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b

CHAPTER!

5

Other discrete probability distributions

a A radiation counter.

b A flush! What are the odds?

a

� "'

"'-�..,,!-.ll >c.',:.i,;::-,� :;,T��,,-to, :'1

138 OTHER DISCRETE PROBABILITY DISTRIBUTIONS

In Chapter 4, we considered the distribution of a discrete variable - namely the binomial

distribution. We consider now three other useful distributions of a discrete random variable, each with its own characteristic properties:

a Hypergeometric distribution b Geometric distribution c Poisson distribution.

5.1 Hypergeometric distribution: sampling without replacement

In the study of probability, there are many situations which deal with sampling without

replacement from a small population consisting of two kinds of items.

Example 1 A bag contains szyel1.lll�S, thr.e.e-of which are white and.f.cru.r_o.f which are red. A random sample o�s_ is drawn from the bag without replacement. Find the probability distribution of the number of white marbles in the sample.

The number of white marbles in the sample is a hypergeometric variable, X, which can assume the values x = 0, 1, 2 or 3. Probability in these situations can be defined as the ratio of the number of favourable selections to the number of possible selections, each of which is equally likely. The number of possible selections is the number of ways of selecting four marbles when we have seven to select from.

This can be done in G), i.e. 35, ways. The number of favourable selections are:

0 white and 4 red which can be selected in (�) (!) ways

1 white and 3 red which can be selected in (i) (1) ways

2 white and 2 red which can be selected in G) (1) ways

3 white and 1 red which can be selected in G) (1) ways

Pr(X = 0) = (�) (!) G)

G) (1)Pr(X = 1) =

G)

Pr(X =2) = G) (1) =

G) G) (1)Pr(X =3)

G)

1 X 1 35

3 X 4 35

3 X 6 35

1 X 4 35

12 35

18 35

4 = 35

OTHER DISCRETE PROBABILITY DISTRIBUTIONS 139

The probability distribution can be tabulated as follows:

X 0 1 2 3

Pr(X = x) 1 12 18 4

- -

35 35 35 35

Example2 A committee of 10 people consists of six women and four men. A subcommittee of four is to be formed. Find the probability distribution of the number of women on the subcommittee.

@ The number of women in the sample of four is a hypergeometric variable, X, which can assume the values x = 0, 1, 2, 3 or 4. Using:

Pr(X = x) = number of favourable selectionsnumber of possible selections

Pr(X = 0) (�) (!) 1 = = 210(�

o

)

Pr(X = 1) (�) (1) 24 = = 210(�

o

)

Pr(X = 2) (�) (1) 90 = = 210(�

o

)

Pr(X = 3) (�) (t) 80

(140)

=210

Pr(X = 4) (� (6) 15 = =210(14

0)

So the probability distribution of the number of women on the subcommittee is:

X 0 1 2 3 4

Pr(X = x) 1 24 90 80 15

- -

210 210 210 210 210

Examples land 2 are problems dealing with the drawing of a sample, without replacement,

from a finite population containing two different kinds of items: white and red marbles (Example 1), women and men (Example 2). Let the size of the population be N, and the size of the sample be n. Suppose also that the population contains D of kind A and, therefore, N - D of kind B. In Example 1, N = 7, n = 4 andD = 3. In Example 2, N = 10; n == 4 andD = 6.

·� '

?


The number, X, of kind A appearing in a sample is a random variable which can assumeintegral values x = 0, l, 2, ... D if n > D, or x = 0, l, 2, ... n if n � D, and is calleda hypergeometric variable. Its probability distribution is called a hypergeometric

distribution. The probability that the sample contains x of kind A is defined by:

Pr(X = x) = (�) (: = i)

(�)

As an illustration, let us use this formula to derive the hypergeometric distribution ofExample l. Put N = 7, n = 4, D = 3, and X can assume the values x = 0, l, 2 or 3.The table below shows the distribution of the.probabilities of 0, 1, 2 and 3 white marbles.

(�) = G) = 35

xwhite C) (4 � x) Pr(X = x)

0 1 1 1

35

1 3 4 12 -

35

2 3 6 18 -

35

3 1 4 4-

35

Sum= 1

Note that � Pr(X = x) = l. This is to be expected. It is certain that the sampleX=O

will contain 0, 1, 2 or 3 white marbles. These are mutually exclusive events and every possibility is exhausted. Students should consider the formula for evaluating the above probabilities as a simple application of probability defined as the ratio of the number of favourable selections to the number of possible selections, each being equally likely. A knowledge of combinations (considered in Chapter 3) is essential in this context, thereby eliminating the use of formulae. The following example illustrates the use of the hypergeometric and binomial distribution in cases of sampling, without replacement and with replacement respectively, from a small population.

Example 3

OTHER DISCRETI: PROBABILITY DISTRIBUTIONS 141

Iman has six tennis balls, four of which are new and the other two have been previously used. A random sample of three balls is selected. Find the probability distribution of the number of new balls in the sample if the sampling is done: a without replacement b with replacement

a The population of six balls is made up of two kinds of items, namely four new and two old. The number of new balls, X, in the sample of three is a hypergeometric variable which can assume values x = 1, 2 or 3. The number of possible selections of three balls from a population of six is (�) , i.e. 20, each of which is equally likely. From four new balls and two old, we can select a sample of three balls consisting of:

1 new, 2 old in (1) . (;) , i.e. 4, different ways with probability 2�.

2 new, 1 old in (i) . (i), i.e. 12, different ways with probability 1�-3 new, 0 old in (!) . (�), i.e. 4, different ways with probability 2�.

b Since sampling is done with replacement, the number of new balls, X, in the sample of three is a binomial variable which can assume values x = 0, 1, 2 or 3.

2 1 P = 3' q = 3' n = 3.

Pr(X = 0) = (½) 3

Pr(X = 1) = (D GY(�)

=-27

=-27

Pr(X = 2) = G) G) (�Y = 1; Pr(X = 3) = (�)

3

= 27 The following table gives the distribution of the number of new balls in the sample for each of the two situations.

X

0

1

2

3

Total

Pr(X = x)

Hypergeometric

O*

4 20 = 0.2

� = 0.620

4 20 = 0.2

1

Pr(X = x)

Binomial

1 27 = 0.0370

6 27 = 0.2222

.!_g_ = 0.444427

8

27 = 0.2963

1

* Note that the hypergeometric distribution in this case is applicable forx = 1, 2 and 3 but not x = 0. Why is Pr(X = 0) = O?

/


Example 4 A customer buys pistons in batches containing 10 pistons. The customer decides whether or not to accept a batch by inspecting two pistons selected at random without replacement.If neither of the sampled items is defective, the customer accepts the batch but, otherwise,rejects it. a If five per cent of all pistons manufactured are defective, calculate the probabilities of

0, 1, 2, .. . 10 defectives in a batch. b Find the probability that batches actually containing 0, 1, 2, . .. 10 defectives are

accepted. c Calculate the overall expected proportion of batches accepted.

a Since five per cent of all pistons manufactured are defective, the probability ofany one piston being defective is p = 0.05. The number of defectives, X, perbatch is a binomial variable:

Pr(X = x) = (�)(0.05)x(0.95) 10 -xfor 0 � X � 10.

X 0 1 2 3 4 5 6 7 8 9 10

Pr(X = x) 0.598 0.315 0.075 0.011 0.001 zO :::::0 ""0 ""0 ""0 ""0

b If the batch actually contained 0, 1, 2 .. . 10 defectives and a random sample of two pistons is selected without replacement, the distribution of the numberof defectives in this case is hypergeometric.

The probability of acceptance = p_robability of 0 defectives

(i) (10; X)

=

(11) wherex = 0, 1, 2, . . . 10

i.e. the probability of selecting �ectives from x defectives and 2.nondefectives from (10 - x) non-defectives.

Number of 0 1 2 3 4 5 6 7 8 defectives

.;\

Pr(acceptance) 1 4 28 7 / .! \\ 2 2 1 1 - -5 45 15 3 I 9 15 15 45

j

9 10

0 0

c Combining the two distributions, we see that 0.5 98 of all batches are expected to have 0 defectives and all of these would be accepted; 0.315 of all batches are

expected to have 1 defective and� of these would be accepted, etc. So the overall

expected proportion of batches accepted would be:

(1)(0.5 98) + (�) (0.315) + (�D (0.075) + (?5) (0.011) + (½) (0.001)

i.e. approximately 90 per cent.


5.2 Mean and variance of a hypergeometric distribution

Example& Let us calculate the mean and variance of the hypergeometric distributions of Examples 1 and 2. For Example 1:

X Pr(X = x) xPr(X = x) (x - µ)

0 1 0 12 35 - 7

1 12 12 5

35 35 -;::;

2 18 36 2 35 35 7

3 4 12 9 35 35 7

Total 1�

µ = qandu2 = �!in whichN= 7,n = 4andD = 3.

For Example 2:

X Pr(X = x) xPr(X = x) (x - µ)

0 1 0 12 210 - 5

1 24 24 7210 210 - 5

2 90 180 2210 210 5

3 80 240 3210 210 5

4 15 60 8 210 210 5

Total 2i

µ = 2� and u2 = �� in which N = 10, n = 4 and D = 6. The mean of the hypergeometric distribution is:

(x - µ)2 Pr(X = x)

144 49.35

300 49.35

72 49.35

324 49.35

Total 2449

(x - µ)2 Pr(X = x)

144 25.210

1176 25.210

360 25.210

720 25.210

960 25.210

Total 1625

the derivation of which is beyond the scope of this book but is verified in the two worked examples above.


It is interesting to compare the mean of the hypergeometric distribution with the mean, np,

of the binomial distribution. We recall that pis the proportion of 'successes'.

Similarly,� is the proportion of kind A in the population.

The variance of the hypergeometric distribution is:

the derivation of which is beyond the scope of the book but is verified in the two worked examples on page 143. Again, it is interesting to compare this with the variance, np(l - p), of the binomial

distribution where p corresponds to tand the factor 1-:.r = 7 approaches 1 when N, the

population size, is very large and n, the sample size, is small. In this case, the binomial is a suitable approximation to the hypergeometric, and this approximation is frequently used in sampling without replacement from a large population. See Example 11 in Chapter 4.

Exercises Sa

1 A carton contains 12 eggs, five of which are brown and the remainder are white. A / pastrycook selects four eggs at random to make a cake. What is the probability

M distribution of the number of brown eggs in the sample?

W From a pack of 52 playing cards, three cards are drawn at random, without replacement. a What is the probability distribution of the number of clubs in the sample? b What is the probability that the sample contains no more than one club?

� A box contains four red and four blue cubes. Three cubes are drawn at random from the box. What is the probability distribution of the number of red cubes in the sample if the sampling is done: a without replacement? b with replacement?

� A group of five is selected at random from five girls and four boys. If X denotes the number of gi_rlsjn the sample, find: a the probability distribution of Xb the mean and variance of X.

5 In a lottery game, three numbers are selected at random without replacement from the numbers 1, 2, 3 ... 10. If X denotes the number of even numbers in the sample, find: a Pr(X = x),x = 0, l, 2, and 3 b the mean and variance of X.

6 A group of 10 students contains six males and four females from whom a random sample of three is selected. Write the probability distribution of the number of females in the group and find the mean and variance.

7 A box contains five black and three red balls. Three balls are drawn at random from the box. Find the probability distribution of the number of black balls in the sample if: a the first ball is replaced before the next is withdrawn b the balls are chosen without replacement.


8 A group of nine people contains three males and six females. A random sample of five is selected. a What is the probability distribution of the number of males in the sample? b What is the probability that the sample contains more males than females?

9 From a set of eight cards numbered 1 to 8, four cards are selected at random. What is the probability distribution of the number of even-numbered cards in the sample if the sampling is done: a with replacement? b without replacement?

10 A carton contains eight electric light globes of which two are defective. If a random · sample of five globes is drawn from the carton without replacement, find: a the probability distribution of the number of defective globes in the sample b the mean and variance of the number of defective globes in the sample.

11 An angler has caught 15 fish of which three are undersized. A random sample of-three fish is drawn, without replacement, by an inspector. If X denotes the number of undersized fish in the sample, find the mean and variance of X.

12 In a group of 10 students, six are girls and four are boys. The students are selected at random. If X denotes the number of girls in the sample of two, find: a Pr(X = 0), Pr(X = 1) and Pr(X = 2) b the mean and variance of X.

13 A box contains five cubes, three of which are white and two of which are black. Two cubes are selected at random without replacement. If X denotes the number of black cubes selected, find Pr(X = x) where x = 0, 1, and 2. Find the mean and variance of X.

-l 14 From a pack of 52 playing cards, a hand of 13 cards is dealt. Give the mean andstandard deviation of the number of clubs in the hand and give limits between which this number will lie with a probability of about 0.95.

15 A bottle contains 100 marbles of which 60 are red and the remainder are blue. A random sample of 40 marbles is drawn without replacement. How many of these would you expect to be red? Give limits between which this number will lie with a probability of about 0.95.

l �, 16 Mass-produced articles are offered for sale to a customer in batches containing �ght articles, some of which may be defective. Th.e customer accepts or rejects a batch by inspecting two articles selected at random without replacement. I( neither of the sampled articles is defective, the customer accepts the batch, otherwise it is rejected. Find the probabilities that batches containing 0, 1, 2, ... 8 defectives are accepted.

17 Transistors are sold to a customer in cartons containing 10 transistors. The customer selects two transistors at random without replacement. If both transistors are defective, the customer rejects the batch. If none is defective, the customer accepts the batch. If one of the transistors is defective, the customer selects two more at random from the remaining eight and rejects the batch if one or both of these is defective, otherwise the batch is accepted. What is the probability that a batch containing four defectives will be accepted?

18 A carton of 12 eggs contains three eggs which have brown shells. If four eggs are selected at random from the carton and X denotes the number of eggs with brown shells, find: a Pr(X = x), x = 0, 1, 2, and 3 b the mean and variance of X.


" 19 A customer buys transistors in batches of eight and decides whether or not to accept a batch according to the following procedure. Two transistors are selected at random, without replacement. The batch is accepted if neither is defective. If one of the inspected transistors is defective the customer selects another at random from the remaining six. If this additional transistor is defective, or if both of the transistors in the first sample are defective, the batch is rejected. Otherwise the batch is accepted. a Find the probabilities, to three decimal places, that batches containing 0, 1, 2, .. . 8

defectives are accepted. � If four per cent of transistors produced are defective, calculate the probabilities, to

three decimal places, that a batch will contain 0, 1, 2, ... 8 defectives. c Calculate the overall expected proportion of batches accepted.

20 A box contains six black balls and four white balls. A batch of five balls is drawn at random from the box. Find the probability distribution of the number of white balls in the batch if the balls are chosen: a without replacement b with replacement.

Other distributions

So far we have considered two probability distributions, namely the binomial and the hypergeometric. They both refer to a discrete random variable which has an upper limit. We consider now two other distributions of a discrete random variable which have, theoretically at least, no upper limit. The variable can assume the values 0, 1, 2, 3, ...

5.3 Geometric distribution

Frequently, when dealing with a sequence of binomial trials we are interested in the number of failures that occur before the first success. A punter, for example, becomes greatly concerned after a succession of losses; the captain of a team is very disappointed when he or she keeps losing the toss. In cases like these, the pattern is a series of failures (F) followed by a success (S):

i.e. FFF ... S

If X denotes the number of failures before the first success in a sequence of binomial trials in which the probability of a success in a single trial is p and the probability of a failure is q, i.e. (1 - p), then the probability of x failures before the first success is given by:

I Pr(X = x) = qxp, X = 0, 1, 2, 3,. .. I The following properties of the geometric distribution should be noted and verified where necessary: a Xis a discrete random variable that can assume non-negative values, x, from zero to any

positive integer, however large. 00

b � Pr(X = x) = 1. x-0

c The mean is given by E(X) = L=....e p.


d The distribution is calle{�;o�;r�ecause, for successive values of the variable, theterms are the terms of a geometric sequence with common ratio q.

Sometimes the geometric distribution is presented in an-alternative fc;um in which Xdenotes the number of trials necessary to achieve t�st su�e� in this case:

Pr(X=x) = qx - 1

p, X= 1,2,3,... �-

Example 6 Police frequently make road checks for cars exceeding the speed limit and find that, on theaverage, one car in every five on a particular stretch of road exceeds the limit. If X denotesthe number of cars traveffingwithin the speed limit before one exceeding the limit is. �---,,_ discovered, find: · -

a Pr(X = 4) b Pr(X :s,;; 4)

p = ½, q = � and Pr(X = x) = (�Y (½), x = 0, 1, 2, ...

a Pr(x = 4) = (�)4 (½) 25 6

= 3125"""0.082

b Pr(X :s,;; 4) = Pr(X = 0) +

Pr(X = 1) + , .. + Pr(X = 4)= ½ + � · ½ + (�Y G) + · · · + (�)

4

G)Th. · · · · h l 4 d 51s 1s a geometric senes wit a = 5, r = 5 an n =

Example 7

Ss =

a(l - ,s)1 - r

½ [1 - (�)5

]4 1 - -5

= 1 - (�)5

= l _ 102 43125

2101= 3125"""0.672

A targetshooter scores a bull's-eye on 60% of occasions. If X denotes the number of trialsnecessary for the first success, find: -a Pr(X = 5) b Pr(2 :s,;; X :s,;; 6)

p = 0.6, q = 0.4 and Pr(X = x) = (0.4t- 1 (0.6), x=l,2,3, ...

a Pr(X = 5) = (0.4)4 (0.6)= 0.0153 6

.,


b Pr(2 � X � 6) = Pr(X = 2) + Pr(X = 3) + . . . + Pr(X = 6) = (0.4)(0.6) + (0.4)2 (0.6) + ... + (0.4) 5 (0.6)

This is a geometric series with a = (0.4)(0.6), r = 0.4 and n = 5.

Ss =

a(l - r4) 1 - r

(0.4)(0.6)[1 - (0.4)4]

=--------

1 - 0.4

::::: 0.3898

Exercises Sb

1 A punter who keeps a record of her achievements finds that, on average, she picks a winner in one race in every five. If X denotes the number of losses before the first win, find: a Pr(X = 5) b Pr(X � 5).

2 A die is rolled until a six turns up. If X denotes the number of trials necessary for a six to turn up, find: a Pr(X = 4) b Pr(2 � X � 6).

00

3 For the geometric distribution defined by Pr(X = x) = qxp, show that � Pr(X = x) = 1.

4 For the geometric distribution defined by Pr(X = x) = qxp, find: a Pr(X = 3) b Pr(X> 3) c Pr(3 � X � 8).

X=O

5 A fair die is tossed until an even number appears. What is the probability that the number of odd numbers appearing uppermost before an even number appears is: a 2? b at least 2? c no more than 2?

6 A fair coin is to be tossed until a head appears. What is the probability that at least three tails will appear before a head appears?

7 An urn contains six cubes, four of which are black and two of which are white. What is the probability that no more than three white cubes are drawn before a black one is drawn, assuming the sampling is done with replacement?

, 8 A fair die is thrown until a six appears. What is the probability that the number of throws necessary is: a exactly three? b at least three? c less than three?


9 A targetshooter finds that, on average, a bull's-eye is scored on three shots out of four. If X denotes the number of misses before a bull's-eye is achieved, find: a Pr(X = 2) b Pr(3 :,;:; X :,;; 5) c the least value of x such that Pr(x :,;; X:,;; x + 2) < 0.05.

10 In a mathematics examination, 75 per cent of students are granted a pass. If X denotes the number of failures �efore a pass is obtained, find: a Pr(X � 1)

b Pr(l :,;:; X:,;; 5)

11 A footballer finds that if he marks the ball within 20 metres of the goals, he scores a goal, on average, three times out of four. Find the probability that, in a particular game, he fails four times before scoring a goal.

"\ 12 A punter finds that on average he is able to pick one winner in every ten races. If X denotes the number of losses before the first win: a show that Pr(X = x) = (0.9Y(0.1) for x = 0, 1, 2, ... b find Pr(4 :,;:; X:,;; 14)

7c find the expected value of X.

13 A die is loaded in such a way that the probability that an even number will appear three times before an odd number appears is nine times the probability that an even number will appear five times before an odd number appears. What is the probability that an even number will appear in a single trial?

14 A punter finds that, in the long run, she is able to pick a winner in one race out of five. If X denotes the number of losses before the first win: a write an expression for Pr(X = x).

b find Pr(5 :;;; X :,;; 10).

15 Police frequently make road checks for unroadworthy cars. If, on average, one car in ten is unroadworthy, find the probability that, in a particular road check, the first five cars are roadworthy and the sixth is unroadworthy.

16 The probability that a newly-born child is a girl is approximately 0.48. If X denotes the number of girls born before a boy is born on any particular day in the maternity section

· of a large hospital, find:a Pr(X = 0)b Pr(X � 1)

c Pr(2 <X < 8).

5.4 Poisson distribution

An approximation to the binomial distribution The binomial distribution assumes a convenient form whenp, the probability of a success in a single trial, is very small and n, the number of independent trials, is very large (n -+ oo), but the average number of successes,µ, which is equal to np, is finite - in which case, the probability of x successes is given by the formula:

e-µ µx Pr(X= x) = -

xi


Consider the probability of, for example, three successes in the binomial distribution.

Pr(X = 3) = (;) qn - 3 p 3

= n(n - it-

2) p 3(1 - p)n - 3 since q = 1 - p.

( 1) ( 2) 3 ( np)n1 - ,i 1 - ,i (np) 1 - r.i

3 ! (1 - p)3

Since n --> oo .! � --> 0 and (1 - .!) and (1 - �) --> 1.'n' n ' n n

Then, noting thatµ = np, we obtain:

µ3 (1 - t!:)"Pr(X = 3) = n

3!(1 - p) 3

But ( 1. - �)" --> e-µ for large values of n, and since p is small, (1 - p) --> I.

e-µ µ3 Pr(X = 3) = 31.

Consider, for example,µ = 2 and so e - 2 = 0.1353 ... using a calculator as follows:

[g)II Ill II

Consider ( 1 - �)" for n = 200, 500 and 1000

When n = 200, ( 1 - �)" = 0. 99200 = 0.133 9 ... using a calculator.

When n = 500, ( 1 - �)" = 0.996500 = 0.1347 .. .

When n = 1000, ( 1 - �)" = 0. 998 1000 = 0.1350 .. .

As n increases, ( 1 - �)" --> e - 2•

Example 8 The records of the maternity ward of a large hospital show that, on average, one birth in every 100 produces twins. What is the probability that, out of a total of 250 births in a given month, there will be at least two sets of twins?

The number of twins per month is a binomial variable withp = lbo and n = 250.

The Poisson approximation in this case would be a reasonable one.

1 µ = np = 250 X lO0 = 2.5

This is the expected number of twins per month. If X denotes the number of twins per month, then:

Pr(X = x) =

e-z.s �2·5f, x = 0, 1, 2, . .. 250. x.

Pr(X � 2) = 1 - [Pr(X = 0) + Pr(X = 1)]

("

Pr(X = 0) = e - 2·5


= 0.0821 (using a calculator) Pr(X = 1) = e- 2

-5 x 2.5

= 0.0821 X 2.5 = 0.2052

Pr(X � 2) � 1 - (0.0821 + 0.2052)= 0.7127

Using the binomial distribution with p = 0.01, q = 0.99 and n = 250, the result is 0.7142. Check this with the aid of your calculator.

Other uses of Poisson distribution

The Poisson distribution must not merely be regarded as an approximation to the binomial distribution. In the binomial distribution, we know the number of times the event occurred and also the number of times it did not occur. If a die is tossed 100 times and a six turns up 60 times, we know then that a six did not turn up 40 times. The Poisson distribution has been found to fit a variety of situations in which the number of times an event occurs can be counted but there is, theoretically at least, no upper limit to the number of times it may occur. The proprietor of a service station can count the number of cars entering the station's driveway per day, the number of flaws on specimens of steel can be counted, the number of telephone calls per hour can be noted, the number of dust particles in standard volumes of the atmosphere can be counted, the number of electricity power failures per week can be noted, the number of radioactive emissions per unit time from a radioactive source can be counted on a Geiger counter, and so on. Many of these examples refer to random occurrences in continuous time, e.g. the number of cars on the driveway, or the number of telephone calls, electricity power failures, or radioactive emissions.

Others refer to random occurrences in length, area or volume, e.g. the number of flaws on specimens of steel, or the number of dust particles in the atmosphere. Using as an example the occurrence of surface defects on specimens of sheet metal, we can indicate the general nature of the process which produces a Poisson probability distribution. The surface area would have some average rate of defects per unit area, for example two defects per specimen. If we consider each specimen to be divided into very small sub-areas, for instance 1 cm 2, the probability of a defect in any one sub-area would be small and the probability of more than one would be practically zero. The number of defects that occurs in unit area does not depend on where that unit area is on the specimen, i.e. the defects are randomly scattered over the specimen. The number of defects that occurs in any subarea does not depend on the number of defects in any other sub-area.

In this example, the sub-interval is a unit of area. Similar conditions would characterise examples in which the sub-interval is a unit of time, length or volume.

In all of these cases, there is, theoretically at least, no upper limit to the number of times an event may occur. They refer to a random variable, X, which can assume any integral value, x, from zero to any positive integer, however large, with a probability given by the formula:

Pr(X = x) =

e-;/'X. x = 0, 1, 2, .. .

and the parameter ),. can be shown to be equal to µ,, the mean value of X.


Example 9 Electricity power failures occur according to the Poisson law with an average of three failures every twenty weeks. Find: a the probability distribution of the �umber of failures over a four-week period b the probability that during a particular four-week period there will be some failures.

Note:

a µ = 3 failures every 20 weeks = 0.6 failures every four weeks

Assuming the Poisson law applies withµ = 0.6, then:

Pr(X = x) = e -o.6 (0.6r x!

,

Pr(X = 0) = e- 0·6

Pr(X= 1) = e-0·6 X 0.6

Pr(X = 2) =

e-0.6 X (0.6) 2

2!

X = 0, 1, 2, .. .

= 0.5488 = 0.3293

= Pr(X = 1) x 0.6= 0.0988 2

Pr(X = 3) = e -o.6 X (0.6) 3

3!

= Pr(X = 2) X 0,6= 0.0198 3

Pr(X = 4) = e -o.6 X (0.6)4

4!

= Pr(X = 3) X 0.6 - = 0.00304

Pr(X = 5) = e -o.6 X (0.6)5

5!

= Pr(X = 4) x 0.6

= 0.0004 5

Pr(X > 5) = 1 - Pr(0 � X � 5) = 0

b The statement 'there will be some failures' means that there will be one or more failures.

Pr(X;;:: I) = I - Pr(X = 0) = I - e -o

.6

= 1 - 0.5488 = 0.4512

(i) The probabilities increase for values of X up to around the mean value of X and thendecrease. This may not be apparent in this example.

(ii) Each probability can be evaluated from the preceding one by multiplying by ______!!:__1.x+

i.e. Pr(X = x + I) = Pr(X = x) X ------1!:...__l • x+

Example 10


The number of surface defects on specimens of steel is known to have a Poisson distribution. On examination of a large number of specimens, it is found that 20 per cent of specimens have one or more surface defects. Calculate the average number of defects per specimen and the proportion of specimens which has only one defect.

Since 20 per cent of specimens have one or more defects, it follows that Pr(X ;;;: 1) = 0.2, where X denotes the number of defects per specimen.

i.e. 1 - Pr(X = 0) = 0.2 i.e. e-11 = 0.8

-µ, = loge 0,8= - 0.2231

µ, = 0.2231 = average number of defects per specimen

Pr(X = 1) = e-11 . µ, = 0.8 X 0.2231 = 0.1785 (18 per cent approximately)

Example 11 Accidents in a particular factory occur randomly throughout the year at an average rate of nine per year. The company offers a bonus of $1000 per month to the employees if no accidents occur during that month, $800 if one accident occurs, $400 if two occur and no bonus if more than two occur. Calculate the amount expected to be paid in bonuses each month. Assume that the number of accidents per month is a Poisson variable.

µ, = 9 accidents per year = 0.75 per month

If X denotes the number of accidents per month, then:

e - o.1s (0.75YPr(X = x) = --�� x = 0, 1, 2, . . .

x! ,

Pr(X= 0) = e-0

·75

Pr(X = 1) = e- 0·75 X 0.75

= 0.4724 = 0.3543

Pr(X = 2) = e-0.

75 X (0.75)2

= 0.1328 2!

The amount, $ Y, paid in bonuses per month is a random variable which can assume the values y = 1000, 800 and 400 with probabilities associated with the number of accidents per month as shown in the following table:

y Pr(Y = y) y Pr(Y = y)

1000 0.4724 472.40

800 0.3543 283.44

400 0.1328 53.12

808.96

/J,y = 808.96 The company expects to pay $808.96 per month.


Below is a program that, for an input value M, the mean value of X will output Pr(X = x)

for x = 0, 1, 2, 3, 4, 5 and> 5 for the Poisson distribution defined by: -MMx

Pr (X = x) = e , in the case M = l x.

using the fact that:

Pr(X = x) = Pr(X = x - l)M

.X

1 O REM Poisson Distribution 35 REM M stores mean

40 REM S stores partial sum of probabilities

45 REM X stores counter 50 REM P stores prob for each X

60 S=O

65 INPUT M

70 PRINT "MEAN OF DISTRIBUTION = "; M

75 PRINT

80 PRINT TAB (3); "X"; TAB ( 12); "PROB"; TAB ( 27); "S" 85 PRINT TAB ( 2);" ---------";TAB ( 9); " --------- "; TAB ( 24)" ---------"90 PRINT

95 P=EXP(-M ):S=S+P

100 PRINT TAB ( 3); X; TAB ( 1 O); P; TAB ( 25); S

1 05 FOR X = 1 TO 5

110 P=P*M/X

115 S=S+P

120 PRINT TAB ( 3); X; TAB ( 1 O); P; TAB ( 25); S

125 NEXT X

130 P= 1 -S

135 PRINT TAB ( 2); ">5"; TAB ( 1 O); P

140 END

X PROB s X

0 0.818730753 0.818730753 0 1 0.163746151 0.982476904 1 2 0.0163746151 0.998851519 2 3 1.091641 E-03 0.99994316 3 4 5.45820502E-05 0.999997742 4 5 2.18328201 E-06 0.999999925 5

>5 7.49714673E-08 >5

PROB

1.23409804E-04 1 . 11 068824E-03 4.99809707E-03 0.0149942912 0.0337371552 0.0607268794 0.884309479

Mean of distribution = 0.2 Mean of distribution = 9

Exercises Sc

s

1.23409804E-04 1.23409804E-03 6.23219511 E-03 0.0212264863 0.549636415 0.115690521

1 A manufacturer of electric light globes finds that, in the long run, 2 per cent of the

globes manufactured are defective. What is the probability that, out of a batch of 200

globes, there will be more than one defective?

2 Certain mass-produced articles which have, on the average one defective in every 200,

are marketed in packets of 50. What percentage of cartons, assuming a Poisson

distribution, would contain:

a no defective articles?

b not more than two defective articles?


3 Assuming that the chance of death of accidental drowning is 30 �OO' use the Poisson

distribution to estimate the probability that, out of a random selection of 1500, there will be at least one death by accidental drowning.

4 During any one year, fatal accidents occur in the mining industry with the small probability of 0.50Jo. Use the Poisson approximation to the binomial distribution to determine the probability of a mine which employs 250 workers having at least one fatal accident in a year.

5 On the average, a typist has to correct one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page.

6 Between the hours of 2 p.m. and 4 p.m. the average number of telephone calls made by a certain company is 60. Calculate the probability that during any period of five minutes there will be 0, 1, 2, or more than 2 telephone calls made.

In a particular clothing store, the number of demands per day for a certain item of clothing is known to have a Poisson distribution with_!11ean 4. The daily supply of this item to the store is 5. What is the probability that on a particular day: a there will be some demand for the item? b the demand will be less than the supply?

8 A radioactive source emits particles at the average rate of 120 per hour. What is the probability that, during a given period of one minute, there will be: a no more than one particle emitted? b at least one particle emitted?

9 During weekdays, the proprietor of a service station counts the number of cars entering the driveway and finds that, on the average, there are 10 cars every·hour. Assuming thePoisson law applies, find:

' \ '

·

a the probability distribution of the number of cars on the driveway per quarter-hour b the probability that during a particular quarter-hour some cars will enter the

/- driveway. � A variable, X, has a Poisson distribution and Pr(X = 2) = Pr(X = 1).

a Find,the mean value of X.

b Find the probability that Xis less than the mean value.

11 a A pastrycook keeps a regular check on production and finds that the curran,t buns contain an average of 2.5 currants per bun. What percentage of currant buns would be expected to contain O currants, assuming the Poisson law in this case?

b If the pastrycook increases the proportion of currants in the mixture so that the mean number of currants per bun is increased-by 10 per cent, calculate the probability that no bun contains fewer than two currants.'

12 Faults occur randomly along the length of a piece of yarn, and the number of faults in bobbins holding a fixed length of yarn may be assumed to have a Poisson distribution. A bobbin is rejected as soon as a fault is found on it, and over a long period 33 per cent of all bobbins inspected have been rejected. Calculate: a the mean number of faults per bobbin b the probability that a rejected bobbin contains only o.ne fault.-

13 The number of surface defects on specimens of steel, polished and examined in a metallurgical laboratory as a routine inspection measure, is known to have a Poisson distribution. a In a large homogeneous collection of specimens, 10 per cent have one or more

defects. Find the percentage having precisely two defects." b Why might you expect the Poisson distribution_to apply to data of this kind, and

what could happen to upset it?

156 OTHER DISCRETE PROBABILITY DiSTRIBUTIONS

14 The number of demands for a certain expensive item of equipment received by a laboratory supplier averages 2.0 per week, varying randomly from week to week in accordance with the Poisson probability law. At the beginning of a particular week the supplier had five items in stock. Determine: a the probability that there is a demand for precisely three items in that week b the probability that there are more demands than can be satisfied out of the supplier's

stock c the probability distribution of the number of items the supplier has in stock at the

end of the week.

15 a Glass sheets have faults, called seeds, which occur at rando� at the rate1ofO.1 _!J1er

square metre. Find the probability that rectangular sheets of glass, of dimensj,dnJ 5 metres x 2 metres, will contain: (i) 0 seeds

(ii) more than one seed.b Sheets with 0 seeds bring a profit of $20, sheets with one seed a profit of $10 and

a rejected sheet incurs a loss o1f $5. Find the overall profit (or loss) per sheet. ,,

16 If Xis a Poisson variable and Pr(X � 1) = 0. 7769, find: a E(X)

b Pr(X> 1 J X<4).

17 In one hour, 1800 alpha-particles are radiated by radioactive material, the amount of unchanged material being not appreciably diminished. If the whole period of one hour is divided into intervals of 10 seconds, find the expected number of intervals during which 0, 1, 2 and 3 alpha-particles are radiated. Assume the Poisson law.

18 Accidents at a large assembly plant are known to follow a Poisson distribution with a rate of 0.8 accidents per month. To encourage safety on the plant the company operates a bonus scheme whereby, each month, the workers share a sum of $2000 if no accidents occur and a sum of $1000 if one or two accidents occur, while no bonuses are paid if more than two accidents occur. Find the mean amount the company has to pay out in bonuses each month.

19 A typist finds that, on average, he makes one mistake every two pages. Assuming the Poisson law, what proportion of pages that contain one or more mistakes will contain exactly one mistake?

1 20

l A hire company has five trucks which it hires out by the day. The manager observes that, on about nineteen days out of every twenty, there is some demand for the trucks. The number of demands per day is a Poisson variable. a What is the expected number of demands per day? b What is the probability that there is a demand for exactly two trucks on a given day? c What is the probability that, if there is some demand on a particular day, that

number does not exceed two? ··

d What is the probability that during a particular week (five days), there will be some demand on at least four days?

21 An observer stood at the intersection of two suburban streets for one hour and counted the number of cars passing through the intersection each minute. The observer noted that on six of these time intervals there were no cars. Assuming that the number of cars per minute is a Poisson variable, estimate: a the mean number of cars per minute b the probability of no more than one car per minute

"c the number of time intervals in which there would be exactly two cars.

OTHER DISCRE'TE PROBABILITY DISTRIBUTIONS 157

22 It is observed over a long period that, in one week in 20, one or more flying saucers are reported over the Northern Territory. The number reported is assumed to have a Poisson distribution. State the probability that none will be reported in a particular week and use your result to obtain an estimate of the mean number of reports per week. Show that the mean number per year is 2.7, correct to one decimal place.

23 In Victoria, in 1987, there were approximately 15 suicides per 100 000 of population. Assuming a Poisson distribution, what is the probability of no more than one suicide in a town of 25 000 people?

24 A manufacturer produces rectangular sheets of glass, measuring 2.5 metres by 2.0 metres, which are subject to flaws at the rate of 0.1 per square metre. a Assuming that the number of flaws per sheet follows a Poisson distribution, calculate

the probability that a sheet chosen at random contains: (i) no flaws(ii) exactly one flaw(iii) more than one flaw.

b If sheets ·with no flaws produce a profit of $5 and sheets with one flaw produce a profit of $2, while sheets with more than one flaw incur a loss of $3, find the expected profit per sheet produced.

5.5 Exponential distribution Many of the examples considered in relation to the Poisson distribution refer to the number of random 'even'ts' in continuous time, e.g. the number of cars on a driveway per uriit time, the number of telephone calls per hour, the number of electricity power failures per week, the number of radioactive emissions per unit time, and so on, in which the variable is l_ discrete and can assume the values 0, 1, 2, 3 ... In these contexts, we can also consider the time lapse between successive 'events', e.g. the time lapse between successive telephone calls, the time lapse between radioactive emissions, and so on. Also, the examples considered in relation to the geometric distribution refer to the number of failures before the first success. In these contexts we can also consider the time lapse before the first success. We are concerned, then, with a continuous variable, 'time', and it can be shown that in these contexts, the variable has an exponential distribution defined by:

f(x) = }..e-xx for all x � 0

This distribution will be considered in Chapter 14.

[:l 5.6 Poisson distribution projectLook through the questions in the set of exercises just completed (Exercises 5c) and observe the situations in �hich we assumed that the Poisson law applied. (Ignore the first five questions in which the Poisson distribution is used as an approximation to the binomial distribution.) The obvious question one.would ask is 'How can we \ make such an assumption?' Obviously, we could collect data, set out these data in the form of a frequency distribution and calculate the mean value of the variable. Then, using this mean as an estimate ofµ, we could calculate the frequencies expected on the assumption of the Poisson law and compare them with the actual frequencies. If the calculated frequencies follow a pattern similar to the actual frequencies, this will justify the assumption of a Poisson distribution.

1d


The most famous example is the data compiled by Quetelet and von Bortkiewiez from the records of 10 Prussian army corps over a 20 year period from 1875 to 1894. The data refer to the number of deaths per army corps per annum resulting from the kick of a horse.

Number of deaths (X) 0 1 2 3 4

Frequency (f) 109 65 22 3 1

X f xf

0 109 0

1 65 65

2 22 44

3 3 9

4 1 4

200 122

The average number of deaths per annum x = 1�� = 0.61.

Using x as an estimate of µ,, then the proportion of corps years during which we would expect the number of deaths, X, to be equal to x, on the assumption of a Poisson distribution, is given by:

e -o.61 (O.61fPr(X = x) = --�-, x = 0, l, 2, ...

x!

Number of corps years during which we would expect x deaths

= 2ooe-o.61 (O.6lf = 200 Pr(X = x)x!

Give x the values 0, 1, 2, 3 and 4.

200 Pr(X = O) = 200e -0·61

200 Pr(X = 1) = 200e -0·61 X 0.61

200 Pr(X = 2) = 200e -0.61 X (0.61)2

2!

200 Pr(X = 3) = 200e -0.61 X (0.61) 3

3!

200 Pr(X = 4) = 200e -0.61 X (0.61)4

4!

108.7 66.3

20.2

= 4.1

= 0.6

The calculated frequencies follow a pattern similar to the actual frequencies and, therefore, justify the assumption of a Poisson distribution.

1 The following frequency distribution shows the number of 'wrong number' telephone calls per day, X, over a period of 100 days, recorded by a telephonist in a city office.

Number of wrong numbers per day (XJ 0 1 2 3 4

Number of days (f) 57 30 10 2 1


Assuming that these 'wrong number' calls are due to random causes, calculate the frequencies expected on the assumption of a Poisson distribution, and compare them with the actual frequencies:

(55, 33, 10, 2, 0)

2 A load of 480 blocks of firewood, of fairly even size, were delivered to me. On splitting them, I found 120 wood grubs in them, three of them in one log, two each in 12 logs, and the others occurring singly. Calculate the frequencies expected on the assumption of a Poisson distribution and compare with those observed:

(374, 93, 12, 1)

3 Find, from whatever source you can, the number of days over a two year period on which there were 0, 1, 2, 3, 4, ... deaths as a result of car accidents on Victorian roads. Test whether the number of deaths per day is a Poisson variable.

4 Your Physics laboratory may have a Geiger counter which counts the number of emissions per unit time from a radioactive source. Use it to determine whether the number of emissions per unit time is a Poisson variable.

5 Count the number of cars which pass a certain point, on a not-too-busy road, every minute over an extended period of an hour or two. Construct a frequency distribution of the number of cars per minute and calculate the frequencies on the assumption of a Poisson distribution.

6 Test some of the following: a The number of errors per page of a book b The number of phone calls received per unit time at home each night over a

period of time c The number of power failures per week in a certain area. (Obtain data from

the SEC.)

5. 7 Probability and matrices: Markov chains

Example 12 In a certain region, each day can be considered as either wet or fine. Records show that, if it is wet one day, there is a probability of 0.4 that it will be wet again the next day. If it is fine one day, there is a probability of 0. 7 it will be fine again the next day. a If it is wet on Saturday, find the probability that it will be:

(i) fine on Sunday· (ii) wet on Monday(;ii)'1 fine on Monday

b If it is.fine on Saturday, find the probability that it will be: (i) wet on Monday(ii) fine on Monday.

I


a The given information can be set out neatly as a 2 x 2 probability matrix P:- W F

p = W [0.4 0.6]F 0.3 0.7

This is called a transition matrix. The elements of the transition matrix representprobabilities and the sum of the elements of each row is 1. Why? The two possible outcomes, W and F, for each day are called states.

W is state 1 (first row), i.e. wet on SaturdayF is state 2 (second row), i.e. fine on Saturday.

If it is wet on Saturday, there is a probability of 0.4 that it will be wet on Sundayand a probability of 0.6 that it will be fine on Sunday. So the answer to a(i) is 0.6.The process must begin with one of these states.In a Markov chain process, a series of trials is conducted. We are given, in parta of this example, that it is wet on Saturday. This is our starting point, day 0. We can then calculate the probability of it being wet or fine on days 1, 2, 3 ... ,i.e. on Sunday, Monday, Tuesday .. . This chain process can be set out neatly with the aid of the following treediagram to answer parts a(ii) and a(iii).

Figure 5-1

w

Day O Day 1 (Sat) (Sun)

w

F

w

F

Day2 (Mon)

Outcomes

WW

WF

FW

FF

This tree diagram applies only if it is wet on Saturday.(ii) Pr(W on Monday) = Pr(WW) + Pr(FW)

= 0.16 + 0.18 = 0.34

(iii) Pr(F on Monday) = Pr(WF) + Pr(FF)= 0.24 + 0.42 = 0.66

Probability

0.16

0.24

0.18

0.42

_J

b We are given that it isjine on Saturday. This is now our starting point for day0. We can now calculate the probability of it being wet or fine on days 1, 2, 3 . . . , i.e. on Sunday, Monday, Tuesday . . .


Outcomes

Figure 5-2

F

Day0 (Sat)

Day 1 (Sun)

w WW

F WF

w FW

F FF

Day 2 (Mon)

This tree diagram applies only if it is.fine on Saturday.(i) Pr(W on Monday) = Pr(WW) + Pr(FW)

= 0.12 + 0.21 = 0.33

(ii) Pr(F on Monday) = Pr(WF) + Pr(FF)= 0.18 + 0.49 = 0.67

Consider again the transition matrix:

Then:

p = [0.4 0.6]0.3 0.7

p2 = [0.4 0.6] [0.4 0.6]0.3 0.7 0.3 0.7

= [0.34 0.66]0.33 0.67

Probability

0.12

0.18

0.21

0.49

The elements in the first row of P2 are the answers to a(ii) and a(iii) and the elementsof the second row are the answers to b(i) and b(ii).

P2 is a probability matrix and its elements represent the probabilities after two trials(in this case, after two days).

p3 = p. p2

= [0.4 0.6] [0.34 0.66]0.3 0.7 0.33 0.67

W F= W [0.334 0.666]F 0.333 0.667

This means that, if it is wet on Saturday, the probability that it will be wet onTuesday is 0.334 and the probability that it will be fine on Tuesday is 0.666. Also, if it is.fine on Saturday, the probability that it will be wet on Tuesday is 0.333and the probability that it will be fine on Tuesday is 0.667. Check that these are correct by extending the tree diagrams in Figure 5-1 andFigure 5-2.


Example 13 Suppose that each mother has exactly one daughter. Suppose that a tall mother has a probability of 0.6 of having a tall daughter, a probability of 0.2 of having a daughter of medium height and a probability of 0.2 of having a short daughter. Suppose a mother of medium height has probabilities of 0.1, 0. 7 and 0.2 of having a tall, medium and short daughter respectively and a short mother has probabilities of 0.2, 0.4 and 0.4 of having a tall, medium and short daughter respectively.

a Find a 3 x 3 transition matrix, P. b What is the probability that the granddaughter of a short mother will be short?

a T M S

T [

0.6 0.2 0.2] P = M 0.1 0. 7 0.2

S 0.2 0.4 0.4

Note that the sum of the probabilities of each row is 1. Why?

b This is a two-stage process extending over two generations from mother to daughter to granddaughter.

[0.6

P2 = 0.1 0.2

0.2 0.7 0.4

T T

[0.42

= M 0.17 S 0.24

0.2] [

0.6 0.2 0.1 0.4 0.2

M s

0.2 0.7 0.4

0.34 0.24] 0.59 0.24

0.48 0.28

0.2] 0.2

0.4

The elements of the first row of this matrix tell us that there is a probability of 0.42 of a tall mother having a tall granddaughter, a probability of 0.34 of her having a medium-height granddaughter .and a probability of 0.24 of her having a short granddaughter. We can give a similar interpretation to the elements of rows two and three.The probability that the granddaughter of a short mother will be short is 0.28 (3rd row, 3rd column). We can, likewise, answer the question with the aid of the following tree diagram.

Figure 5-3

s

Short mother

0.4

s

Daughter

Outcomes Probabilities

T

M

S - TS 0.04

T

Grand

-MS 0.08

- ss 0.16

daughter

Pr(granddaughter is short) = Pr(TS) + Pr(MS) + Pr(SS) = 0.04 + 0.08 + 0.16 = 0.28


Exercises 5d

1 In a small town, two supermarkets compete for business. Each week IOOJo of thecustomers of A change to B, while 200Jo of the customers of B change to A.

a If a customer shops at A this week, what is the probability that he/ she shops at Ain two weeks time?

b If a customer shops at B this week, what is the probability that he/she shops at Bfor each of the next two weeks?

2 Renata finds that she is late for work on 50Jo of occasions if she is on time the previousday, and on IOOJo of occasions if she is late the previous day. Given that she was on time on Monday: a find the transition matrix for this Markov chainb find the probability that she is on time on Wednesday.

3 An officeworker either takes a bus or drives to work each day. The officeworker nevertakes the bus two days in a row, but if she drives to work there is a 50-50 chance that she will drive to work again the next day. a Find the transition matrix. b If she takes the bus on Monday, what is the probability that she will take the bus

on Wednesday? c If she drives on Wednesday, what is the probability she will take a bus on Friday?

4 State which of the following matrices are probability matrices:

a [i �] b u n C [ ! -1 1]

d [l I

!] e u 1] 5 The so-called 'drift' to the cities has been of concern in recent years. Suppose the

transition matrix for yearly movement between city and country is as shown below:

CityCity [0.99Country 0.05

a Interpret this matrix.

Country0.01]0.95

b Find the probability that a person who is now living in the country will be living inthe city in two years time.

6 Market analysis in a certain region has established that, on average, a new car ispurchased every three years. With respect to those changing cars, the buying patternsare described by the matrix:

Large SmallA = Large [600Jo 400Jo]Small 250Jo 750Jo

Rewrite A as a probability matrix and find the following:a the probability that a person who now owns a large car will own a large car in eight

years' time. b the probability that a person will be driving the same kind of car in eight years' time

as he/ she does now.


7 Suppose that a rumour is circulated concerning whether person A did or did not make

a certain statement. Suppose that each time the rumour is passed on there is a 0.01

probability that the teller reverses the form in which he/she heard it.

a Write the transition matrix for the two states corresponding to 'A made the

statement' and 'A did not make the statement'.

b What will the population believe about A after the rumour has been circulated twice?

8 A country has a policy whereby two airlines compete for the domestic market. A study

of customer satisfaction shows that 75% of those currently flying with airline A will

do so again next time, while 25% will change to airline B for their next trip. It is found

that 80% currently using airljne B are satisfied to remain with the airline but that 20%

will fly with airline A on their next trip.

a Write a transition matrix for the system.

b If, at the present time, customers are shared equally by the airlines, what is the

probability that a particular person will fly with airline A on her third trip?

9 In a certain country there are three classes of voter - conservative (C), progressive (P)

and undecided (U). Among the conservatives of one generation, 80% of their next

generation are also conservatives, 10% become progressive, while 10% are undecided

and vote according to prevailing influences.The corresponding figures for the

progressives are 20% (C), 60% (P) and 20% (U), while for those undecided, the figures

are 20% (C), 40% (P) and 40% (U).

a Set out the data in matrix form.

b Calculate the probability that a conservative will have a grandchild who is a

progressive.

10 Transmission of a code consisting of a series of zeros and ones is subject to error. If

0 is sent, there is a probability of 0.01 of error (i.e. that a 1 is received instead of a 0).

If a 1 is sent, the probability of error is 0.05.

a Construct a Markov chain to represent the process of transmission, and find the

matrix of transition probabilities.

b If the initial transmission is a zero, what is the probability that a zero is received after

the third stage?

11 Gabriel and Neumann (1962) developed a Markov Chain model for daily rainfall

occurrence in Tel Aviv during the wet season. (The probabilities were based on data

for a 27-year period.) The statistics for the month of November are:

Previous Day Actual Day Wet

Wet 11 7 out of 1 95

Dry 80 out of 615

Construct a transition matrix for these data and find the probability that, if a given

Sunday is wet, the following Wednesday will be dry.

12 Three boys Aldo, Brian and Carlos throw a ball to each other. Aldo always throws the

ball to Brian and Brian always throws to Carlos. However Carlos is just as likely to

throw the ball to Aldo as he is to throw it t,J Brian.

a Find the 3 x 3 transition matrix.

b If Carlos has the ball, what is the probability that, after two throws, the ball is back

to him?


13 Assume that a person's profession can be described as professional, trade or labour.

Assume that of the children of professional people, 700/o are professional, 20% are

tradespeople and 100/o labourers. For the children of tradespeople, the figures are 40%,

40% and 200/o and, for the children of labourers, the figures are 200/o, 40% and 40%.

a Set up a 3 x 3 transition matrix.

b Find the probability that the grandchild of a labourer is a professional person.

14 If P is a probability matrix, then P2 is a probability matrix. Prove this by showing that

the square of a 2 x 2 matrix whose rows add to one is another matrix whose rows add

to one.

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