers among the 10 who choose oversize one. Then X ~ Bin(10, 0.6).
a. P(X ≥ 6) = 1 – P(X ≤ 5)
= 1 – B(5; 10, 0.6)
= 1 – 0.367
= 0.633
b. µ = np = (10(0.6) = σ = √[10(0.6)(0.4)] = 1.55
→ µ – σ = 4.45, and µ + σ = 7.55
→ P(4.45  X  7.55) = P(5  X  7) = P(X≤7) – P(X≤4)
= B(7; 10, 0.6) – B(4; 10, 0.6) = – = 0.667

82. Exercise Problem 54
cont’d
c. All 10 customers can get what they want iff 3 ≤ X ≤ 7.
→ P(3  X  7)
= P(X≤7) – P(X≤2)
= B(7; 10, 0.6) – B(2; 10, 0.6)
= – 0.012
= 0.821

83. Hypergeometric and Negative
Binomial Distributions
3.5

84. The Hypergeometric Distribution
The assumptions leading to the hypergeometric distribution are as follows:
1. The population to be sampled consists of N elements (a finite population). 2. Each element is characterized either as a success (S) or a failure (F), and there are M successes in the population. 3. A sample of n elements is randomly selected without replacement.

85. The Hypergeometric Distribution
X = # of successes in the random sample of n elements.
Then X is a rv, called hypergeometric rv.
Its value range is
{t, t+1, t+2, …, min(n, M)}, where t = max{0, n – N + M}.
pmf: h(x; n, M, N) = P(X = x). See Eq. (3.15)

86. Example 3.35
We received 20 service calls for printer problems (8 laser printers and 12 inkjet printers).
A random sample of 5 is selected for inclusion in a customer satisfaction survey.
P(exactly 0, 1, 2, 3, 4, or 5 inkjet printers are selected) = ?
Let X = # of inkjet printers in the sample of 5 printers.
Then X is a hypergeometric rv.

87. Example 3.35
cont’d
N = 20, n = 5, and M = 12.
Value range: {0, 1, 2, 3, 4, 5}
Consider X = 2.
P(X = 2) =
= 77/323 = 0.238

88. Example 3.35 If M =18 (not 12), P(X = 2) = ? Sample space is given by
cont’d
If M =18 (not 12), P(X = 2) = ?
Sample space is given by
{t, t+1, t+2, …, min(n, M)}, where t = max{0, n – N + M}.
N = 20, n = 5, and M = 18.
→ t = max(0, ) = 3, and min(n, M) = 5
→ Sample space: {3, 4, 5}
P(X = 2) = 0.

89. Example 3.35 If the sample size is increased to n = 15, P(X = 14) = ?
cont’d
If the sample size is increased to n = 15, P(X = 14) = ?
Value range is given by
{t, t+1, t+2, …, min(n, M)}, where t = max{0, n – N + M}.
N = 20, n = 15, and M = 12.
→ t = 0 and min(n, M) = min(15, 12) = 12
→ Value range: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
P(X = 14) = 0.

90. The Hypergeometric Distribution
For hypergeometric rv X with pmf h(x; n, M, N), the mean and variance are
In Example 3.35, N = 20, n = 5, and M = 12.
→ E(X) = 5(12)/20 = 3.0
V(X) =

91. The Negative Binomial Distribution
The negative binomial rv and distribution are based on an experiment satisfying the following conditions:
1. The experiment consists of a sequence of independent trials. 2. Each trial can result in a success (S) or a failure (F). 3. The probability of success is constant from all trial.
4. The experiment continues until a total of r successes observed.

92. The Negative Binomial Distribution
Let X = # of failures that precede the rth success. Then
X is called a negative binomial random variable,
Its sample space is {0, 1, 2, 3, 4, 5, 6, …}, and
Its pmf is written as nb(x; r, p), computed by

93. Example 3.38
A pediatrician wishes to recruit 5 couples, each of whom is expecting their first child, to participate in a new natural childbirth regimen. Let
p = P(a randomly selected couple agrees to participate).
If p = 0.2, P(15 couples must be asked) = ?
With S = {agrees to participate}, r = 5, p = 0.2, and x = 10,
P(15 couples must be asked)
= = 0.034

94. Example 3.38 P(at most 10 F’s are observed)
cont’d
P(at most 10 F’s are observed)
= P(at most 15 couples are asked) =

95. The Negative Binomial Distribution
Remark. Some books define the negative binomial rv as the # of trials, rather than the # of failures.
A special case: r = 1.
The pmf is
The random variable X = # of failures (or trials) until one success is observed is called a geometric random variable, and the pmf in Eq. (3.17) is called a geometric distribution.
(3.17)

96. The Negative Binomial Distribution
For a geometric random variable X (defined as # of failures until 1 success observed in our Textbook),
E(X) = (1 – p)/p, and V(X) = (1–p)/p2.
If a geometric random variable X is defined as # of trials until 1 success observed, then
E(X) = 1/p, and V(X) = (1–p)/p2.

97. The Negative Binomial Distribution
For a negative binomial rv X with pmf nb(x; r, p), its mean and variance are
E(X) = r(1 – p)/p, and
V(X) = r(1–p)/p2.

98. Exercise Problem 74
Inspector checks 10 firms for possible violation of regulations.
Let X = # of firms among 10 checked that actually violate.
a. If there are 50 firms, & 15 actually violate,
What if the pmf of X?
b. If there are 500 firms, and 150 actually violate,
What is the pmf of X? An approximate pmf?
c. Find E(X) and V(X) for both pmfs in Part (b).

99. Exercise Problem 74
cont’d
a. X is Hypergeometric with n = 10, M = 15, and N = 50.
Its pmf: p(x) = h(x; 10, 15, 50), x = 0, 1, 2, …, 10.
b. X is Hypergeometric with n = 10, M = 150, and N = 500.
Its pmf: p(x) = h(x; 10, 150, 500), x = 0, 1, 2, …, 10.
Since N is very large relative to n, X can be approximated as a binomial rv with n = 10 and p = M/N. That is,
h(x; 10, 150, 500) ≈ b(x; 10, 0.3) Note 0.3 = 150/500
c. For h(x; 10, 150, 500), E(X) = 10(150/500) = 3, V(X) = 2.06
For b(x; 10, 0.3), E(X) = 10(0.3) = 3, V(X) = 10x0.7x0.7 = 2.1

100. Exercise Problem 76
A family decides to have children until 3 same sex children.
Let X = # of children in the family.
With P(B) = P(G) = 0.5, What is the pmf of X?

101. Exercise Problem 76
cont’d
This question relates to negative binomial distribution.
But we cannot use it directly.
Clearly, X cannot be 0, 1 or 2, and X < 6 → 3 ≤ X ≤ 5.
P(X=3) = P(BBB U GGG) = P(BBB) + P(GGG) = 2(0.5)3 = 0.25
P(X=4)
= P(GBBB U BGBB U BBGB U BGGG U GBGG U GGBG)
= 6(0.5)4 = 0.375
P(X=5) = 1 – P(X=3) – P(X=4) = 0.375

102. The Poisson Probability Distribution
3.6
The Poisson Probability Distribution

103. The Poisson Probability Distribution
Definition
A discrete random variable X is said to have a Poisson distribution with parameter  ( > 0) if the pmf of X is
P(X=x) =
Its cdf is F(x, µ) = P(X ≤ x).
X is called a Poisson random variable.

104. Example 3.39
Let X denote the number of creatures of a particular type captured in a trap during a given time period.
Suppose that X has a Poisson distribution with  = 4.5, so on average traps will contain 4.5 creatures.
P(a trap contains exactly five creatures) is

105. Example 3.39
cont’d
P(a trap has at most five creatures) is

106. The Poisson Distribution as a Limit
Proposition
Suppose that in the binomial pmf b(x; n, p), we let n  and p  0 in such a way that np approaches a value  > 0. Then b(x; n, p)  p(x; ).
By this proposition, if n is large and p is small, then
b(x; n, p)  p(x; ), where  = np.
As a rule of thumb, this approximation can safely be applied if n > 50 and np < 5.

107. Example 3.40
If (i) P(any given page containing at least typo) = and
(ii) typos are independent from page to page,
P(a 400-page book has exactly one page with typos) = ?
P(it has at most three pages with typos) = ?
Let S be a page containing at least one error and F be an error-free page. Then X = # of pages having at least one typo is a binomial rv with n = 400 and p = 0.005, so np = 2.

108. Example 3.40 P(a 400-page book has exactly one page with typos)
cont’d
P(a 400-page book has exactly one page with typos)
= P(X = 1) = b(1; 400, .005)  p(1; 2)
The binomial value b(1; 400, 0.005) = , so the approximation is very good.

109. Example 3.40 Similarly, P(X  3)
cont’d
Similarly,
P(X  3) = =
This again is very close to the binomial value
P(X  3) =

110. The Poisson Distribution as a Limit
Table below shows the Poisson distribution for  = 3 along with three binomial distributions with np = 3.

111. The Poisson Distribution as a Limit
Table A.2 gives the cdf F(x; ) for  = 0.1, 0.2, ,1, 2, . . ., 10, 15, and 20.
Consider  = 2. From Table A.2 on page A-5,
P(X  3) = F(3; 2) =
P(X = 3) = F(3; 2) – F(2; 2) =

112. The Mean and Variance of X
Proposition
For a Poisson distribution X with parameter ,
E(X) = V(X) = .

113. Example 3.41 Example 39 continued…
Both the expected number of creatures per trap and the variance of the number trapped equal 4.5, and
X =
= 2.12.

114. The Poisson Process

115. The Poisson Process
A very important application of the Poisson distribution arises in connection with the occurrence of events of some type over time.
Vehicles (driving east) passing USF main entrance
Visits to a particular website
Pulses of some sort recorded by a counter
messages sent to a particular address
Accidents in an industrial facility
Cosmic ray showers observed by astronomers

116. The Poisson Process
Given an interval, events occur at random throughout the interval. Partition the interval into subintervals of small enough length such that
1. P(more than one event in a subinterval) = The probability of one event in a subinterval is the same for all subintervals & proportional to the length of the subinterval. 3. The event in each subinterval is independent of other subintervals.
The experiment of observing random events in an interval is called a Poisson process.

117. The Poisson Process
X = # of events in the interval is a Poisson rv with
pmf p(x; µ),
where µ is the expected number of events in the interval.
Remark If events occur at a rate of α per unit interval,
then µ = α·t for a t-unit interval.

118. Example 3.42
Suppose pulses arrive at a counter at an average rate of 6 per minute, so that  = 6.
P(at least one pulse is received in a 0.5-min interval) = ?
The interval length t = 0.5 minutes → µ = α·t = 3
X = # of pulses received in a 30-sec interval is a Poisson rv.

119. Exercise Problem 85
Small aircrafts arrive according to a Poisson process with a rate of 8 per hour.
a. Let X = # of aircrafts that arrive in 1 hour.
P(X = 6) = ? P(X ≥ 6) = ? P(X ≥10) = ?
b. Let X = # of aircrafts that arrive in a 90-min period.
E(X) = ? V(X) = ?
c. Let X = # of aircrafts that arrive in a 2.5-hour period.

120. Exercise Problem 85 a. E(X) = 8 aircrafts/hour.
cont’d
a. E(X) = 8 aircrafts/hour.
P(X = 6) = e–886/6! = 0.122
P(X ≥ 6) = 1 – F(5; 8) = 1 – = 0.809
P(X ≥10) = 1 – F(9; 8) = 1 – = 0.283
b. E(X) = 8(1.5) = 12 aircrafts arriving in 90 minutes.
P(X = 6) = e–12126/6! =
P(X ≥ 6) = 1 – F(5; 8) = 1 – =
P(X ≥10) = 1 – F(9; 8) = 1 – = 0.758
E(X) = V(X) = 12

121. Exercise Problem 85
cont’d
c. E(X) = 8(2.5) = 20 aircrafts arriving in 2.5 hours.
P(X = 6) = e–20206/6! =
P(X ≥ 6) = 1 – F(5; 20) = 1 – =
P(X ≥10) = 1 – F(9; 20) = 1 – = 0.995
µ P(X = 6) P(X ≥ 6) P(X ≥10) t hr
hrs
hrs

122. Exercise Problem 92
cont’d
Vehicles arrive according to a Poisson process at a rate of α = 10 vehicles per hour. Suppose that
p = P(an arriving vehicle has no equipment violations) = 0.5.
a. P(exactly 10 arrive in 1 hr & none has violations) = ?
b. P(y arrive in 1 hr & 10 of the y have no violations) = ? y≥10.
c. P(10 no-violation vehicles arrive in the next hour) = ?
[Hint: Sum the probabilities in part (b) from y=10 to ∞]

123. Exercise Problem 92
cont’d
a. Let Y = # of cars arriving in the hour. → Y ~ Poisson(10).
P(Y = 10 ∩ no violations)
= P(Y=10)P(no violations│Y=10)
= [e–101010/10!][(0.5)10] =
b. P(Y = y ∩ 10 have no violations)
= P(Y=y)P(10 have no violations│Y=y)
= [e–1010y/y!][C10,y (0.5)10 (0.5)y–10]
= e–105y/[10!(y-10)!]

124. Exercise Problem 92 c. P(exactly 10 without violations)
cont’d
c. P(exactly 10 without violations)
= ∑y=10~∞ e–105y/[10!(y-10)!]
= [e–10510/10!] ∑y=10~∞ [5y–10 /(y-10)!]
= [e–10510/10!] ∑u=0~∞ [5u /(u!] u=y–10
= [e–10510/10!] e5
= e–5510/10!
= p(10; 5)
→ X = # of vehicles without violations per hour. Then
X ~ Poisson(µ), with µ = αp = 10(0.5) = 5 vehicles/hour.

125. Chapter Summary Concepts and Probability Distributions Discussed
Random Variables
Bernoulli Random Variables
Probability Mass Functions (pmf)
Cumulative Distribution Functions (cdf)
Expected Values E(X). E[h(X)] & Variances V(X), V[h(X)]
Binomial Distribution
Hypergeometric Distribution
Negative Binomial Distribuion
Geometric Distribution
Poisson Distribution