Download Homework 1 Solutions

1206/DCP1206 Probability, Fall 2014
9-Oct-2014
Homework 1 Solutions
Instructor: Prof. Wen-Guey Tzeng
1. A box contains three red and five blue balls. Define a sample space for the experiment of recording the colors of three balls that are drawn from the box, one by one,
with replacement.
Answer. S = {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB }.
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2. Define a sample space for the experiment of putting three different books on a shelf in
random order. If two of these three books are a two-volume dictionary, describe the
event that these volumes stand in increasing order side-by-side (i.e., volume I precedes
volume II).
Answer. Denote the dictionaries by d1, d2; the third book by a.
The answers are {d1d2a, d1ad2, d2d1a, d2ad1, ad1d2, ad2d1} and {d1d2a, ad1d2}.
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3. Define a sample space for the experiment of drawing two coins from a purse that contains two quarters, three nickels, one dime, and four pennies. For the same experiment
describe the following events:
(a) drawing 26 cents;
(b) drawing more than 9 but less than 25 cents;
(c) drawing 29 cents.
Answer. S={QQ,QN,QP,QD,DN,DP,NP,NN,PP}
(a){QP}
(b){DN,DP,NN}
(c)φ.
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4. Define a sample space for the experiment of putting in a random order seven different
books on a shelf. If three of these seven books are a three-volume dictionary, describe
the event that these volumes stand in increasing order side by side (i.e., volume I
precedes volume II and volume II precedes volume III).
Answer. Let a1 , a2 , and a3 be the first, the second, and the third volumes of the
dictionary. Let a4 , a5 , a6 , and a7 be the remaining books. Let A = {a1 , a2 , ..., a7 }; the
answers are
S = {x1 x2 x3 x4 x5 x6 x7 : xi ∈ A, 1 ≤ i ≤ 7, xi 6= xj If i 6= j} and
{x1 x2 x3 x4 x5 x6 x7 ∈ S : xi xi+1 xi+2 = a1 a2 a3 f or somei, 1 ≤ i ≤ 5} respectively.
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5. Suppose that the probability that a driver is a male, and has at least one motor
vehicles accident during a one-year period, is 0.12. Suppose that the corresponding
1-1
probability for a female is 0.06. What is the probability of a randomly selected driver
having at least one accident during the next 12 months?
Answer. Let A be the event of a randomly selected driver having anaccident during
the next 12 months. Let B be the event that the person is male. By Theorem 1.7,
the desired probability is
P (A) = P (AB) + P (AB c ) = 0.12 + 0.06 = 0.18.
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6. A company has only one position with three highly qualified applicants: John, Barbara, and Marty. However, because the company has only a few women employees,
Barbara’s chance to be hired is 20% higher than John’s and 20% higher than Marty’s.
Find the probability that Barbara will be hired.
Answer. P (B) = 2/10 + P (J), P (B) = 2/10 + P (M ) ,P (J) + P (B) + P (M ) = 1
P (B) − 2/10 + P (B) + P (B) − 2/10 = 1, So P (B) = 7/14
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7. Which of the following statements is true? If a statement is true, prove it. If it is
false, give a counterexample.
(a) If P (A) + P (B) + P (C) = 1, then the events A,B, and C are mutually exclusive.
(b) If P (A ∪ B ∪ C) = 1, then A, B, and C are mutually exclusive events.
Answer. (a) False; toss a die and let A = {1,2}, B = {2,3}, and C = {1,3}.
(b) False; toss a die and let A = {1,2,3,4}, B = {1,2,3,4,5}, C = {1,2,3,4,5,6}.
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8. Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these chairs
at random. What is the probability that chair 5 is occupied by a boy?
Answer. Since the probability of each chair is equally likely, therefore 7/11.
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9. For a Democratic candidate to win an election, she must win districts I, II, and III.
Polls have shown that the probability of winning I and III is 0.55, losing II but not I is
0.34, and losing II and III but not I is 0.15. Find the probability that this candidate
will win all three
V districts. (Draw a Venn diagram.)
Answer.
P
r[I
II] = 0.55
V
c
P r[II V I] = V
0.34
P r[II c III cV I] = 0.15
V
V
V
2
P r[?] = P r[I II] − (P r[II c I] − P r[II c III c I]) = 0.36
10. Mr. Smith has 12 shirts, eight pairs of slacks, eight ties, and four jackets. Suppose
that four shirts, three pairs of slacks, two ties, and two jackets are blue. (a) What is
the probability that an all-blue outfit is the result of a random selection? (b) What
is the probability that he wears at least one blue item tomorrow?
4∗3∗2∗2
Answer. (a)= 12∗8∗8∗4
= 1/64
8∗5∗6∗2
(b)=1 − 12∗8∗8∗4 = 27/32
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1-2
Figure 1: Veen diagram
11. A campus telephone extension has four digits. How many different extensions with
no repeated digits exist? Of these, (a) how many do not start with a 0; (b) how many
do not have 01 as the first two digits?
Answer. In the begining we have 10 numbers, after we chose the first one, there are
9 numbers left and so on. 10 ∗ 9 ∗ 8 ∗ 7 = 5040.
(a) Since the first digit shouldn’t be ziro, so we have 9 numbers for the first digit. For
the second digit we have 8 numbers left plus ziro, so together we have 9 numbers to
choose, and so on. 9 ∗ 9 ∗ 8 ∗ 7 = 4536;
(b) In this case we first consider all combinations that start with ”01” and then
calculate the complement. You can think that the first two digits are already chosen.
5040 − 1 ∗ 1 ∗ 8 ∗ 7 = 4984.
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12. In a large town, Kennedy Avenue is a long north-south avenue with many intersections. A drunken man is wandering along the avenue and does not really know
which way he is going. He is currently at an intersection O somewhere in the middle
of the avenue. Suppose that, at the end of each block, he either goes north with
probability 1/2, or he goes south with probability 1/2. Draw a tree diagram to find
the probability that, after walking four blocks, (a) he is back at intersection O; (b)
he is only one block away from intersection O.
Answer. He is at 0 first,next he goes to 1 or 1. If at 1,then he goes to 0 or 2. If at
1 ,then he goes to 0 or 2, and so on. Draw a tree diagram. You will find that after
walking 4 blocks, he is at one of the points 4, 2, 0, 2, or 4. There are 16 possible cases
altogether. Of these 6 end up at 0, none at 1, and none at 1. Therefore, the answer
to (a) is 6/16 and the answer to (b) is 0.
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1-3
13. At various yard sales, a woman has acquired five forks, of which no two are alike. The
same applies to her four knives and seven spoons. In how many different ways can
three place settings be chosen if each place setting consists of exactly one fork, one
knife, and one spoon? Assume that the arrangement of the place settings on the table
is unimportant.
Answer. Since the permutation of three places are the same, we have:
(5∗4∗7)∗(4∗3∗6)∗(3∗2∗5)
= 50400
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3!
14. A town has six parks. On a Saturday, six classmates, who are unaware of each others
decision, choose a park at random and go there at the same time. What is the
probability that at least two of them go to the same park?
Answer. we first calculate the probaility of each friend to choose different park so
that none of them are in the same park, 66!6 . Therefore, the probability that at least
two of them go to the same park is the complement of the first part, 1 − 66!6 .
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15. Lili has 20 friends. Among them are Kevin and Gerry, who are husband and wife. Lili
wants to invite six of her friends to her birthday party. If neither Kevin nor Gerry
will go to a party without the other, how many choices does Lili have?
Answer. The
trick
is to choose Kevin and Gerry and then invite the remaining
18
2
+
friends , 18
6 = 21624
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16. A history professor who teaches three sections of the same course every semester
decides to make several tests and use them for the next 10 years (20 semesters) as
final exams. The professor has two policies: (1) not to give the same test to more
than one class in a semester, and (2) not to repeat the same combination of three
tests for any two semesters. Determine the minimum number of different tests that
the professor should prepare.
Answer. The answer is the solution of the equation x3 = 20. This equation is
equivalent to x(x − 1)(x − 2) = 120and its solution is x = 6.
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17. In a closet there are 10 pairs of shoes. If six shoes are selected at random, what is the
probability of (a) no complete pairs; (b) exactly one complete pair; (c) exactly two
complete pairs; (d) exactly three complete pairs?
Answer. Suppose we have 10 objects (pairs of shoes), everytime we choose one
1-4
object, there is 2 different ways to choose one single shoe (L or R).
(10)26
(a)= 620
(6)
(10
)(9)24
(b)= 1 204
(6)
(10
)(8)22
(c)= 2 202
(6)
(10
3)
(d)= 20
(6)
1-5
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