Math/Stat 394 Homework 2
Due Wednesday Jan 18
1. Six awards are to be given out among 40 students. How many ways
can the awards be given out if
(a) one student will win exactly five awards.
(b) one student will win exactly four awards.
(c) no student can win more than three awards.
(a) There are 40 choices for which student wins five awards, 65
choices for which awards they win and 39 choices for who wins
the remaining award. Thus there are
6
40
39
5
ways the awards can be given out such that a student can win
exactly five awards.
(b) There are 40 choices for which student wins five awards, 65
choices for which awards they win and 39 choices for who wins
the remaining award. Thus there are
6
40
392
4
ways the awards can be given out such that a student can win
exactly four awards.
(c) There are 406 total number of choices of how to give out the
six awards. We subtract off the mumber of ways that one student wins six awards (40), that one student wins five awards
1
(40 65 (39)) and the number of ways one student can win four
awards 40 64 392 to get
6
6
6
40 − 40 − 40
(39) − 40
392
5
4
ways the awards can be given out such that a student can win at
most three awards.
2. A bridge hand consists of 13 cards from a standard 52 card deck. How
many bridge hands are there with
(a) exactly 5 spades?
(b) 5 cards if one suit, three cards of two suits and two cards of the
fourth suit?
(c) 10 cards of one suit?
(a) There are 13
5 choices of the 5 spades and
other cards. Thus there are
13 39
8
5
39
8
choices of the 8
total hands with exactly 5 spades.
4
.
(b) There are twelve choices for which suits get which numbers 2,1,1
13 13
13
13
and 2 choices for which ranks are
There are 5
3
3
chosen in the different suits. Thus there are
13 13 13 13
12
5
3
3
2
total choices.
(c) There are 4 choices of the suit which gets 10 cards, 13
choices of
10
39
the cards in that suit and 3 choices of the remaining 3 cards.
Thus there are
13 39
4
10
3
total choices.
and from chapter 2 of Ross problems 3,5,6,9,12,17 and 37 (which are
copied below.)
2
3. Two dice are thrown. Let E be the event that the sum of the dice is
odd; let F be the event that at least one of the dice lands on 1; and let
G be the event that the sum is five. Describe the events EF , E ∪ F ,
F G, EF C and EF G
EF is the event that one dice lands on 1 and the other lands on an
even number so the sum is odd. E ∪ F is the event that either the
sum of the two dice is odd or one die lands on 1 and the other die is
odd. F G is the event that one die lands on 1 and the sum is five so
the other lands on 4. EF C is the event that the sum of the dice is odd
and neither lands on 1. EG is the same as G (if the sum is five then
it is odd) so EF G = F G which is the event that one die lands on 1
while the other lands on 4.
5. A system is composed of 5 components, each of which is either working
or failed. Consider an experiment that consists of observing the status
of each component, and let the outcome of the experiment be given
by the vector (x1 , x2 , x3 , x4 , x5 ), where xi is equal to 1 if component i
is working and is equal to 0 if component i is failed.
(a) How many outcomes are in the sample space.
(b) Suppose that the system will work if components 1 and 2 are
both working, or if components 3 and 4 are both working, or if
components 1,3 and 5 are all working. Let W be the event that
the system will work. Specify all the outcomes in W .
(c) Let A be the event that components 4 and 5 are both failed. how
many outcomes are contained in the event A?
(d) Write out all the outcomes in the event AW .
(a) 32. Each of five components can be either 0 or 1.
(b) W consists of all eight vectors where both components 1 and
2 are working (1, 1, 0, 0, 0) (1, 1, 0, 0, 1) (1, 1, 0, 1, 0) (1, 1, 0, 1, 1)
(1, 1, 1, 0, 0) (1, 1, 1, 0, 1) (1, 1, 1, 1, 0) and (1, 1, 1, 1, 1). It also consists of six additional elements where both components 3 and
4 are working (0, 0, 1, 1, 0) (0, 0, 1, 1, 1) (0, 1, 1, 1, 0) (0, 1, 1, 1, 1)
(1, 0, 1, 1, 0) (1, 0, 1, 1, 1) and one element where components 1,3
and 5 are working. (1, 0, 1, 0, 1)
(c) 8. A is the event that both components 4 and 5 have failed so
components 1,2 and 3 are free to be either working or failed.
(d) The outcomes in the event AW are (1, 1, 0, 0, 0) and (1, 1, 1, 0, 0).
3
6. A hospital administrator codes incoming patients suffering gunshot
wounds according to whether they have insurance (coding 1 if they
do and 0 if they do not) and according to their condition, which is
rated as good (g), fair (f) or serious (s). Consider an experiment that
consists of determining the type of the coding of such a patient.
(a) Give the sample space of the experiment.
(b) Let A be the event that the patient is in serious condition. Specify
the outcomes in A.
(c) Let B be the event that the patient is uninsured. Specify the
outcomes in B.
(d) Give all the outcomes in the event B C ∩ A.
(a) S = {(i, j) : i ∈ {0, 1} and j ∈ {g, f, s}}.
(b) A = {(0, s), (1, s)}
(c) B = {(0, g), (0, f ), (0, s)}
(d) B C ∩ A = {(1, s)}
9. A retail establishment accepts either the American Express or the
VISA credit card. A total of 24 percent of its customers carry an
American Express card, 61 percent carry a VISA card, and 11 percent
carry both. What percent of its customers carry a credit card the
establishment will accept?
Let V be the event that the customer carries VISA and A be the event
that they carry American Express. We want to find P (A ∪ V ). From
the problem we see that P (A) = .24, P (V ) = .61 and P (A ∩ V ) = .11
and plug it into the equation
P (A ∪ V ) = P (A) + P (V ) − P (A ∩ V )
to get
P (A ∪ V ) = .24 + .61 − .11 = .74.
12. An elementary school is offering 3 language classes: one in Spanish,
one in French and one in German. These classes are open to any of
the 100 students in the school. There are 28 students in the Spanish
class, 26 in the French class, and 16 in the German class. There are
12 students in both Spanish and French, 4 that are in both Spanish
and German, and 6 that are in both French and German. In addition,
there are 2 students taking all 3 classes.
4
(a) If a student is chosen randomly, what is the probability that he
or she is not in any of these classes?
(b) If a student is chosen randomly, what is the probability that he
or she is taking exactly one language class?
(c) If 2 students are chosen randomly, what is the probability that
at least one is taking a language class.
Let S be the event that a student is taking Spanish, let G be the event
that a student is taking German and let F be the event that a student
is taking French.
(a) We can use inclusion exclusion to find the probability a student
is taking at least one class.
P (S ∪ G ∪ F ) = P (S) + P (G) + P (F ) − P (SG) − P (SF ) − P (F G) + P (SF G)
= .28 + .26 + .16 − .04 − .12 − .06 + .02
= .5.
Then P ((SF G)C ) = 1 − P (SF G) = .5 so there are 50 taking at
least one class.
(b) There are 10 students in Spanish and French but not German, 2
students in Spanish and German but not French, and 4 students
in German and French but not Spanish. Thus there are 16 students taking exactly two classes and two taking all three. Thus
there are 32 students taking exactly one language class and
P (a student is taking exactly one class) = .32.
have both
(c) There are 100
choices of two students. Of those 50
2
2
100
50
students not taking a language class so 2 − 2 have at least
one student taking a language class. Thus the probability that
at least one of the two is taking a language class is
100
50
2 − 2
.
100
2
17. If eight castles (that is rooks) are randomly placed on a chessboard
compute the probability that none of the rooks can capture any of the
others. That is, compute the probability that no row or file contains
more than one rook.
5
If we have already placed down i rooks with no two rooks in the same
row or column then there are 8−i possible choices for the row and 8−i
possible choices for the column so there are (8 − i)2 possible choices
for where to put the i + 1st rook. There are 64 possible squares to
place the first rook, 49 places to place the second, 36 for the third, 25
for the fourth all the way to 1 square for the eighth. Thus there are
(64)(49)(36)(25)(16)(9)(4)(1) ordered ways to arrange the rooks. (The
problem isn’t specific we will assume that no two rooks are allowed on
the same space.) There are (64)(63)(62)(61)(60)(59)(58)(57) possible
ordered ways to arrange the rooks. Thus the probability that no two
rooks are in the same row or column is
(64)(49)(36)(25)(16)(9)(4)(1)
(64)(63)(62)(61)(60)(59)(58)(57)
37. An instructor gives her class a set of 10 problems with the information
that the final exam will consist of a random selection of 5 of them. If
a student has figured out how to do 7 of the problems, what is the
probability that he or she will answer correctly
(a) all five problems;
(b) at least four of the problems?
(a) There are 10
5 possible exams the instructor can give. There are
7
5 exams that consist only of the seven problems the student has
figured out. Thus
7
P (the student has figured out all 5 problems) =
5
10 .
5
(b) First we calculate the number of possible exams where
the student
7
has figured out exactly four problems. There are 4 choices of 4
problems the student has figured out. For each of these there 3
choices of the problem the student hasn’t figured out. Adding in
the 75 possible exams that the student figured out five problems
we get that there are
7
7
3
+
4
5
possible exams where the student has figured out at least four
problems. Thus
3 74 + 75
.
P (the student has figured out at least 4 problems) =
10
5
6