[Ans. .3; .4; .3.]

14. Assume that chest X-rays for detecting tuberculosis have the fol-

lowing properties. For people having tuberculosis the test will

detect the disease 90 out of every 100 times. For people not hav-

ing the disease the test will in 1 out of every 100 cases diagnose

the patient incorrectly as having the disease. Assume that the

incidence of tuberculosis is 5 persons per 10,000. A person is

selected at random, given the X-ray test, and the radiologist re-

ports the presence of tuberculosis. What is the probability that

the person in fact has the disease?

4.8 Independent trials with two outcomes

In the preceding section we developed a way to determine a probabil-

ity measure for any sequence of chance experiments where there are

only a ¬nite number of possibilities for each experiment. While this

provides the framework for the general study of stochastic processes,

it is too general to be studied in complete detail. Therefore, in proba-

bility theory we look for simplifying assumptions which will make our

probability measure easier to work with. It is desired also that these as-

sumptions be such as to apply to a variety of experiments which would

occur in practice. In this book we shall limit outselves to the study of

two types of processes. The ¬rst, the independent trials process, will

be considered in the present section. This process was the ¬rst one to

be studied extensively in probability theory. The second, the Markov

131

4.8. INDEPENDENT TRIALS WITH TWO OUTCOMES

chain process, is a process that is ¬nding increasing application, par-

ticularly in the social and biological sciences, and will be considered in

Section 4.13.

A process of independent trials applies to the following situation.

Assume that there is a sequence of chance experiments, each of which

consists of a repetition of a single experiment, carried out in such a way

that the results of any one experiment in no way a¬ect the results in

any other experiment. We label the possible outcome of a single ex-

periment by a1 , . . . , ar . We assume that we are also given probabilities

p1 , . . . , pr for each of these outcomes occurring on any single experi-

ment, the probabilities being independent of previous results. The tree

representing the possibilities for the sequence of experiments will have

the same outcomes from each branch point, and the branch probabili-

ties will be assigned by assigning probability pj to any branch leading

to outcome aj . The tree measure determined in this way is the measure

of an independent trials process. In this section we shall consider the

important case of two outcomes for each experiment. The more general

case is studied in Section 4.11.

In the case of two outcomes we arbitrarily label one outcome “suc-

cess” and the other “failure”. For example, in repeated throws of a

coin we might call heads success, and tails failure. We assume there is

given a probability p for success and a probability q = 1 ’ p for failure.

The tree measure for a sequence of three such experiments is shown in

Figure 4.13. The weights assigned to each path are indicated at the end

of the path. The question which we now ask is the following. Given an

independent trials process with two outcomes, what is the probability

of exactly x successes in n experiments? We denote this probability by

f (n, x; p) to indicate that it depends upon n, x, and p.

Assume that we had a tree for this general situation, similar to the

tree in Figure 4.13 for three experiments, with the branch points labeled

S for success and F for failure. Then the truth set of the statement

“Exactly x successes occur” consists of all paths which go through x

branch points labeled S and n ’ x labeled F . To ¬nd the probability

of this statement we must add the weights for all such paths. We are

helped ¬rst by the fact that our tree measure assigns the same weight

to any such path, namely px q n’x . The reason for this is that every

branch leading to an S is assigned probability p, and every branch

leading to F is assigned probability q, and in the product there will

be x p™s and (n ’ x) q™s. To ¬nd the desired probability we need only

¬nd the number of paths in the truth set of the statement “Exactly x

132 CHAPTER 4. PROBABILITY THEORY

Figure 4.13: ™¦

successes occur”. To each such path we make correspond an ordered

partition of the integers from 1 to n which has two cells, x elements in

the ¬rst and n ’ x in the second. We do this by putting the numbers

of the experiments on which success occurred in the ¬rst cell and those

for which failure occurred in the second cell. Since there are n such

x

partitions there are also this number of paths in the truth set of the

statement considered. Thus we have proved:

In an independent trials process with two outcomes the

probability of exactly x successes in n experiments is given by

n x n’x

f (n, x; p) = pq .

x

Example 4.17 Consider n throws of an ordinary coin. We label heads

1

“success” and tails “failure”, and we assume that the probability is 2

for heads on any one throw independently of the outcome of any other

throw. Then the probability that exactly x heads will turn up is

1 n 1n

f (n, x; ) = ( ).

2 x2

For example, in l00 throws the probability that exactly 50 heads will

133

4.8. INDEPENDENT TRIALS WITH TWO OUTCOMES

turn up is

1 100 1 100

f (100, 50; ) = () ,

2 50 2

which is approximately .08. Thus we see that it is quite unlikely that

exactly one-half of the tosses will result in heads. On the other hand,

suppose that we ask for the probability that nearly one-half of the tosses

will be heads. To be more precise, let us ask for the probability that

the number of heads which occur does not deviate by more than l0 from

1

50. To ¬nd this we must add f (100, x; 2 ) for x = 40, 41, . . . , 60. If this

is done, we obtain a probability of approximately .96. Thus, while it

is unlikely that exactly 50 heads will occur, it is very likely that the

number of heads which occur will not deviate from 50 by more than l0.

™¦

Example 4.18 Assume that we have a machine which, on the basis

of data given, is to predict the outcome of an election as either a Re-

publican victory or a Democratic victory. If two identical machines are

given the same data, they should predict the same result. We assume,

however, that any such machine has a certain probability q of reversing

the prediction that it would ordinarily make, because of a mechanical

or electrical failure. To improve the accuracy of our prediction we give

the same data to r identical machines, and choose the answer which

the majority of the machines give. To avoid ties we assume that r is

odd. Let us see how this decreases the probability of an error due to a

faulty machine.

Consider r experiments, where the jth experiment results in success

if the jth machine produces the prediction which it would make when

operating without any failure of parts. The probability of success is

then p = 1 ’ q. The majority decision will agree with that of a per-

fectly operating machine if we have more than r/2 successes. Suppose,

for example, that we have ¬ve machines, each of which has a proba-

bility of .1 of reversing the prediction because of a parts failure. Then

the probability for success is .9, and the probability that the majority

decision will be the desired one is

f (5, 3; 0.9) + f (5, 4; 0.9) + f (5, 5; 0.9)

which is found to be approximately .991 (see Exercise 3).

Thus the above procedure decreases the probability of error due to

machine failure from .1 in the case of one machine to .009 for the case

™¦

of ¬ve machines.

134 CHAPTER 4. PROBABILITY THEORY

Exercises

1. Compute for n = 4, n = 8, n = 12, and n = 16 the probability of

1

obtaining exactly 2 heads when an ordinary coin is thrown.

[Ans. .375; .273; .226; .196.]

2. Compute for n = 4, n = 8, n = 12, and n = 16 the probability

that the fraction of heads deviates from 2 by less than 1 .

1

5

[Ans. .375; .711, .854; .923.]

3. Verify that the probability .991 given in Example 4.18 is correct.

4. Assume that Peter and Paul match pennies four times. (In match-

ing pennies, Peter wins a penny with probability 1 , and Paul wins

2

1

a penny with probability 2 .) What is the probability that Peter

wins more than Paul? Answer the same for ¬ve throws. For the

case of 12,917 throws.

511

[Ans. ; ; .]

16 2 2

5. If an ordinary die is thrown four times, what is the probability

that exactly two sixes will occur?

6. In a ten-question true-false exam, what is the probability of get-

ting 70 per cent or better by guessing?

11

[Ans. .]

64

7. Assume that, every time a batter comes to bat, he or she has

probability .3 for getting a hit. Assuming that hits form an in-

dependent trials process and that the batter comes to bat four

times, what fraction of the games would he or she expect to get

at least two hits? At least three hits? Four hits?

[Ans. .348; .084; .008.]

8. A coin is to be thrown eight times. What is the most probable

number of heads that will occur? What is the number having

the highest probability, given that the ¬rst four throws resulted

in heads?

135

4.8. INDEPENDENT TRIALS WITH TWO OUTCOMES

9. A small factory has ten workers. The workers eat their lunch at

one of two diners, and they are just as likely to eat in one as in the

other. If the proprietors want to be more than .95 sure of having

enough seats, how many seats must each of the diners have?

[Ans. Eight seats.]

10. Suppose that ¬ve people are chosen at random and asked if they

favor a certain proposal. If only 30 per cent of the people favor

the proposal, what is the probability that a majority of the ¬ve

people chosen will favor the proposal?

11. In Example 4.18, if the probability for a machine reversing its

answer due to a parts failure is .2, how many machines would

have to be used to make the probability greater than .89 that the

answer obtained would be that which a machine with no failure

would give?

[Ans. Three machines.]

12. Assume that it is estimated that a torpedo will hit a ship with

1

probability 3 . How many torpedos must be ¬red if it is desired

that the probability for at least one hit should be greater than

.9?

13. A student estimates that, if he or she takes four courses, he or

she has probability .8 of passing each course. If he or she takes

¬ve courses, he or she has probability .7 of passing each course,

and if he or she takes six courses he or she has probability .5 for

passing each course. The student™s only goal is to pass at least

four courses. How many courses should he or she take for the

best chance of achieving this goal?

[Ans. 5.]

Supplementary exercises.

14. In a certain board game players move around the board, and each

turn consists of a player™s rolling a pair of dice. If a player is on

the square Park Bench, he or she must roll a seven or doubles

before being allowed to move out.

136 CHAPTER 4. PROBABILITY THEORY

(a) What is the probability that a player stuck on Park Bench

will be allowed to move out on the next turn?

1

[Ans. .]

3

(b) How many times must a player stuck on Park Bench roll

before the chances of getting out exceed 3 .

4

[Ans. 4.]

15. A restaurant orders ¬ve pieces of apple pie and ¬ve pieces of

cherry pie. Assume that the restaurant has ten customers, and

3

the probability that a customer will ask for apple pie is 4 and for

1

cherry pie is 4 .

(a) What is the probability that the ten customers will all be

able to have their ¬rst choice?

(b) What number of each kind of pie should the restaurant order

if it wishes to order ten pieces of pie and wants to maximize

the probability that the ten customers will all have their ¬rst

choice?

16. Show that it is more probable to get at least one ace with 4 dice

than at least one double ace in 24 throws of two dice.

17. A thick coin, when tossed, will land “heads” with a probability of

5 5

, “tails” with a probability of 12 , and will land on edge with a

12

1

probability of 6 . If it is tossed six times, what is the probability

that it lands on edge exactly two times?

[Ans. .2009.]

18. Without actually computing the probabilities, ¬nd the value of x

for which f (20, x; .3) is largest.

2

19. A certain team has probability of winning whenever it plays.

3

(a) What is the probability the team will win exactly four out

of ¬ve games?

80

[Ans. .]

243

(b) What is the probability the team will win at most four out

of ¬ve games?

137

4.9. A PROBLEM OF DECISION

211

[Ans. .]

243

(c) What is the probability the team will win exactly four games

out of ¬ve if it has already won the ¬rst two games of the

¬ve?

4

[Ans. .]

9

4.9 A problem of decision

In the preceding sections we have dealt with the problem of calculating

the probability of certain statements based on the assumption of a given

probability measure. In a statistics problem, one is often called upon

to make a decision in a case where the decision would be relatively

easy to make if we could assign probabilities to certain statements, but

we do not know how to assign these probabilities. For example, if a

vaccine for a certain disease is proposed, we may be called upon to

decide whether or not the vaccine should be used. We may decide that

we could make the decision if we could compare the probability that

a person vaccinated will get the disease with the probability that a

person not vaccinated will get the disease. Statistical theory develops

methods to obtain from experiments some information which will aid

in estimating these probabilities, or will otherwise help in making the

required decision. We shall illustrate a typical procedure.

Smith claims to have the ability to distinguish ale from beer and

has bet Jones a dollar to that e¬ect. Now Smith does not mean that

he or she can distinguish beer from ale every single time, but rather a

1

proportion of the time which is signi¬cantly greater than 2 .

Assume that it is possible to assign a number p which represents

the probability that Smith can pick out the ale from a pair of glasses,

one containing ale and one beer. We identify p = 1 with having no

2

1 1

ability, p > 2 with having some ability, and p < 2 with being able to

distinguish, but having the wrong idea which is the ale. If we knew

the value of p, we would award the dollar to Jones if p were ¤ 1 , and

2

1

to Smith if p were > 2 . As it stands, we have no knowledge of p and

thus cannot make a decision. We perform an experiment and make a

decision as follows.

Smith is given a pair of glasses, one containing ale and the other

beer, and is asked to identify which is the ale. This procedure is re-

peated ten times, and the number of correct identi¬cations is noted. If

138 CHAPTER 4. PROBABILITY THEORY

the number correct is at least eight, we award the dollar to Smith, and

if it is less than eight, we award the dollar to Jones.

We now have a de¬nite procedure and shall examine this procedure

both from Jones™s and Smith™s points of view. We can make two kinds of

errors. We may award the dollar to Smith when in fact the appropriate

1

value of p is ¤ 2 , or we may award the dollar to Jones when the

1

appropriate value for p is > 2 There is no way that these errors can

be completely avoided. We hope that our procedure is such that each

bettor will be convinced that, if he or she is right, he or she will very

likely win the bet.

1

Jones believes that the true value of p is 2 . We shall calculate the

probability of Jones winning the bet if this is indeed true. We assume

that the individual tests are independent of each other and all have the

1

same probability 2 for success. (This assumption will be unreasonable if

the glasses are too large.) We have then an independent trials process

with p = 1 to describe the entire experiment. The probability that

2

Jones will win the bet is the probability that Smith gets fewer than

eight correct. From the table in Figure 4.14 we compute that this

probability is approximately .945. Thus Jones sees that, if he or she

is right, it is very likely that he or she will win the bet.

Smith, on the other hand, believes that p is signi¬cantly greater

than 1 . If Smith believes that p is as high as .9, we see from Figure

2

4.14 that the probability of Smith™s getting eight or more correct is

.930. Then both parties will be satis¬ed by the bet.

Suppose, however, that Smith thinks the value of p is only about

.75. Then the probability that Smith will get eight or more correct and

thus win the bet is .526. There is then only an approximately even

chance that the experiment will discover Smith™s abilities, and Smith

probably will not be satis¬ed with this. If Smith really thinks his or her

ability is represented by a p value of about 3 , we would have to devise

4

a di¬erent method of awarding the dollar. We might, for example,

propose that Smith win the bet if he or she gets seven or more correct.

3

Then, if Smith has probability 4 of being correct on a single trial, the

1

probability that Smith will win the bet is approximately .776. If p = 2

the probability that Jones will win the bet is about .828 under this new

arrangement. Jones™s chances of winning are thus decreased, but Smith

may be able to convince him or her that it is a fairer arrangement than

the ¬rst procedure.

In the above example, it was possible to make two kinds of er-

rors. The probability of making these errors depended on the way we

139

4.9. A PROBLEM OF DECISION