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Problems on Probability Systems

Module by: Paul E Pfeiffer. E-mail the author

Exercise 1

Let Ω consist of the set of positive integers. Consider the subsets

A = { ω : ω 12 } B = { ω : ω < 8 } C = { ω : ω is even } A = { ω : ω 12 } B = { ω : ω < 8 } C = { ω : ω is even }

D = { ω : ω is a multiple of 3 } E = { ω : ω is a multiple of 4 } D = { ω : ω is a multiple of 3 } E = { ω : ω is a multiple of 4 }

Describe in terms of A,B,C,D, E A,B,C,D, E

 
and their complements the following sets:

  1. {1,3,5,7}{1,3,5,7}
  2. {3,6,9}{3,6,9}
  3. {8,10}{8,10}
  4. The even integers greater than 12.
  5. The positive integers which are multiples of six.
  6. The integers which are even and no greater than 6 or which are odd and greater than 12.

Solution

a=BCc,b=DAEc,c=CABcDc,d=CAc,e=CD,f=BCAcCca=BCc,b=DAEc,c=CABcDc,d=CAc,e=CD,f=BCAcCc

Exercise 2

Let Ω be the set of integers 0 through 10. Let A={5,6,7,8}A={5,6,7,8}, B=B= the odd integers in Ω, and C=C= the integers in Ω which are even or less than three. Describe the following sets by listing their elements.

  1. ABAB
  2. ACAC
  3. ABcCABcC
  4. ABCcABCc
  5. ABcABc
  6. ABCcABCc
  7. ABCABC
  8. AcBCcAcBCc

Solution

  1. AB={5,7}AB={5,7}
  2. AC={6,8}AC={6,8}
  3. ABCC=CABCC=C
  4. ABCc=ABABCc=AB
  5. ABc={0,2,4,5,6,7,8,10}ABc={0,2,4,5,6,7,8,10}
  6. ABC=ABC=
  7. AcBCc={3,9}AcBCc={3,9}

Exercise 3

Consider fifteen-word messages in English. Let A=A= the set of such messages which contain the word “bank” and let B=B= the set of messages which contain the word “bank” and the word “credit.” Which event has the greater probability? Why?

Solution

BABA implies P(B)P(A)P(B)P(A).

Exercise 4

A group of five persons consists of two men and three women. They are selected one-by-one in a random manner. Let Ei be the event a man is selected on the ith selection. Write an expression for the event that both men have been selected by the third selection.

Solution

A=E1E2E1E2cE3E1cE2E3A=E1E2E1E2cE3E1cE2E3

Exercise 5

Two persons play a game consecutively until one of them is successful or there are ten unsuccessful plays. Let Ei be the event of a success on the ith play of the game. Let A,B,CA,B,C be the respective events that player one, player two, or neither wins. Write an expression for each of these events in terms of the events Ei, 1i101i10.

Solution

A = E 1 E 1 c E 2 c E 3 E 1 c E 2 c E 3 c E 4 c E 5 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 c E 9 A = E 1 E 1 c E 2 c E 3 E 1 c E 2 c E 3 c E 4 c E 5 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 c E 9
(1)

B = E 1 c E 2 E 1 c E 2 c E 3 c E 4 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 c E 9 c E 10 B = E 1 c E 2 E 1 c E 2 c E 3 c E 4 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 E 1 c E 2 c E 3 c E 4 c E 5 c E 6 c E 7 c E 8 c E 9 c E 10

C = i = 1 10 E i c C = i = 1 10 E i c

Exercise 6

Suppose the game in Exercise 5 could, in principle, be played an unlimited number of times. Write an expression for the event D that the game will be terminated with a success in a finite number of times. Write an expression for the event F that the game will never terminate.

Solution

Let F0=ΩF0=Ω and Fk=i=1kEicFk=i=1kEic for k1k1. Then

D = n = 1 F n - 1 E n and F = D c = i = 1 E i c D = n = 1 F n - 1 E n and F = D c = i = 1 E i c
(2)

Exercise 7

Find the (classical) probability that among three random digits, with each digit (0 through 9) being equally likely and each triple equally likely:

  1. All three are alike.
  2. No two are alike.
  3. The first digit is 0.
  4. Exactly two are alike.

Solution

Each triple has probability 1/103=1/10001/103=1/1000

  1. Ten triples, all alike: P=10/1000P=10/1000.
  2. 10×9×810×9×8 triples all different: P=720/1000P=720/1000.
  3. 100 triples with first one zero: P=100/1000P=100/1000
  4. C(3,2)=3C(3,2)=3 ways to pick two positions alike; 10 ways to pick the common value; 9 ways to pick the other. P=270/1000P=270/1000.

Exercise 8

The classical probability model is based on the assumption of equally likely outcomes. Some care must be shown in analysis to be certain that this assumption is good. A well known example is the following. Two coins are tossed. One of three outcomes is observed: Let ω1 be the outcome both are “heads,” ω2 the outcome that both are “tails,” and ω3 be the outcome that they are different. Is it reasonable to suppose these three outcomes are equally likely? What probabilities would you assign?

Solution

P({ω1})=P({ω2})=1/4,P({ω3}=1/2P({ω1})=P({ω2})=1/4,P({ω3}=1/2.

Exercise 9

A committee of five is chosen from a group of 20 people. What is the probability that a specified member of the group will be on the committee?

Solution

C(20,5)C(20,5) committees; C(19,4)C(19,4) have a designated member.

P = 19 ! 4 ! 15 ! 5 ! 15 ! 20 ! = 5 / 20 = 1 / 4 P = 19 ! 4 ! 15 ! 5 ! 15 ! 20 ! = 5 / 20 = 1 / 4
(3)

Exercise 10

Ten employees of a company drive their cars to the city each day and park randomly in ten spots. What is the (classical) probability that on a given day Jim will be in place three? There are n!n! equally likely ways to arrange n items (order important).

Solution

10! permutations. 1×9!1×9! permutations with Jim in place 3. P=9!/10!=1/10P=9!/10!=1/10.

Exercise 11

An extension of the classical model involves the use of areas. A certain region L (say of land) is taken as a reference. For any subregion A, define P(A)=area(A)/area(L)P(A)=area(A)/area(L). Show that P()P() is a probability measure on the subregions of L.

Solution

Additivity follows from additivity of areas of disjoint regions.

Exercise 12

John thinks the probability the Houston Texans will win next Sunday is 0.3 and the probability the Dallas Cowboys will win is 0.7 (they are not playing each other). He thinks the probability both will win is somewhere between—say, 0.5. Is that a reasonable assumption? Justify your answer.

Solution

P(AB)=0.5P(AB)=0.5is not reasonable. It must no greater than the minimum of P(A)=0.3P(A)=0.3 and P(B)=0.7P(B)=0.7.

Exercise 13

Suppose P(A)=0.5P(A)=0.5 and P(B)=0.3P(B)=0.3. What is the largest possible value of P(AB)P(AB)? Using the maximum value of P(AB)P(AB), determine P(ABc)P(ABc), P(AcB)P(AcB), P(AcBc)P(AcBc) and P(AB)P(AB). Are these values determined uniquely?

Solution

Draw a Venn diagram, or use algebraic expressions P(ABc)=P(A)-P(AB)=0.2P(ABc)=P(A)-P(AB)=0.2

P ( A c B ) = P ( B ) - P ( A B ) = 0 P ( A c B c ) = P ( A c ) - P ( A c B ) = 0 . 5 P ( A B ) = 0 . 5 P ( A c B ) = P ( B ) - P ( A B ) = 0 P ( A c B c ) = P ( A c ) - P ( A c B ) = 0 . 5 P ( A B ) = 0 . 5
(4)

Exercise 14

For each of the following probability “assignments”, fill out the table. Which assignments are not permissible? Explain why, in each case.

Table 1
P ( A ) P ( A ) P ( B ) P ( B ) P ( A B ) P ( A B ) P ( A B ) P ( A B ) P ( A B c ) P ( A B c ) P ( A c B ) P ( A c B ) P ( A ) + P ( B ) P ( A ) + P ( B )
0.3 0.7 0.4        
0.2 0.1 0.4        
0.3 0.7 0.2        
0.3 0.5 0        
0.3 0.8 0        

Solution

Table 2
P ( A ) P ( A ) P ( B ) P ( B ) P ( A B ) P ( A B ) P ( A B ) P ( A B ) P ( A B c ) P ( A B c ) P ( A c B ) P ( A c B ) P ( A ) + P ( B ) P ( A ) + P ( B )
0.3 0.7 0.4 0.6 -0.1 0.3 1.0
0.2 0.1 0.4 -0.1 -0.2 -0.3 0.3
0.3 0.7 0.2 0.8 0.1 0.5 1.0
0.3 0.5 0 0.8 0.3 0.5 0.8
0.3 0.8 0 1.1 0.3 0.8 1.1

Only the third and fourth assignments are permissible.

Exercise 15

The class {A,B,C}{A,B,C} of events is a partition. Event A is twice as likely as C and event B is as likely as the combination A or C. Determine the probabilities P(A),P(B),P(C)P(A),P(B),P(C).

Solution

P(A)+P(B)+P(C)=1P(A)+P(B)+P(C)=1, P(A)=2P(C)P(A)=2P(C), and P(B)=P(A)+P(C)=3P(C)P(B)=P(A)+P(C)=3P(C), which implies

P ( C ) = 1 / 6 , P ( A ) = 1 / 3 , and P ( B ) = 1 / 2 P ( C ) = 1 / 6 , P ( A ) = 1 / 3 , and P ( B ) = 1 / 2
(5)

Exercise 16

Determine the probability P(ABC)P(ABC) in terms of the probabilities of the events A,B,CA,B,C and their intersections.

Solution

P(ABC)=P(AB)+P(C)-P(ACBC)P(ABC)=P(AB)+P(C)-P(ACBC)

= P ( A ) + P ( B ) - P ( A B ) + P ( C ) - P ( A C ) - P ( B C ) + P ( A B C ) = P ( A ) + P ( B ) - P ( A B ) + P ( C ) - P ( A C ) - P ( B C ) + P ( A B C )
(6)

Exercise 17

If occurrence of event A implies occurrence of B, show that P(AcB)=P(B)-P(A)P(AcB)=P(B)-P(A).

Solution

P(AB)=P(A)P(AB)=P(A) and P(AB)+P(AcB)=P(B)P(AB)+P(AcB)=P(B) implies P(AcB)=P(B)-P(A)P(AcB)=P(B)-P(A).

Exercise 18

Show that P(AB)P(A)+P(B)-1P(AB)P(A)+P(B)-1.

Solution

Follows from P(A)+P(B)-P(AB)=P(AB)1P(A)+P(B)-P(AB)=P(AB)1.

Exercise 19

The set combination AB=ABcAcBAB=ABcAcB is known as the disjunctive union or the symetric difference of A and B. This is the event that only one of the events A or B occurs on a trial. Determine P(AB)P(AB) in terms of P(A)P(A), P(B)P(B), and P(AB)P(AB).

Solution

A Venn diagram shows P(AB)=P(ABc)+P(ABc)=P(A)+P(B)-2P(AB)P(AB)=P(ABc)+P(ABc)=P(A)+P(B)-2P(AB).

Exercise 20

Use fundamental properties of probability to show

  1. P ( A B ) P ( A ) P ( A B ) P ( A ) + P ( B ) P ( A B ) P ( A ) P ( A B ) P ( A ) + P ( B )
  2. P j = 1 E j P ( E i ) P j = 1 E j j = 1 P ( E j ) P j = 1 E j P ( E i ) P j = 1 E j j = 1 P ( E j )

Solution

ABAABABAAB implies P(AB)P(A)P(AB)=P(A)+P(B)-P(AB)P(A)+P(B)P(AB)P(A)P(AB)=P(A)+P(B)-P(AB)P(A)+P(B). The general case follows similarly, with the last inequality determined by subadditivity.

Exercise 21

Suppose P1,P2P1,P2 are probability measures and c1,c2c1,c2 are positive numbers such that c1+c2=1c1+c2=1. Show that the assignment P(E)=c1P1(E)+c2P2(E)P(E)=c1P1(E)+c2P2(E) to the class of events is a probability measure. Such a combination of probability measures is known as a mixture. Extend this to

P ( E ) = i = 1 n c i P i ( E ) , where the P i are probabilities measures, c i > 0 , and i = 1 n c i = 1 P ( E ) = i = 1 n c i P i ( E ) , where the P i are probabilities measures, c i > 0 , and i = 1 n c i = 1
(7)

Solution

Clearly P(E)0P(E)0. P(Ω)=c1P1(Ω)+c2P2(Ω)=1P(Ω)=c1P1(Ω)+c2P2(Ω)=1.

E = i = 1 E i implies P ( E ) = c 1 i = 1 P 1 ( E i ) + c 2 i = 1 P 2 ( E i ) = i = 1 P ( E i ) E = i = 1 E i implies P ( E ) = c 1 i = 1 P 1 ( E i ) + c 2 i = 1 P 2 ( E i ) = i = 1 P ( E i )
(8)

The pattern is the same for the general case, except that the sum of two terms is replaced by the sum of n terms ciPi(E)ciPi(E).

Exercise 22

Suppose {A1,A2,,An}{A1,A2,,An} is a partition and {c1,c2,,cn}{c1,c2,,cn} is a class of positive constants. For each event E, let

Q ( E ) = i = 1 n c i P ( E A i ) i = 1 n c i P ( A i ) Q ( E ) = i = 1 n c i P ( E A i ) i = 1 n c i P ( A i )
(9)

Show that Q()Q() us a probability measure.

Solution

Clearly Q(E)0Q(E)0 and since AiΩ=AiAiΩ=Ai we have Q(Ω)=1Q(Ω)=1. If

E = k = 1 E k , then P ( E A i ) = k = 1 P ( E k A i ) i E = k = 1 E k , then P ( E A i ) = k = 1 P ( E k A i ) i
(10)

Interchanging the order of summation shows that Q is countably additive.

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