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˜ ˜
(a) X ∪ Y = (X © Y ) ∪ (X © Y ) ∪ (X © Y ).
˜ ˜
(b) m(X ∪ Y ) = m(X © Y ) + m(X © Y ) + m(X © Y ).
(c) m(X ∪ Y ) = m(X) + m(Y ) ’ m(X © Y ).

23. If X, Y , and Z are any three sets, prove that, for any probability

m(X ∪ Y ∪ Z) = m(X) + m(Y ) + m(Z)
’m(X © Y ) ’ m(Y © Z) ’ m(X © Z)
+m(X © Y © Z).

24. Translate the result of Exercise 23 into a result concerning three
statements p, q, and r.

25. A man o¬ers to bet “dollars to doughnuts” that a certain event
will take place. Assuming that a doughnut costs a nickel, what
must the probability of the event be for this to be a fair bet?

[Ans. .]

26. Show that the inclusion-exclusion formula 3.2 is true is n is re-
placed by m. Apply this result to

Pr[p1 ∨ p2 ∨ . . . ∨ pn ].

4.3 The equiprobable measure
We have already seen several examples where it was natural to assign
the same weight to all possibilities in determining the appropriate prob-
ability measure. The probability measure determined in this manner
is called the equiprobable measure. The measure of sets in the case of
the equiprobable measure has a very simple form. In fact, if U has n
elements and if the equiprobable measure has been assigned, then for
any set X, m(X) is r/n, where r is the number of elements in the set
X. This is true since the weight of each element in X is 1/n, and hence
the sum of the weights of elements of X is r/n.
The particularly simple form of the equiprobable measure makes
it easy to work with. In view of this, it is important to observe that
a particular choice for the set of possibilities in a given situation may
lead to the equiprobable measure, while some other choice will not. For
example, consider the case of two throws of an ordinary coin. Suppose
that we are interested in statements about the number of heads which
occur. If we take for the possibility set the set U = {HH, HT, TH, TT}
then it is reasonable to assign the same weight to each outcome, and
we are led to the equiprobable measure. If, on the other hand, we
were to take as possible outcomes the set U = {no H, one H, two H},
it would not be natural to assign the same weight to each outcome,
since one head can occur in two di¬erent ways, while each of the other
possibilities can occur in only one way.

Example 4.4 Suppose that we throw two ordinary dice. Each die can
turn up a number from 1 to 6; hence there are 6 · 6 possibilities. We
assign weight 36 to each possibility. A prediction that is true in j cases
will then have probability j/36. For example, “The sum of the dice is
5” will be true if we get 1 + 4, 2 + 3, 3 + 2, or 4 + 1. Hence the
probability that the sum of the dice is 5 is 36 = 1 . The sum can be 12
in only one way, 6 + 6. Hence the probability that the sum is 12 is 36 .

Example 4.5 Suppose that two cards are drawn successively from a
deck of cards. What is the probability that both are hearts? There
are 52 possibilities for the ¬rst card, and for each of these there are
51 possibilities for the second. Hence there are 52 · 51 possibilities for
the result of the two draws. We assign the equiprobable measure. The
statement “The two cards are hearts” is true in 13 · 12 of the 52 · 51

possibilities. Hence the probability of this statement is 13·12/(52·51) =

Example 4.6 Assume that, on the basis of a predictive index applied
to students A, B, and C when entering college, it is predicted that after
four years of college the scholastic record of A will be the highest, C the
second highest, and B the lowest of the three. Suppose, in fact, that
these predictions turn out to be exactly correct. If the predictive index
has no merit at all and hence the predictions amount simply to guessing,
what is the probability that such a prediction will be correct? There
are 3! = 6 orders in which the men might ¬nish. If the predictions were
really just guessing, then we would assign an equal weight to each of the
six outcomes. In this case the probability that a particular prediction is
true is 6 . Since this probability is reasonably large, we would hesitate
to conclude that the predictive index is in fact useful, on the basis
of this one experiment. Suppose, on the other hand, it predicted the
order of six men correctly. Then a similar analysis would show that,
1 1
by guessing, the probability is 6! = 720 that such a prediction would be
correct. Hence, we might conclude here that there is strong evidence
that the index has some merit.

1. A letter is chosen at random from the word “random”. What is
the probability that it is an n? That it is a vowel?

[Ans. 6 ; 3 .]

2. An integer between 3 and 12 inclusive is chosen at random. What
is the probability that it is an even number? That it is even and
divisible by three?

3. A card is drawn at random from a pack of playing cards.

(a) What is the probability that it is either a heart or the king
of clubs?
[Ans. .]

(b) What is the probability that it is either the queen of hearts
or an honor card (i.e., ten, jack, queen, king, or ace)?

[Ans. .]

4. A word is chosen at random from the set of words

U = {men, bird, ball, ¬eld, book}.

Let p, q, and r be the statements:

p: The word has two vowels.

q: The ¬rst letter of the word is b.

r: The word rhymes with cook.

Find the probability of the following statements.

(a) p.
(b) q.
(c) r.
(d) p § q.
(e) (p ∨ q) § ¬r.
(f) p ’ q.
[Ans. .]

5. A single die is thrown. Find the probability that

(a) An odd number turns up.
(b) The number which turns up is greater than two.
(c) A seven turns up.

6. In the Primary voting example of Section 2.1, assume that all 36
possibilities in the elections are equally likely. Find

(a) The probability that candidate A wins more states than ei-
ther B or C.
[Ans. .]

(b) That all the states are won by the same candidate.
[Ans. .]

(c) That every state is won by a di¬erent candidate.

[Ans. 0.]

7. A single die is thrown twice. What value for the sum of the two
outcomes has the highest probability? What value or values of
the sum has the lowest probability of occurring?

8. Two boys and two girls are placed at random in a row for a
picture. What is the probability that the boys and girls alternate
in the picture?

[Ans. .]

9. A certain college has 500 students and it is known that
300 read French.
200 read German.
50 read Russian.
20 read French and Russian.
30 read German and Russian.
20 read German and French.
10 read all three languages.
If a student is chosen at random from the school, what is the
probability that the student

(a) Reads two and only two languages?
(b) Reads at least one language?

10. Suppose that three people enter a restaurant which has a row
of six seats. If they choose their seats at random, what is the
probability that they sit with no seats between them? What is
the probability that there is at least one empty seat between any
two of them?

11. Find the probability of obtaining each of the following poker
hands. (A poker hand is a set of ¬ve cards chosen at random
from a deck of 52 cards.)

(a) Royal ¬‚ush (ten, jack, queen, king, ace in a single suit).
[Ans. 4/ = .0000015.]

(b) Straight ¬‚ush (¬ve in a sequence in a single suit, but not a
royal ¬‚ush).

[Ans. (40 ’ 4)/ = .000014.]

(c) Four of a kind (four cards of the same face value).
[Ans. 624/ = .00024.]

(d) Full house (one pair and one triple of the same face value).
[Ans. 3744/ = .0014.]

(e) Flush (¬ve cards in a single suit but not a straight or royal
[Ans. (5148 ’ 40)/ = .0020.]

(f) Straight (¬ve cards in a row, not all of the same suit).
[Ans. (10, 240 ’ 40)/ = .0039.]

(g) Straight or better.
[Ans. .0076.]

12. If ten people are seated at a circular table at random, what is
the probability that a particular pair of people are seated next to
each other?

[Ans. .]

13. A room contains a group of n people who are wearing badges
numbered from 1 to n. If two people are selected at random, what
is the probability that the larger badge number is a 3? Answer
this problem assuming that n = 5, 4, 3, 2.

[Ans. ; ; ; 0.]

14. In Exercise 13, suppose that we observe two men leaving the room
and that the larger of their badge numbers is 3. What might we
guess as to the number of people in the room?

15. Find the probability that a bridge hand will have suits of

(a) 5, 4, 3, and 1 cards.
[Ans. = .129.]
5 4 3 1

(b) 6, 4, 2, and 1 cards.
[Ans. .047.]
(c) 4, 4, 3, and 2 cards.
[Ans. .216.]
(d) 4, 3, 3, and 3 cards.
[Ans. .105.]

16. There are 52 = 6.5 — 1011 possible bridge hands. Find the
probability that a bridge hand dealt at random will be all of one
suit. Estimate roughly the number of bridge hands dealt in the
entire country in a year. Is it likely that a hand of all one suit
will occur sometime during the year in the United States?

Supplementary exercises.

17. Find the probability of not having a pair in a hand of poker.

18. Find the probability of a “bust” hand in poker. [Hint: A hand
is a “bust” if there is no pair, and it is neither a straight nor a

[Ans. .5012.]

19. In poker, ¬nd the probability of having

(a) Exactly one pair.
[Ans. .4226.]
(b) Two pairs.
[Ans. .0475.]
(c) Three of a kind.
[Ans. .0211.]

20. Verify from Exercises 11, 18, 19 that the probabilities for all pos-
sible poker hands add up to one (within a rounding error).

21. A certain French professor announces that he or she will select
three out of eight pages of text to put on an examination and that
each student can choose one of these three pages to translate.

(a) What is the maximum number of pages that a student should
prepare in order to be certain of being able to translate a
page that he or she has studied?
(b) Smith decides to study only four of the eight pages. What
is the probability that one of these four pages will appear on
the examination?

4.4 Two nonintuitive examples
There are occasions in probability theory when one ¬nds a problem for
which the answer, based on probability theory, is not at all in agreement
with one™s intuition. It is usually possible to arrange a few wagers that
will bring one™s intuition into line with the mathematical theory. A
particularly good example of this is provided by the matching birthdays
Assume that we have a room with r people in it and we propose
the bet that there are at least two people in the room having the same
birthday, i.e., the same month and day of the year. We ask for the
value of r which will make this a fair bet. Few people would be willing
to bet even money on this wager unless there were at least 100 people
in the room. Most people would suggest 150 as a reasonable number.
However, we shall see that with 150 people the odds are approximately
4,100,000,000,000,000 to 1 in favor of two people having the same birth-
day, and that one should be willing to bet even money with as few as
23 people in the room.
Let us ¬rst ¬nd the probability that in a room with r people, no two
have the same birthday. There are 365 possibilities for each person™s
birthday (neglecting February 29). There are then 365r possibilities
for the birthdays of r people. We assume that all these possibilities
are equally likely. To ¬nd the probability that no two have the same
birthday we must ¬nd the number of possibilities for the birthdays
which have no day represented twice. The ¬rst person can have any of
365 days for his or her birthday. For each of these, if the second person
is to have a di¬erent birthday, there are only 364 possibilities for his
or her birthday. For the third person, there are 363 possibilities if he
or she is to have a di¬erent birthday than the ¬rst two, etc. Thus the
probability that no two people have the same birthday in a group of r
people is
365 · 364 · . . . · (365 ’ r + 1)
qr = .

Figure 4.1: ™¦

The probability that at least two people have the same birthday is
then pr = 1 ’ qr . In Figure 4.1 the values of pr and the odds for a fair
bet, pr : (1 ’ pr ) are given for several values of r.
We consider now a second problem in which intuition does not lead
to the correct answer. A hat-check clerk has checked n hats, but they
have become hopelessly scrambled. The clerk hands back the hats at
random. What is the probability that at least one head gets its own
hat? For this problem some people™s intuition would lead them to
guess that for a large number of hats this probability should be small,
while others guess that it should be large. Few people guess that the
probability is neither large nor small and essentially independent of the
number of hats involved.

Let pj be the statement “the jth head gets its own hat back”. We
wish to ¬nd Pr[p1 ∨ p2 ∨ . . . ∨ pn ]. We know from Exercise 26 that
a probability of this form can be found from the inclusion-exclusion
formula. We must add all probabilities of the form Pr[pi ], then subtract
the sum of all probabilities of the form Pr[pi § pj ], then add the sum of
all probabilities of the form Pr[pi § pj § pk ], etc.
However, each of these probabilities represents the probability that
a particular set of heads get their own hats back. These probabilities
are very easy to compute. Let us ¬nd the probability that out of n
heads some particular m of them get back their own hats. There are n!
ways that the hats can be returned. If a particular m of them are to get

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