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a house, what are the probabilities of having taken each job?

[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

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

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
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 .

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

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
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.

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

[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
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.
[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?
[Ans. .]

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?

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
probability 3 . How many torpedos must be ¬red if it is desired
that the probability for at least one hit should be greater than

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.

(a) What is the probability that a player stuck on Park Bench
will be allowed to move out on the next turn?
[Ans. .]

(b) How many times must a player stuck on Park Bench roll
before the chances of getting out exceed 3 .

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

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

(a) What is the probability the team will win exactly four out
of ¬ve games?
[Ans. .]

(b) What is the probability the team will win at most four out
of ¬ve games?

[Ans. .]

(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
[Ans. .]

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
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
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
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

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
value of p is ¤ 2 , or we may award the dollar to Jones when the
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.
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
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
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
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
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.
Then, if Smith has probability 4 of being correct on a single trial, the
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

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