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Figure 4.14: ™¦

designed the experiment and the method we used for the required de-
cision. In some cases we are not too worried about the errors and can
make a relatively simple experiment. In other cases, errors are very im-
portant, and the experiment must be designed with that fact in mind.
For example, the possibility of error is certainly important in the case
that a vaccine for a given disease is proposed, and the statistician is
asked to help in deciding whether or not it should be used. In this case
it might be assumed that there is a certain probability p that a person
will get the disease if not vaccinated, and a probability r that the per-
son will get it if he or she is vaccinated. If we have some knowledge
of the approximate value of p, we are then led to construct an experi-
ment to decide whether r is greater than p, equal to p, or less than p.
The ¬rst case would be interpreted to mean that the vaccine actually
tends to produce the disease, the second that it has no e¬ect, and the
third that it prevents the disease; so that we can make three kinds of
errors. We could recommend acceptance when it is actually harmful,
we could recommend acceptance when it has no e¬ect, or ¬nally we
could reject it when it actually is e¬ective. The ¬rst and third might
result in the loss of lives, the second in the loss of time and money of
those administering the test. Here it would certainly be important that
the probability of the ¬rst and third kinds of errors be made small. To
see how it is possible to make the probability of both errors small, we
return to the case of Smith and Jones.

Suppose that, instead of demanding that Smith make at least eight
correct identi¬cations out of ten trials, we insist that Smith make at
least 60 correct identi¬cations out of 100 trials. (The glasses must now
be very small.) Then, if p = 1 , the probability that Jones wins the
bet is about .98; so that we are extremely unlikely to give the dollar to
Smith when in fact it should go to Jones. (If p < 2 it is even more likely
that Jones will win.) If p > 1 we can also calculate the probability that
Smith will win the bet. These probabilities are shown in the graph in
Figure 4.15. The dashed curve gives for comparison the corresponding
probabilities for the test requiring eight out of ten correct. Note that
with 100 trials, if p is 4 , the probability that Smith wins the bet is
nearly 1, while in the case of eight out of ten, it was only about 2 .
Thus in the case of 100 trials, it would be easy to convince both Smith
and Jones that whichever one is correct is very likely to win the bet.

Thus we see that the probability of both types of errors can be made
small at the expense of having a large number of experiments.

Figure 4.15: ™¦

1. Assume that in the beer and ale experiment Jones agrees to pay
Smith if Smith gets at least nine out of ten correct.

(a) What is the probability of Jones paying Smith even though
Smith cannot distinguish beer and ale, and guesses?

[Ans. .011.]

(b) Suppose that Smith can distinguish with probability .9. What
is the probability of not collecting from Jones?

[Ans. .264.]

2. Suppose that in the beer and ale experiment Jones wishes the
probability to be less than .1 that Smith will be paid if, in fact,
Smith guesses. How many of ten trials must Jones insist that
Smith get correct to achieve this?

3. In the analysis of the beer and ale experiment, we assume that
the various trials were independent. Discuss several ways that
error can enter, because of the nonindependence of the trials, and
how this error can be eliminated. (For example, the glasses in
which the beer and ale were served might be distinguishable.)

4. Consider the following two procedures for testing Smith™s ability
to distinguish beer from ale.

(a) Four glasses are given at each trial, three containing beer and
one ale, and Smith is asked to pick out the one containing
ale. This procedure is repeated ten times. Smith must guess
correctly seven or more times. Find the probability that
Smith wins by guessing.

[Ans. .003.]

(b) Ten glasses are given to Smith, and Smith is told that ¬ve
contain beer and ¬ve ale, and asked to name the ¬ve that
contain ale. Smith must choose all ¬ve correctly. Find the
probability that Smith wins by guessing.

[Ans. .004.]

(c) Is there any reason to prefer one of these two tests over the

5. A testing service claims to have a method for predicting the order
in which a group of freshmen will ¬nish in their scholastic record
at the end of college. The college agrees to try the method on
a group of ¬ve students, and says that it will adopt the method
if, for these ¬ve students, the prediction is either exactly correct
or can be changed into the correct order by interchanging one
pair of adjacent students in the predicted order. If the method is
equivalent to simply guessing, what is the probability that it will
be accepted?

[Ans. .]

6. The standard treatment for a certain disease leads to a cure in 4
of the cases. It is claimed that a new treatment will result in a
cure in 3 of the cases. The new treatment is to be tested on ten
people having the disease. If seven or more are cured, the new
treatment will be adopted. If three or fewer people are cured, the
treatment will not be considered further. If the number cured
is four, ¬ve, or six, the results will be called inconclusive, and
a further study will be made. Find the probabilities for each of
these three alternatives under the assumption ¬rst, that the new
treatment has the same e¬ectiveness as the old, and second, under
the assumption that the claim made for the treatmnent is correct.

7. Three students debate the intelligence of Springer spaniels. One
claims that Springers are mostly (say 90 per cent of them) in-
telligent. A second claims that very few (say 10 per cent) are
intelligent, while a third one claims that a Springer is just as
likely to be intelligent as not. They administer an intelligence
test to ten Springers, classifying them as intelligent or not. They
agree that the ¬rst student wins the bet if eight or more are in-
telligent, the second if two or fewer, the third in all other cases.
For each student, calculate the probability that he or she wins
the bet, if he or she is right.

[Ans. .930, .930, .890.]

8. Ten students take a test with ten problems. Each student on each
question has probability 2 of being right, if he or she does not
cheat. The instructor determines the number of students who get
each problem correct. If instructor ¬nds on four or more problems
there are fewer than three or more than seven correct, he or she
considers this convincing evidence of communication between the
students. Give a justi¬cation for the procedure. [Hint: The table
in Figure 4.14 must be used twice, once for the probability of fewer
than three or more than seven correct answers on a given problem,
and the second time to ¬nd the probability of this happening on
four or more problems.]

4.10 The law of large numbers
In this section we shall study some further properties of the indepen-
dent trials process with two outcomes. In Section 4.8 we saw that the
probability for x successes in n trials is given by

n x n’x
f (n, x; p) = pq .

In Figure 4.16 we show these probabilities graphically for n = 8 and
p = 3 . In Figure 4.17 we have done similarly for the case of n = 7 and
p = 4.
We see in the ¬rst case that the values increase up to a maximum
value at x = 6 and then decrease. In the second case the values increase
up to a maximum value at x = 5, have the same value for x = 6, and

Figure 4.16: ™¦

Figure 4.17: ™¦

then decrease. These two cases are typical of what can happen in
Consider the ratio of the probability of x + 1 successes in n trials to
the probability of x successes in n trials, which is
px+1 q n’x’1 n’x p
n x+1 q
px q n’x

This ratio will be greater than one as long as (n ’ x)p > (x + 1)q
or as long as x < np ’ q. If np ’ q is not an integer, the values
n x n’x
pq increase up to a maximum value, which occurs at the ¬rst
integer greater than np ’ q, and then decrease. In case np ’ q is an
integer, the values n px q n’x increase up to x = np ’ q, are the same
for x = np ’ q and x = np ’ q + 1, and then decrease.
Thus we see that, in general, values near np will occur with the
largest probability. It is not true that one particular value near np is
highly likely to occur, but only that it is relatively more likely than
a value further from np. For example, in 100 throws of a coin, np =
100 · 2 = 50. The probability of exactly 50 heads is approximately .08.
The probability of exactly 30 is approximately .00002.
More information is obtained by studying the probability of a given
deviation of the proportion of successes x/n from the number p; that
is, by studying for > 0,
’ p| < ].
For any ¬xed n, p, and , the latter probability can be found by
adding all the values of f (n, x; p) for values of x for which the inequality
p ’ < x/n < p + is true. In Figure 4.18 we have given these
probabilities for the case p = .3 with various values for and n. In the
¬rst column we have the case = .1. We observe that as n increases,
the probability that the fraction of successes deviates from .3 by less
than .1 tends to the value 1. In fact to four decimal places the answer
is 1 after n = 400. In column two we have the same probabilities for
the smaller value of = .05. Again the probabilities are tending to 1
but not so fast. In the third column we have given these probabilities
for the case = .02. We see now that even after 1000 trials there is still
a reasonable chance that the fraction x/n is not within .02 of the value
of p = .3. It is natural to ask if we can expect these probabilities also
to tend to 1 if we increase n su¬ciently. The answer is yes and this is

Figure 4.18: ™¦

Figure 4.19: ™¦

assured by one of the fundamental theorems of probability called the
law of large numbers. This theorem asserts that, for any > 0,
’ p| < ]
tends to 1 as n increases inde¬nitely.
It is important to understand what this theorem says and what it
does not say. Let us illustrate its meaning in the case of coin tossing.
We are going to toss a coin n times and we want the probability to be
very high, say greater than .99, that the fraction of heads which turn
up will be very close, say within .00l of the value .5. The law of large
numbers assures us that we can have this if we simply choose n large
enough. The theorem itself gives us no information about how large n
must be. Let us however consider this question.
To say that the fraction of the times success results is near p is the
same as saying that the actual number of successes x does not deviate
too much from the expected number np. To see the kind of deviations
which might be expected we can study the value of Pr[|x ’ np| ≥ d]. A
table of these values for p = .3 and various values of n and d are given
in Figure 4.19. Let us ask how large d must be before a deviation as
large as d could be considered surprising. For example, let us see for
each n the value of d which makes Pr[|x ’ np| ≥ d] about .04. From
the table, we see that d should be 7 for n = 50, 9 for n = 80, 10 for
n = 100, etc. To see deviations which might be considered more typical
we look for the values of d which make Pr[|x ’ np| ≥ d] approximately
. Again from the table, we see that d should be 3 or 4 for n = 50, 4
or 5 for n = 80, 5 for n = 100, etc. The answers to these two questions

are given in the last two columns of the table. An examination of these
numbers shows us that deviations which we would consider surprising

are approximately √ while those which are more typical are about
one-half as large or n/2.

This suggests that n, or a suitable multiple of it, might be taken
as a unit of measurement for deviations. Of course, we would also have
to study how Pr[|x ’ np| ≥ d] depends on p. When this is done, one

¬nds that npq is a natural unit; it is called a standard deviation. It
can be shown that for large n the following approximations hold.

Pr[|x ’ np| ≥ npq] ≈ .3174

Pr[|x ’ np| ≥ 2 npq] ≈ .0455

Pr[|x ’ np| ≥ 3 npq] ≈ .0027
That is, a deviation from the expected value of one standard devi-
ation is rather typical, while a deviation of as much as two standard
deviations is quite surprising and three very surprising. For values of p
√ 1
not too near 0 or 1, the value of pq is approximately 2 . Thus these
approximations are consistent with the results we observed from our
√ √
For large n, Pr[x ’ np ≥ k npq] or Pr[x ’ np ¤ ’k npq] can be
shown to be approximately the same. Hence these probabilities can be
estimated for k = 1, 2, 3 by taking 2 the values given above.

Example 4.19 In throwing an ordinary coin 10,000 times, the ex-
pected number of heads is 5000, and the standard deviation for the
number of heads is 10, 000( 2 )( 2 ) = 50. Thus the probability that
the number of heads which turn up deviates from 5000 by as much as
one standard deviation, or 50, is approximately .317. The probability
of a deviation of as much as two standard deviations, or 100, is ap-
proximately .046. The probability of a deviation of as much as three
standard deviations, or 150, is approximately .003.

Example 4.20 Assume that in a certain large city, 900 people are
chosen at random and asked if they favor a certain proposal. Of the
900 asked, 550 say they favor the proposal and 350 are opposed. If, in
fact, the people in the city are equally divided on the issue, would it be
unlikely that such a large majority would be obtained in a sample of 900
of the citizens? If the people were equally divided, we would assume
that the 900 people asked would form an independent trials process

1 1
with probability 2 for a “yes” answer and 2 for a “no” answer. Then
the standard deviation for the number of “yes” answers in 900 trials is
900( 2 )( 1 ) = 15. Then it would be very unlikely that we would obtain
a deviation of more than 45 from the expected number of 450. The
fact that the deviation in the sample from the expected number was
100, then, is evidence that the hypothesis that the voters were equally
divided is incorrect. The assumption that the true proportion is any
value less than 1 would also lead to the fact that a number as large
as 550 favoring in a sample of 900 is very unlikely. Thus we are led to
suspect that the true proportion is greater than 2 . On the other hand,
if the number who favored the proposal in the sample of 900 were 465,
we would have only a deviation of one standard deviation, under the
assumption of an equal division of opinion. Since such a deviation is
not unlikely, we could not rule out this possibility on the evidence of
the sample.

Example 4.21 A certain Ivy League college would like to admit 800
students in their freshman class. Experience has shown that if they
admit 1250 students they will have acceptances from approximately
800. If they admit as many as 50 too many students they will have
to provide additional dormitory space. Let us ¬nd the probability that
this will happen assuming that the acceptances of the students can be
considered to be an independent trials process. We take as our estimate
for the probability of an acceptance p = 1250 . Then the expected num-
ber of acceptances is 800 and the standard deviation for the number of

acceptances is 1250 · .64 · .36 ≈ 17. The probability that the number
accepted is three standard deviations or 51 from the mean is approx-
imately .0027. This probability takes into account a deviation above
the mean or below the mean. Since in this case we are only interested
in a deviation above the mean, the probability we desire is half of this
or approximately .0013. Thus we see that it is highly unlikely that the
college will have to have new dormitory space under the assumptions
we have made.

We ¬nish this discussion of the law of large numbers with some ¬nal
remarks about the interpretation of this important theorem.
Of course no matter how large n is we cannot prevent the coin from
coming up heads every time. If this were the case we would observe a
fraction of heads equal to 1. However, this is not inconsistent with the
theorem, since the probability of this happening is ( 2 )n which tends to

0 as n increases. Thus a fraction of 1 is always possible, but becomes
increasingly unlikely.
The law of large numbers is often misinterpreted in the following
manner. Suppose that we plan to toss the coin 1000 times and after
500 tosses we have already obtained 400 heads. Then we must obtain
less than one-half heads in the remaining 500 tosses to have the fraction
come out near 1 . It is tempting to argue that the coin therefore owes
us some tails and it is more likely that tails will occur in the last 500
tosses. Of course this is nonsense, since the coin has no memory. The
point is that something very unlikely has already happened in the ¬rst

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