Probability Distributions for Discrete Random Variables

3.2 Probability Distributions for Discrete Random Variables

  Probabilities assigned to various outcomes in S in turn determine probabilities asso- ciated 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 poss ible

  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. Possible X values are then 0, 1, 2, 3, and 4. The probability distribution will tell us how the probability of 1 is subdivided among these five possible values—how much probability is associated with the X value 0, how much is apportioned to the X value 1, and so on. We will use the following notation for the probabilities in the distribution:

  p (0) 5 the probability of the X value 0 5 P(X 5 0) p (1) 5 the probability of the X value 1 5 P(X 5 1)

  and so on. In general, p(x) will denote the probability assigned to the value x.

  100 Chapter 3 Discrete random Variables and probability Distributions

  ExamplE 3.7 The Cal Poly Department of Statistics has a lab with six computers reserved for sta tistics majors. Let X denote the number of these computers that are in use at a particular time of day. Suppose that the probability distribution of X is as given in the following table; the first row of the table lists the possible X values and the second row gives the probability of each such value.

  We can now use elementary probability properties to calculate other probabilities of interest. For example, the probability that at most 2 computers are in use is

  P (X 2) 5 P(X 5 0 or 1 or 2) 5 p(0) 1 p(1) 1 p(2) 5 .05 1 .10 1 .15 5 .30 Since the event at least 3 computers are in use is complementary to at most 2 com-

  puters are in use,

  P (X 3) 5 1 2 P(X 2) 5 1 2 .30 5 .70

  which can, of course, also be obtained by adding together probabilities for the values

  3, 4, 5, and 6. The probability that between 2 and 5 computers inclusive are in use is

  P (2 X 5) 5 P(X 5 2, 3, 4, or 5) 5 .15 1 .25 1 .20 1 .15 5 .75 whereas the probability that the number of computers in use is strictly between 2 and 5 is

  The probability distribution or probability mass function (pmf) of a discrete rv is defined for every number x by p(x) 5 P(X 5 x) 5 P(all v [ S: X(v) 5 x).

  In words, for every possible value x of the random variable, the pmf specifies the probability of observing that value when the experiment is performed. The con- ditions p(x) 0 and o all possible x p(x) 5 1 are required of any pmf.

  The pmf of X in the previous example was simply given in the problem description. We now consider several examples in which various probability proper- ties are exploited to obtain the desired distribution.

  ExamplE 3.8 Six boxes of components are ready to be shipped by a certain supplier. The number of defective components in each box is as follows:

  Box

  Number of defectives

  One of these boxes is to be randomly selected for shipment to a particular customer. Let X be the number of defectives in the selected box. The three possible X val- ues are 0, 1, and 2. Of the six equally likely simple events, three result in X 5 0, one in X 5 1, and the other two in X 5 2. Then

  p(0) 5 P(X 5 0) 5 P(box 1 or 3 or 6 is sent) 5 5 .500

  1 p(1) 5 P(X 5 1) 5 P(box 4 is sent) 5 5 .167

  p(2) 5 P(X 5 2) 5 P(box 2 or 5 is sent) 5 5 .333

  3.2 probability Distributions for Discrete random Variables

  That is, a probability of .500 is distributed to the X value 0, a probability of .167 is placed on the X value 1, and the remaining probability, .333, is associated with the X value 2. The values of X along with their probabilities collectively specify the pmf. If this exper- iment were repeated over and over again, in the long run X 5 0 would occur one-half of the time, X 5 1 one-sixth of the time, and X 5 2 one-third of the time.

  n

  ExamplE 3.9 Consider whether the next person buying a computer at a certain electronics store buys a laptop or a desktop model. Let

  1 if the customer purchases a desktop computer

  5 0 if the customer purchases a laptop computer

  X5

  If 20 of all purchasers during that week select a desktop, the pmf for X is

  p(0) 5 P(X 5 0) 5 P(next customer purchases a laptop model) 5 .8 p(1) 5 P(X 5 1) 5 P(next customer purchases a desktop model) 5 .2

  p(x) 5 P(X 5 x) 5 0 for x ± 0 or 1 An equivalent description is

  if x 5 0

  5 0 if x ± 0 or 1

  p (x) 5 .2

  if x 5 1

  Figure 3.2 is a picture of this pmf, called a line graph. X is, of course, a Bernoulli rv and p(x) is a Bernoulli pmf.

  p (x) 1

  x

  Figure 3.2 The line graph for the pmf in Example 3.9 n

  ExamplE 3.10 In a group of five potential blood donors—a, b, c, d, and e—only a and b have type O-positive blood. Five blood samples, one from each individual, will be typed in ran- dom order until an O1 individual is identified. Let the rv Y 5 the number of typings necessary to identify an O1 individual. Then the pmf of Y is

  p(1) 5 P(Y 5 1) 5 P(a or b typed first) 5

  p(2) 5 P(Y 5 2) 5 P(c, d, or e first, and then a or b)

  5 P(c, d, or e first) ? P(a or b next u c, d, or e first) 5 ? 5 .3

  p(3) 5 P(Y 5 3) 5 P(c, d, or e first and second, and then a or b)

  p(4) 5 P(Y 5 4) 5 P(c, d, and e all done first) 5

  p(y) 5 0 if y Þ 1, 2, 3, 4

  102 Chapter 3 Discrete random Variables and probability Distributions

  In tabular form, the pmf is y

  where any y value not listed receives zero probability. Figure 3.3 shows a line graph of the pmf.

  p (y) .5

  y 0 1 2 3 4

  Figure 3.3 The line graph for the pmf in Example 3.10 n

  The name “probability mass function” is suggested by a model used in physics for a system of “point masses.” In this model, masses are distributed at various loca- tions x along a one-dimensional axis. Our pmf describes how the total probability mass of 1 is distributed at various points along the axis of possible values of the random variable (where and how much mass at each x).

  Another useful pictorial representation of a pmf, called a probability histogram, is similar to histograms discussed in Chapter 1. Above each y with p(y) . 0, construct a rectangle centered at y. The height of each rectangle is proportional to p(y), and the base width is the same for all rectangles. When possible values are equally spaced, the base width is frequently chosen as the distance between successive y values (though it could

  be smaller). Figure 3.4 shows two probability histograms.

  Figure 3.4 Probability histograms: (a) Example 3.9; (b) Example 3.10

  It is often helpful to think of a pmf as specifying a mathematical model for a discrete population.

  ExamplE 3.11 Consider selecting a household in a certain region at random and let X 5 the number of individuals in the selected household. Suppose the pmf of X is as follows:

  [this is very close to the household size distribution for rural Thailand given in the arti-

  cle “The Probability of Containment for Multitype Branching Process Models for

  Emerging Epidemics” (J. of Applied Probability, 2011: 173–188) , which modeled

  influenza transmission.]

  Suppose this is based on 1 million households. One way to view this situation is to think of the population as consisting of 1 million households, each with its own

  X value; the proportion with each X value is given by p(x) in the above table. An alter- native viewpoint is to forget about the households and think of the population itself

  3.2 probability Distributions for Discrete random Variables

  as consisting of X values: 14 of these values are 1, 17.5 are 2, and so on. The pmf then describes the distribution of the possible population values 1, 2, … , 10.

  n

  Once we have such a population model, we will use it to compute values of population characteristics (e.g., the mean m) and make inferences about such characteristics.

  A Parameter of a Probability distribution

  The pmf of the Bernoulli rv X in Example 3.9 was p(0) 5 .8 and p(1) 5 .2 because 20 of all purchasers selected a desktop computer. At another store, it may be the case that p(0) 5 .9 and p(1) 5 .1. More generally, the pmf of any Bernoulli rv can be expressed in the form p(1) 5 a and p(0) 5 1 2 a, where

  0 , a , 1. Because the pmf depends on the particular value of a, we often write p(x; a) rather than just p(x):

  12a if x 5 0

  5 0 otherwise

  p (x; a) 5

  a if x 5 1 (3.1)

  Then each choice of a in Expression (3.1) yields a different pmf.

  DEFINITION

  Suppose p(x) depends on a quantity that can be assigned any one of a number of possible values, with each different value determining a different probabil- ity distribution. Such a quantity is called a parameter of the distribution. The collection of all probability distributions for different values of the parameter is called a family of probability distributions.

  The quantity a in Expression (3.1) is a parameter. Each different number

  a between 0 and 1 determines a different member of the Bernoulli family of distributions.

  ExamplE 3.12 Starting at a fixed time, we observe the gender of each newborn child at a certain hospital until a boy (B) is born. Let p 5 P(B), assume that successive births are independent, and define the rv X by x 5 number of births observed. Then

  p(1) 5 P(X 5 1) 5 P(B) 5 p

  p(2) 5 P(X 5 2) 5 P(GB) 5 P(G) ? P (B) 5 (1 2 p)p

  and

  p (3) 5 P(X 5 3) 5 P(GGB) 5 P(G) ? P(G) ? P(B) 5 (1 2 p) 2 p

  Continuing in this way, a general formula emerges:

  (1 2 p) x2 1 px5

  5 0 otherwise

  The parameter p can assume any value between 0 and 1. Expression (3.2) describes the family of geometric distributions. In the gender scenario, p 5 .51 might be appropriate, but if we were looking for the first child with Rh-positive blood, then it might be the case that p 5 .85.

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  104 Chapter 3 Discrete random Variables and probability Distributions

  the cumulative distribution Function

  For some fixed value x, we often wish to compute the probability that the observed value of X will be at most x. For example, let X be the number of number of beds occupied in

  a hospital’s emergency room at a certain time of day; suppose the pmf of X is given by

  Then the probability that at most two beds are occupied is P (X 2) 5 p(0) 1 p(1) 1 p(2) 5 .75

  Furthermore, since X 2.7 if and only if X 2, we also have P(X 2.7) 5 .75, and sim- ilarly P(X 2.999) 5 .75. Since 0 is the smallest possible X value, P(X 21.5) 5 0, P (X 210) 5 0, and in fact for any negative number x, P(X x) 5 0. And because 4 is the largest possible value of X, P(X 4) 5 1, P(X 9.8) 5 1, and so on.

  Very importantly,

  P (X , 2) = p(0) 1 p(1) 5 .45 , .75 5 P(X 2)

  because the latter probability includes the probability mass at the x value 2 whereas the former probability does not. More generally, P(X , x) , P(X x) whenever x is a possible value of X. Furthermore, P(X x) is a well-defined and computable probability for any number x.

  DEFINITION

  The cumulative distribution function (cdf) F(x) of a discrete rv variable X with pmf p(x) is defined for every number x by

  F (x) 5 P(X x) 5

  y : o y x

  p (y) (3.3)

  For any number x, F(x) is the probability that the observed value of X will be at most x.

  ExamplE 3.13 A store carries flash drives with either 1 GB, 2 GB, 4 GB, 8 GB, or 16 GB of mem- ory. The accompanying table gives the distribution of Y 5 the amount of memory in

  a purchased drive: y

  Let’s first determine F(y) for each of the five possible values of Y:

  F(1) 5 P(Y 1) 5 P(Y 5 1) 5 p(1) 5 .05 F(2) 5 P(Y 2) 5 P(Y 5 1 or 2) 5 p(1) 1 p(2) 5 .15 F(4) 5 P(Y 4) 5 P(Y 5 1 or 2 or 4) 5 p(1) 1 p(2) 1 p(4) 5 .50 F(8) 5 P(Y 8) 5 p(1) 1 p(2) 1 p(4) 1 p(8) 5 .90

  F(16) 5 P(Y 16) 5 1 Now for any other number y, F(y) will equal the value of F at the closest possible

  value of Y to the left of y. For example,

  F(2.7) 5 P(Y 2.7) 5 P(Y 2) 5 F(2) 5 .15 F(7.999) 5 P(Y 7.999) 5 P(Y 4) 5 F(4) 5 .50

  3.2 probability Distributions for Discrete random Variables

  If y is less than 1, F(y) 5 0 [e.g. F(.58) 5 0], and if y is at least 16, F(y) 5 1 [e.g.,

  F (25) 5 1]. The cdf is thus

  0 y, 1 .05 1y,2 .15 2y,4

  5 1 16 y

  F (y) 5 .50 4y,8

  .90 8 y , 16

  A graph of this cdf is shown in Figure 3.5.

  Figure 3.5 A graph of the cdf of Example 3.13 n

  For X a discrete rv, the graph of F(x) will have a jump at every possible

  value of X and will be flat between possible values. Such a graph is called a step function.

  ExamplE 3.14

  The pmf of X 5 the number of births up to and including that of the first boy had

  (Example 3.12

  the form

  continued)

  (1 2 p) x2 1 px5 1, 2, 3,…

  5 0 otherwise

  p (x) 5

  For any positive integer x,

  y o x y o 51 y o 50

  (1 2 p) y (3.4)

  To evaluate this sum, recall that the partial sum of a geometric series is

  y5 o 0 12a

  k1 1

  Using this in Equation (3.4), with a 5 1 2 p and k 5 x 2 1, gives

  1 2 (1 2 p) x

  F (x) 5 p ?

  5 1 2 (1 2 p) x x a positive integer

  1 2 (1 2 p)

  Since F is constant in between positive integers,

  5 1 2 (1 2 p) [x]

  0 x, 1

  F (x) 5

  x 1

  106 Chapter 3 Discrete random Variables and probability Distributions

  where [x] is the largest integer x (e.g., [2.7] 5 2). Thus if p 5 .51 as in the birth example, then the probability of having to examine at most five births to see the first

  boy is F(5) 5 1 2 (.49) 5 1 2 .0282 5 .9718, whereas F(10) < 1.0000. This cdf

  is graphed in Figure 3.6.

  F (x) 1.0

  x 0 1 2 3 4 5 50 51

  Figure 3.6 A graph of F(x) for Example 3.14

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  In examples thus far, the cdf has been derived from the pmf. This process can

  be reversed to obtain the pmf from the cdf whenever the latter function is available. For example, consider again the rv of Example 3.7 (the number of computers being used in a lab); possible X values are 0, 1, … , 6. Then

  p(3) 5 P(X 5 3)

  5 [p(0) 1 p(1) 1 p(2) 1 p(3)] 2 [p(0) 1 p(1) 1 p(2)]

  5 P(X 3) 2 P(X 2)

  5 F(3) 2 F(2)

  More generally, the probability that X falls in a specified interval is easily obtained from the cdf. For example,

  P(2 X 4) 5 p(2) 1 p(3) 1 p(4)

  5 [p(0) 1 … 1 p(4)] 2 [p(0) 1 p(1)]

  5 P(X 4) 2 P(X 1)

  5 F(4) 2 F(1)

  Notice that P(2 X 4) ± F(4) 2 F(2). This is because the X value 2 is included in 2 X 4, so we do not want to subtract out its probability. However, P (2 , X 4) 5 F(4) 2 F(2) because X 5 2 is not in the interval 2 , X 4.

  pROpOSITION For any two numbers a and b with a b, P (a X b) 5 F(b) 2 F(a2)

  where “a2” represents the largest possible X value that is strictly less than a. In particular, if the only possible values are integers and if a and b are integers, then

  P(a X b) 5 P(X 5 a or a 1 1 or… or b)

  5 F(b) 2 F(a 2 1)

  Taking a 5 b yields P(X 5 a) 5 F(a) 2 F(a 2 1) in this case.

  3.2 probability Distributions for Discrete random Variables

  The reason for subtracting F(a2) rather than F(a) is that we want to include P (X 5 a); F(b) 2 F(a) gives P(a , X b). This proposition will be used exten- sively when computing binomial and Poisson probabilities in Sections 3.4 and 3.6.

  ExamplE 3.15 Let X 5 the number of days of sick leave taken by a randomly selected employee of

  a large company during a particular year. If the maximum number of allowable sick days per year is 14, possible values of X are 0, 1, … , 14. With F(0) 5 .58,

  F (1) 5 .72, F(2) 5 .76, F(3) 5 .81, F(4) 5 .88, and F(5) 5 .94,

  P (2 X 5) 5 P(X 5 2, 3, 4, or 5) 5 F(5) 2 F(1) 5 .22

  and

  P (X 5 3) 5 F(3) 2 F(2) 5 .05

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  ExERcisEs Section 3.2 (11–28)

  11. Let X be the number of students who show up for a pro-

  x

  fessor’s office hour on a particular day. Suppose that the pmf of X is p(0) 5 .20, p(1) 5 .25, p(2) 5 .30, p(3) 5

  .15, and p(4) 5 .10.

  Calculate the probability of each of the following events.

  a. Draw the corresponding probability histogram.

  a. {at most three lines are in use}

  b. What is the probability that at least two students

  b. {fewer than three lines are in use}

  show up? More than two students show up?

  c. {at least three lines are in use}

  c. What is the probability that between one and three

  d. {between two and five lines, inclusive, are in use}

  students, inclusive, show up?

  e. {between two and four lines, inclusive, are not in use}

  d. What is the probability that the professor shows up?

  f. {at least four lines are not in use}

  12. Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the

  14. A contractor is required by a county planning department to

  random variable Y as the number of ticketed passengers who

  submit one, two, three, four, or five forms (depending on the

  actually show up for the flight. The probability mass func-

  nature of the project) in applying for a building permit. Let

  tion of Y appears in the accompanying table.

  Y5 the number of forms required of the next applicant. The probability that y forms are required is known to be

  y

  proportional to y—that is, p(y) 5 ky for y 5 1, . . . , 5. p 5 (y) .05 .10 .12 .14 .25 .17 .06 .05 .03 .02 .01 a. What is the value of k? [Hint: o y5 1 p(y) 5 1]

  b. What is the probability that at most three forms are

  a. What is the probability that the flight will accommo-

  required?

  date all ticketed passengers who show up?

  c. What is the probability that between two and four

  b. What is the probability that not all ticketed passen-

  forms (inclusive) are required?

  gers who show up can be accommodated?

  d. Could p(y) 5 y 2 y50 for y 5 1,…, 5 be the pmf of Y?

  c. If you are the first person on the standby list (which

  15. Many manufacturers have quality control programs that

  means you will be the first one to get on the plane if

  include inspection of incoming materials for defects.

  there are any seats available after all ticketed passen-

  Suppose a computer manufacturer receives circuit boards

  gers have been accommodated), what is the probabil-

  in batches of five. Two boards are selected from each

  ity that you will be able to take the flight? What is

  batch for inspection. We can represent possible outcomes

  this probability if you are the third person on the

  of the selection process by pairs. For example, the pair

  standby list?

  (1, 2) represents the selection of boards 1 and 2 for

  13. A mail-order computer business has six telephone lines.

  inspection.

  Let X denote the number of lines in use at a specified

  a. List the ten different possible outcomes.

  time. Suppose the pmf of X is as given in the accompa-

  b. Suppose that boards 1 and 2 are the only defective

  nying table.

  boards in a batch. Two boards are to be chosen at

  108 Chapter 3 Discrete random Variables and probability Distributions

  random. Define X to be the number of defective

  couple or individual arrives late is .4 (a couple will

  boards observed among those inspected. Find the

  travel together in the same vehicle, so either both people

  probability distribution of X.

  will be on time or else both will arrive late). Assume that

  c. Let F (x) denote the cdf of X. First determine F(0) 5

  different couples and individuals are on time or late

  P (X 0), F(1), and F(2); then obtain F(x) for all

  independently of one another. Let X 5 the number of

  other x.

  people who arrive late for the seminar.

  16. Some parts of California are particularly earthquake-

  a. Determine the probability mass function of X. [Hint:

  prone. Suppose that in one metropolitan area, 25 of all

  label the three couples 1, 2, and 3 and the two

  homeowners are insured against earthquake damage.

  individuals 4 and 5.]

  Four homeowners are to be selected at random; let X

  b. Obtain the cumulative distribution function of X, and

  denote the number among the four who have earthquake

  use it to calculate P(2 X 6).

  insurance.

  21. Suppose that you read through this year’s issues of the

  a. Find the probability distribution of X. [Hint: Let S

  New York Times and record each number that appears in

  denote a homeowner who has insurance and F one

  a news article—the income of a CEO, the number of

  who does not. Then one possible outcome is SFSS,

  cases of wine produced by a winery, the total charitable

  with proba bility (.25)(.75)(.25)(.25) and associated X

  contribution of a politician during the previous tax year,

  value 3. There are 15 other outcomes.]

  the age of a celebrity, and so on. Now focus on the lead-

  b. Draw the corresponding probability histogram.

  ing digit of each number, which could be 1, 2, … , 8, or 9.

  c. What is the most likely value for X?

  Your first thought might be that the leading digit X of a

  d. What is the probability that at least two of the four

  randomly selected number would be equally likely to be

  selected have earthquake insurance?

  one of the nine possibilities (a discrete uniform distribu- tion). However, much empirical evidence as well as

  17. A new battery’s voltage may be acceptable (A) or unac-

  some theoretical arguments suggest an alternative prob-

  ceptable (U). A certain flashlight requires two batteries,

  ability distribution called Benford’s law:

  1 x 2

  so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 90

  x1 1 p (x) 5 P(1st digit is x) 5 log 10 x5

  1, 2,…, 9

  of all batteries have acceptable voltages. Let Y denote the

  number of batteries that must be tested.

  a. Without computing individual probabilities from this

  a. What is p(2), that is, P(Y 5 2)?

  formula, show that it specifies a legitimate pmf.

  b. What is p(3)? [Hint: There are two different out-

  b. Now compute the individual probabilities and compare

  comes that result in Y 5 3.]

  to the corresponding discrete uniform distribution.

  c. To have Y 5 5, what must be true of the fifth battery

  c. Obtain the cdf of X.

  selected? List the four outcomes for which Y 5 5 and

  d. Using the cdf, what is the probability that the leading

  then determine p(5).

  digit is at most 3? At least 5?

  d. Use the pattern in your answers for parts (a)–(c) to

  [Note: Benford’s law is the basis for some auditing pro-

  obtain a general formula for p(y).

  cedures used to detect fraud in financial reporting—for

  18. Two fair six-sided dice are tossed independently. Let

  example, by the Internal Revenue Service.]

  M5 the maximum of the two tosses (so M(1,5) 5 5,

  22. Refer to Exercise 13, and calculate and graph the cdf

  M (3,3) 5 3, etc.).

  F (x). Then use it to calculate the probabilities of the

  a. What is the pmf of M? [Hint: First determine p(1),

  events given in parts (a)–(d) of that problem.

  then p(2), and so on.]

  b. Determine the cdf of M and graph it.

  23. A branch of a certain bank in New York City has six ATMs. Let X represent the number of machines in use at

  19. A library subscribes to two different weekly news maga-

  a particular time of day. The cdf of X is as follows:

  zines, each of which is supposed to arrive in Wednesday’s mail. In actuality, each one may arrive on Wednesday,

  5 1 6x

  0 x, 0

  Thursday, Friday, or Saturday. Suppose the two arrive inde-

  .06 0 x , 1

  pendently of one another, and for each one P(Wed.) 5 .3,

  .19 1 x , 2

  P (Thurs.) 5 .4, P(Fri.) 5 .2, and P(Sat.) 5 .1. Let

  .39 2 x , 3

  Y5 the number of days beyond Wednesday that it takes

  F (x) 5

  .67 3 x , 4

  for both magazines to arrive (so possible Y values are 0, 1,

  2, or 3). Compute the pmf of Y. [Hint: There are 16 possible

  .92 4 x , 5

  outcomes; Y(W,W) 5 0, Y(F,Th) 5 2, and so on.]

  .97 5 x , 6

  20. Three couples and two single individuals have been

  invited to an investment seminar and have agreed to

  Calculate the following probabilities directly from the

  attend. Suppose the probability that any particular

  cdf:

  3.3 expected Values 109

  a. p(2), that is, P(X 5 2)

  b. P(X . 3)

  1 y3. In this way, Alvie continues to visit friends until he

  c. P(2 X 5)

  d. P(2 , X , 5)

  returns home.

  24. An insurance company offers its policyholders a num-

  A B

  ber of different premium payment options. For a ran- domly selected policyholder, let X 5 the number of

  months between successive payments. The cdf of X is as 0 follows:

  D C 0 x, 1 a. Let X 5 the number of times that Alvie visits a

  5 friends at B and D. If Z 5 the number of visits to

  .30 1 x , 3

  friend. Derive the pmf of X.

  .40 3 x , 4

  b. Let Y 5 the number of straight-line segments that

  F (x) 5

  .45 4 x , 6

  Alvie traverses (including those leading to and from

  .60 6 x , 12

  0). What is the pmf of Y?

  1 12 x

  c. Suppose that female friends live at A and C and male

  a. What is the pmf of X?

  female friends, what is the pmf of Z?

  b. Using just the cdf, compute P(3 X 6) and

  27. After all students have left the classroom, a statistics

  P (4 X).

  professor notices that four copies of the text were left

  25. In Example 3.12, let Y 5 the number of girls born before

  under desks. At the beginning of the next lecture, the

  the experiment terminates. With p 5 P(B) and

  professor distributes the four books in a completely ran-

  1 2 p 5 P(G), what is the pmf of Y? [Hint: First list the

  dom fashion to each of the four students (1, 2, 3, and 4)

  possible values of Y, starting with the smallest, and pro-

  who claim to have left books. One possible outcome is

  ceed until you see a general formula.]

  that 1 receives 2’s book, 2 receives 4’s book, 3 receives his or her own book, and 4 receives 1’s book. This out-

  26. Alvie Singer lives at 0 in the accompanying diagram and

  come can be abbreviated as (2, 4, 3, 1).

  has four friends who live at A, B, C, and D. One day

  a. List the other 23 possible outcomes.

  Alvie decides to go visiting, so he tosses a fair coin twice

  b. Let X denote the number of students who receive

  to decide which of the four to visit. Once at a friend’s

  their own book. Determine the pmf of X.

  house, he will either return home or else proceed to one of the two adjacent houses (such as 0, A, or C when at B),

  28. Show that the cdf F(x) is a nondecreasing function; that

  with each of the three possibilities having probability

  is, x 1 , x 2 implies that F(x 1 ) F(x 2 ). Under what con-

  dition will F(x 1 ) 5 F(x 2 )?