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Introduction to Probability Summer 2021

Homework Assignments Fan Xue

Contents

Instructions 1

Homework Assignment 1 2

Homework Assignment 2 6

Homework Assignment 3 0

Homework Assignment 4 0

Homework Assignment 5 5

Instructions

Replace the highlighted portions of the text with your specific information. Remove the highlighting as you do so.

Homework Assignment 1

If there are LATEXerrors, please take care of them.

Set 1, problem 1

How many plate numbers can be created in California format DLLLDDD, where D stands for digit and L stands for letter? What if no repetitions are allowed in the digits?

Solution: In the plate numbers, there are 4 digits and 3 letters. As we know that no repetitions are allowed in digits. the answer should be 10∗9∗8∗7∗26∗26∗26 = 88583040 So, there are 88583040 plate numbers can be created.

You only answered the second question. 2.5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 1, problem 2

A zip code consists of 5 digits. If the first cannot be the digit 0, how many zip codes can be generated?

Solution: There are ten possible digits(0 to 9). If they can be repeated, and the first cannot be the digit 0. Then, the answer should be 9 ∗ 10 ∗ 10 ∗ 10 ∗ 10 = 90000 There are 90000 zip codes can be generated. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 1, problem 6

If you need to split a class of 32 students in 8 groups of 4, in how many ways can you do it?

Solution: 32! 4!

28! 4!

24! 4!

20! 4!

16! 4!

12! 4!

8! 4!

4! 4!

8! = 5.92 ∗ 1019

there are 5.92 ∗ 1019 ways. Please explain how you get these numbers. You may mean the correct answer but the number you given in the numerator is incorrect. 2/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 1, problem 7

If a student is required to solve 8 out of 10 assigned problems, how many choices does she have? What if the first 4 problems are mandatory?

Solution:

a. 10! 8!2!

= 45 ( 10 8

) b. 6!

4!2! = 15

( 6 4

) She has 45 choices to solve 8 out of 10 assigned problems. If the first 4 problems are mandatory, she has 15 choices.

Please explain how you get these numbers. 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 2, problem 1

An urn contains 3 balls: 1 red, 1 green, and 1 blue. Consider an experiment that consists of taking 1 ball from the urn, replacing it in the box, and drawing a second ball. Describe the sample space of this experiment and for the experiment obtained not replacing the first ball in the urn.

Solution: The event consists of drawing of total of 2 balls from the box since the ball is replaced after taking, we take any of the 3 balls during the second attempt and the sample space will be:{

(r,r)(r,b)(r,g)(b,r)(b,b)(b,g)(g,r)(g,b)(g,g) }

the first color indicated the ball taken during the first attempt, and the second color indicated the ball taken after replacing the ball in the box. Now, if the ball is not replaced in the box after taking it, only two balls will be left if we take 1 ball, hence only 2 possibilities will be there in the second attempt, so, the sample space will be:

{(red,blue)(red,green)(blue,red)(blue,green)(green,red)(green,blue)}

5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 2, problem 4

A hospital administrator codes incoming patients suffering heart disease according to whether they have insurance (code 1 if they do and 0 if they do not) and based on their condition, which can be rated as good G, fair F, or serious S. Consider an experiment that consists of producing codes for such a patient.

a. Give the sample space of this experiment. b. Let S be the event that the patient is in serious condition and list all outcomes in S c. Let U be the event that the patient is uninsured and all outcomes in U d. List the outcomes in Sc ∪U .

Solution: a. Sample space consists all possible outcomes of this experiment. That is, all possible combinations of insurance coding and condition coding

S = {(1,f), (1,g), (1,s), (o,f), (0,g), (0,s)}= {0, 1}×{f,g,s}

b. Given that a person is in serious condition. There are only two possible outcomes in S: insured or not insured

S = {(1,s), (0,s)}

c. Given that a person does not have insurance, there are only three possible outcomes: serious, good or fair

U = {(0,g), (0,f), (0,s)}

d. Bc∪A is the event that the patient has insurance(does not have insurance) or is in serious condition

S = Bc = {(1,g), (1,f), (1,s)}

Bc ∪A = (1,g), (1,f), (1,s), (0,s)

what are all these new sets A,B? And what is S now? 4/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 2, problem 5

Suppose that two event E, Fare such that E ∩F = ∅ and satisfy P(E)=0.3 and P(F)=0.5. What are the probabilities that a. either E or F occurs? b. E occurs but F does not? c. both E and F occur? What happens if you don’t assume that E ∩F = ∅?

Solution:

a. P(either E or F) = P(E) + P(F) = 0.3 + 0.5 = 0.8 use mathematical notation and why?

b. P(E occurs but F does not) = P(E ∩Fc) = P(E) - P(E ∩F) = 0.3 − 0 = 0.3 c. P(both E and F) = P(E ∩F) = 0 d. if we don’t assume that E ∩F = ∅, we cannot find the probabilities.

Set 2, problem 6

At a summer camp three activities are offered: tennis, swimming, and archery. The courses are open to the 120 summer camp participants. There are 31, 24, and 17 in these activities, respectively. There are 4 participants in all activities, 11 in tennis and swimming, 9 in tennis and archery, 7 in swimming and archery. a. If a participant is chosen at random, what is the probability that he or she is taking part exactly one activity? b. If a participant is chosen at random, what is the probability that he or she is taking part in no activity? c. If two participants are chosen at random, what is the probability that they are both taking part in exactly one activity?

Solution: Total = 120 Tennis = 31 Swimming = 24 Archery = 17 Tennis and Swimming = 11 Tennis and Archery = 9 Swimming and Archery = 7 All three activities = 4

Only Tennis = 31 - (9 - 4 + 4 + 7) = 15

Only Swimming = 24 - (7-4 + 4 + 7) = 10

Only Archery = 17 - (9 - 4 + 4 + 7 - 4) = 5

Those not taking any part = 120 - (15 + 7 + 10 + 5 + 4 + 3 + 5) = 120 - 49 = 71

a. P(Exactly 1 activity) = (15 + 10 + 5) / 120 = 30 / 120 = 0.25

b. P(No activity) = 71 / 120 = 0.592

c. P(two will be in exactly 1 activity) = 0.25 * 0.25 = 0.0625

Explanations are needed. Use set notation: T for tennis, S for swimming, and A for archery. Then we have for instance that only tennis is given by the set = T ∩Sc ∩Ac. If you provide your work, I will

give you full credit. 0/5

Homework Assignment 2

Set 3, problem 1

If 8 identical computers are to be divided among 4 lab rooms, how many divisions are possible? How many if each room must receive at least 1 computer?

Solution: We need to divide 9 identical computers to 4 rooms. This problem type of dividing identical can be visualized as dividing n identical 0′s into r groupings. For example 00101000100 can represent dividing 2 computers to first, 1 computer to second 3 to third and 2 to fourth lab. 8 zeros and division represented by 1′s. n be total and r be groups. Then the problem becomes selecting r − 1 1′s to divide from total of n + r− 1. So (n + r− 1)Cr−1 is the number of ways we can divide n identical things into r groups. Here n = 8, 4 = 4

11C3 = 11 ∗ 10 ∗ 9/(3 ∗ 2) = 165

possible ways When each room must receive 1 computer the formula changes to (n − 1)Cr−1. This is because in the other formula we assign one to each and then assign the remaining n−r + r−1 = n−1.

7C3 = 7 ∗ 6 ∗ 5/(3 ∗ 2) = 35

ways.

You need to take a closer look at how to use LATEXsince your formulæ do not display correctly. 5/5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 3, problem 2

Suppose that 10 fish are caught in a pond that contains 5 distinct types of fish. a. How many different outcomes are possible, where an outcome consists in the number of fish caught

of each of the 5 types? b. How many outcomes are possible if 3 of the 10 fish caught are trout? c. How many when at least 2 of the 10 are trout?

Solution:a. So the number of possible combinations can be thought of as dividing 10 identical items into 5 groups since all combinations are possible n = 10,r = 5.

14C4 = 14 ∗ 13 ∗ 12 ∗ 11/(1 ∗ 2 ∗ 3 ∗ 4) = 1001

b. 3 of 10 fish are trought that means remaining 7 fish to be assigned to 4 groups of fish. n = 7,r = 4

10C3 = 10 ∗ 9 ∗ 8/(3 ∗ 2) = 120possibilities

c. It says at least 2 trouts but it can be more so we assign 2 to trouts and then divide the remaining 8 into 5 types. n = 8,r = 5.

12C4 = 12 ∗ 11 ∗ 10 ∗ 9/(2 ∗ 3 ∗ 4) = 495possibilities

Check your LATEX. Your formulæ do not show correctly! 5/5

Set 3, problem 8

Two cards are randomly selected from an ordinary playing deck (52 cards). What is the probability that one of the cards is an ace and the other one is either a ten, a jack, a queen, or a king? This combinations is called a black jack.

Solution:we have to find the probability that they form a back jack. overall outcomes = 52C2. From two cards, first card should be an ace, and the total number of ways = 4 ways. The second card should be a either ”ten” or ”jack” or ”queen” or ”king”. So, the total number of ways = 4∗4 = 16 ways. The overall favourable outcomes = 1stcard∗ 2ndcard = 4 ∗ 16 = 64 ways. To find the probability to have a blackjack.

Probabilitytohaveablackjack = overallfavourableoutcomes

overalloutcomes

= 64

52C2

= 64

26 ∗ 51

= 64

1326

= 0.04826

Thus, probability to have a blackjack = 0.04826.

Check your LATEX. 5/5

Set 3, problem 7

A retail establishment accepts either the American Express or the VISA credit card. A total of 24% of its customers carry an American Express card, 61% carry a VISA card, and 11% carry both cards. What percentage of its customers carry a credit card that the establishment will accept?

Solution: Let A be the event the customer has an American Express card, and let V be the event the customer has a VISA card. Then P(A) = 0.24,P(V ) = 0.61, and P(AV ) = 0.11. The question asks for P(A∪V ). Using the Inclusion-Exclusion formula

P(A∪V ) = P(A) + P(V ) −P(AV ) = 0.74

. 5/5

Set 4, problem 1

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i for i = 1, . . . , 12?

If you use copy and paste, do it from the source code of the HTML page.

Solution: Let A be the event that the first die is a six and let Bi be the event that the sum of the two die is i. There are 6 possible outcomes for A.

A = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Use the command \{ if you want to generate a {.

All of these have a sum of at least 7, so P(ABi) = 0 for all i ≤ 6. For every i ≥ 7. Thus P(ABi) = 136 for all i ≥ 7. Also for all i ≥ 7 there are 13 − i possible outcomes in Bi so P(Bi) = 13−i36 for all i ≥ 7. for

i = 2; P(A|Bi) = 0 for

i = 3; P(A|Bi) = 0 for

i = 4; P(A|Bi) = 0 for

i = 5; P(A|Bi) = 0 for

i = 6; P(A|Bi) = 0 for

i = 7; P(A|Bi) = 1

6 for

i = 8; P(A|Bi) = 1

5 for

i = 9; P(A|Bi) = 1

4 for

i = 10; P(A|Bi) = 1

3 for

i = 11; P(A|Bi) = 1

2 for

i = 12; P(A|Bi) = 1

Set 4, problem 4

In a certain community, 36% of the families own a dog and 22% of the families that own a dog also own a cat. In addition, 30% of the families own a cat. What is

a. the probability that a randomly selected family owns both a dog and a cat? b. the conditional probability that a randomly selected family owns a dog given that it owns a cat?

Solution: Let A be the event that family owns a dog, and let B be the event that family owns a cat

P(A) = 0.36,P(B|A) = 0.22,P(B) = 0, 3

a. P(A∩B) = P(A) ∗P(B|A) = 0.36 ∗ 0.22 = 0.0792

P(A|B) = P(A∩B) P(B)

= 0.0792

0.3 = 0.264

Set 4, problem 2

Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls?

Solution: The total number of balls in the urn is 12. Out of 12 balls 8 balls are white and 4 balls are black. Let A be the event of the first and third balls are white, and let B be the event that the sample drawn contain exactly 3 white balls. The conditional probability is P(A|B) = P(AB)

P(B) .

A∩B : The first and third balls are white and the sample contains exactly 3 white balls. The possibil- ities are (WBWW) and (WWWB).

B : The sample contains exactly 3 white balls. The possibilities are (BWWW)(WBWW)(WWBW)(WWWB)

P(A|B) = P(AB)

P(B)

= (( 8

12 )( 4

12 )( 8

12 ) + ( 8

12 )( 8

12 )( 4

12 ))

4(( 8 12

)( 4 12

)( 8 12

))

= (( 2048

20736 ) + ( 2048

20736 ))

4( 2048 20736

)

= ( 4096 20736

)

(4 2048 20736

)

= 4096

8192 = 0.5

Therefore, the conditional probability that the first and third balls drawn will be white given that the sample drawn contains exactly 3 white balls in the case of with replacement is 0.5

The problem is also asking the same question when balls are drawn with no replacement. 2.5/5

Set 4, problem N4

Urn A contains 2 white and 4 red balls, whereas urn B contains 1 white and 1 red ball. A ball is ran- domly chosen from urn A and put into urn B, and a ball is then randomly selected from urn B. What is

a. the probability that the ball selected from urn B is white? b. the conditional probability that the transferred ball was white given that a white ball is selected

from urn B?

Solution: Let W be white balls, and R be red balls. A ball is randomly chosen from urnA and put into urnB, and a ball is then randomly selected from urnB a.

case1 :

If ball chosen from urn A is white. Probability of chosing a white ball from urn A is P = 2 6

urn B already contains 1W and 1R. After a white ball from urn A put into urn B, then the number of white balls in urn B will be 2. Now, the probability of selecting a white ball from urn B is P(whiteballfromurnB) = 2

3 .

Hence, the probability that the ball selected from urn B is white and the ball chosen from urn A is white is P(W1,W2) =

2 6 ∗ 2

3 = 2

9 case2 :

if ball chosen from urn A is red. Probability of choosing a red ball from urn is P = 4 6 . Then, we place

the red ball in urn B. Now, urn B contains 1W and 2R, now the probability of selecting a white ball from urn B is P(whiteballfromurnB) = 1

3 . Hence, the probability that a ball selected from urn B is

white and the ball chosen from urn A is red is P(R1,W2) = 4 6 ∗ 1

3 = 2

9 . We add these two mutually

exclusive cases to get the required probability is

P(ballselectedfromurnBwhite) = 2

9 +

9 =

. b. The conditional probability that the transferred ball was white. Given that a white ball is selected from urn B is

P( transferredballiswhite

whiteballisselectedfromurnB )

= P(transferredballiswhite∩whiteballisselectedfromurnB)

P(whiteballisselectedfromurnB)

= 4 18 4 9

= 4

18 ∗

4 =

2 = 0.5

You can formulate your solutions in much simpler terms using Bayes’ formula. Also check your usage of LATEX.

5/5

Homework Assignment 3

Set 5, problem 1

A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80% chance that she will get the job if she receives a strong recommendation, a 40% chance if she receives a moderately good recommendation, and a 10% chance if she receives a weak recommendation. She further estimates that the probabilities that the recommendation will be strong, moderate, and weak are 0.7, 0.2, and 0.1, respectively. a. How certain is she that she will receive the new job offer?

b. Given that she does receive the offer, how likely should she feel that she received a strong recom- mendation? A moderate recommendation? A weak recommendation?

c. Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation? A moderate recommendation? A weak recommendation?

Solution: Let O be get-offer, NO be non-offer, ST be strong recommendation, G be good recommenda- tion, and W be week recommendation. From given problem data, we can get

P(O ∩ST) = P(O|ST)P(ST) = (0.8)(0.7) = 0.56

P(O ∩G) = P(O|G)P(G) = (0.4)(0.2) = 0.08

P(O ∩W) = P(O|W)P(W) = (0.1)(0.1) = 0.01

a. P(O) = P(O ∩ST) + P(O ∩G) + P(O ∩W) = 0.56 + 0.08 + 0.01 = 0.65

P(ST|O) = P(ST ∩O) P(O)

= 0.56

0.65 = 0.86

P(G|O) = P(G∩O) P(O)

= 0.08

0.65 = 0.12

P(W |O) = P(W ∩O) P(O)

= 0.01

0.65 = 0.02

c. from the data, we can get

P(NO ∩ST) = 0.7 −P(O ∩ST) = 0.7 − 0.56 = 0.14

P(NO ∩G) = 0.2 −P(O ∩G) = 0.2 − 0.08 = 0.12

P(NO ∩W) = 0.1 −P(O ∩W) = 0.1 − 0.01 = 0.09

P(NO) = P(NO ∩ST) + P(NO ∩G) + P(NO ∩W) = 0, 14 + 0.12 + 0.09 = 0.35

P(ST|NO) = P(ST ∩NO) P(NO)

= 0.14

0.35 = 0.4

P(G|NO) = P(G∩NO) P(NO)

= 0.12

0.35 = 0.34

P(W |NO) = P(W ∩NO) P(NO)

= 0.09

0.35 = 0.26

Set 5, problem 2

Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability P1, while each shot of Dianne’s hits it with probability P2. Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that

a. both shots hit the duck? -b. Barbara’s shot hit the duck?

Solution: Let B be the Barbara hits wooden duck, D be Diana hits wooden duck, and H be the duck is hit

P(B) = P1,P(D) = P2

a. P(Bothhitduck|duckishit) = P(BD|H)

= P(BDH)

= P(H|BD)P(BD)

P(H|BD) = 1,P(BD) = P1P2 P(H) = P(B ∪D) = P(B) + P(D) −P(BD) = P1 + P2 −P1P2

Probability = (1)(P1)(P2)

(P1 + P2) − (P1)(P2) b.

P(B|H) = P(B ∩H) P(H)

= P(H|B)P(B)

P(H) =

1(P1)

(P1 + P2) − (P1P2)

P(B|H) = P1

(P1 + P2) − (P1P2) Both shoot at the duck at the same time and their hits are independent 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 5, problem 3

In successive rolls of a pair of fair dice, what is the probability of getting 2 sums of seven before 6 even numbers?

Solution: Let A, B and C be the events that get a seven, and even number, or neither on a given roll, respectively. Thus

P(A) = 6

36 =

6 ,P(B) =

36 =

2 ,P(C) = 1 −P(A) −P(B) =

We have P(2 sevens before 6 evens)

= 5∑ i=0

P(i evens before 2 sevens)

= 5∑ i=0

i∑ j=0

P(ievensbefore2sevens|jevensbeforefirstseven)P(jevensbeforefirstseven)

= 5∑ i=0

i∑ j=0

P(i− jevensbeforefirstseven)P(jevensbeforefirstseven)

For each non-negative integer j let Ej be the event that you roll j evens before the first seven, We have

P(E0) = P(E0|sevenonthefirstroll)P(sevenonthefirstroll)

+P(E0|evenonthefirstroll)P(evenonthefirstroll)

+P(E0|neithersevennorevenonthefirstroll)P(neithersevennorevenonthefirstroll)

= 1P(A) + 0P(B) + P(E0)P(C)

P(E0) = P(A)

1 −P(C) =

For j > 0 we have

P(Ej) = P(Ej|sevenonthefirstroll)P(sevenonthefirstroll)

+P(Ej) = P(Ej|evenonthefirstroll)P(evenonthefirstroll)

+P(Ej) = P(Ej|neithersevennorevenonthefirstroll)P(neithersevennorevenonthefirstroll)

= 0P(A) + P(Ej=1)P(B) + P(Ej)P(C)

which implies that (1 −P(C))P(Ej) = P(Ej−1)P(B)

P(Ej) = ( P(B)

1 −P(C) )j,P(E0) = (

4 )j

So the desired probability is

5∑ i=0

i∑ j=0

( 3

4 )i=j

4 ( 3

4 )j

4 = (

4 )2

i∑ j=0

(i + 1)( 3

4 )i =' 0.5550

The initial identity in red is incorrect as stated. The presentation needs to be improved. If you correct your solution and improve your presentation, send me a message and I will take another look and give you some credit

0/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 5, problem 4

An urn contains 12 balls, of which 4 are white. Three players (A, B, and C) successively draw from the urn in that order repeatedly. The winner is the first one to draw a white ball. Find the probability of winning for each player if

a. each ball is replaced after it is drawn. b. the balls that are withdrawn are not replaced.

Solution: a. There are 4 white and 8 non-white balls, The probability of winning for with replacement 4 12

= 1 3

and probability that A does not win is 8 12

= 2 3 . Now, probability that A does not win in first

draw is 2 3 . If it happens then B is exactly in the same situation as A was at beginning. SO now chance

of B to win is 2 3 A. Similarly if player A and B does not win then C is in the same situation so probability

of C wins is ( 2 3 )2A. The sum of three probabilities is 1.

A + ( 2

3 )A + (

3 )2A = 1

A(1 + ( 2

3 ) + (

3 )2) = 1

A(( 5

3 ) + (

9 )) = 1

A = 9

So P(Awins) = 0.4737

P(Bwins) = ( 2

3 )(

19 ) = 0.3158

P(Cwins) = ( 2

3 )2(

19 ) = 0.2105

Your solution has some merit since your explanation is reasonable. Your presentation, however, is very poor and does not make explicit use of the formulæ you use. Again, if you give a cleaner proof, I can give you credit.

b. The balls are withdrawn without replacement. The probabilities that 1st white ball is drawn at 1st, 2nd, 3rd, 4th, ......, 9th draw is

1st = 1

2nd = 8

11 =

3rd = 8

10 =

165

4th = 8

9 =

495

5th = 8

8 =

6th = 8

7 =

7th = 8

6 =

8th = 8

5 =

495

9th = 8

4 =

495

Now we know balls are drawn by first A, then B, then C, then A and so on.

P(Awins) = P(whiteball1stdraw) + P(whiteballat4thdraw) + P(whiteballat7thdraw) = 1

3 +

495 +

P(Awins) = 0.4667

P(Bwins) = 8

33 +

99 +

495 =

165

P(Bwins) = 0.3212

P(Cwins) = 28

165 +

99 +

495 =

P(Cwins) = 0.2121

I gave you full credit for part b. 2.5/5

Set 6, problem 2

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X ?

Solution: A coin is tossed six times and X represents the difference between the number of heads and the number of tails. Let H be the event of heads, and T be the event of tails. then, the following are the possible values:

X(6H, 0T) = |6 − 0| = 6

X(5H, 1T) = |5 − 1| = 4

X(4H, 2T) = |4 − 2| = 2

X(3H, 3T) = |3 − 3| = 0

X(2H, 4T) = |2 − 4| = 2

X(1H, 5T) = |1 − 5| = 4

X(0H, 6T) = |0 − 6| = 6

Thus, the possible values of X are 0, 2, 4, 6 5/5

Set 6, problem 3

Suppose that two teams play a series of games that ends when one of them has won i games. Suppose that each game played is, independently, won by team A with probability p. Find the expected number of games that are played when i = 2 and when i = 3. Also, show that this number is maximized when p = 0.5 in both cases.

Solution: Let X denote the number of games played a. When i =2

E(X) = 2(p2 + (1 −p)2) + 3(2p2(1 −p) + 2p(1 −p)2)

= −2p2 + 2p + 2

By letting d

dp EX = 0

we can get p = 1 2

at which EX is maximized.( d 2

dp2 EX|p= 1

2 < 0)

b. When i = 3

E(X) = 3(p3 + (1 −p)3) + 4(3p3(1 −p) + 3p(1 −p)3) + 5((6p3(1 −p)2) + 6p2(1 −p)3)

= 6p4 − 12p3 + 3p2 + 3p + 3

By letting d

dp EX = 0

we can get p = 1 2

at which EX is maximized.( d 2

dp2 EX|p= 1

2 < 0)

Set 6, problem 6

You have $1000, and a certain commodity presently sells for $2 per ounce. Suppose that after one week the commodity will sell for either $1 or $4 an ounce, with these two possibilities being equally likely.

a. If your objective is to maximize the expected amount of money that you possess at the end of the week, what strategy should you employ?

b. If your objective is to maximize the expected amount of the commodity that you possess at the end of the week, what strategy should you employ?

Solution: Suppose that you buy x ounces during the first week, and then you will have 1000 − 2x dollars until next week a. Let M denote the amount of money that you possess at the end of the week.

EM = (1000 − 2x + x) 1

2 + (1000 − 2x + 4x)

= 1000 + 1

2 x

The expected amount of money would be larger if you bought more during the first week. Therefore, the best strategy would be to use your all money to get 500 ounces of the commodity and sell them after one week. b. Let N denote the amount of the commodity

EN = (x + 1000 − 2x) 1

2 + (x +

1000 − 2x 4

2 )

= − 1

4 x + 625

The expected amount of the commodity would be smaller if you bought more during the first week. Therefore, the best strategy would be not to buy anything during the first week, and spend all the money after one week.

Set 6, problem 8

If X has distribution function FX, what is the distribution function FY of Y = e X?

Solution: Let Y = eX. Since we know that X has a distribution function F,

FY (y) = P(Y ≤ y) = P(eX ≤ y)

= P(X ≤ ln(y))

= F(ln(y))

5/5

Homework Assignment 4

Set 7, problem 1

If E[X] = 1 and V ar(X) = 5 find a. E[(2 + X)2] b. V ar(4 + 3X)

Solution: From the lecture we got the following properties:

E(X + a) = E(X) + a

E(aX) = aE(X)

V (aX) = a2V (X)

a. From the information, the population mean E(X) = 1 and the variance of V ar(X) = 5

V ar(x) = E(X2) −E(X)2

E[(2 + X)2] = E[22 + 2(2)X + X2]

= E[4 + 4X + X2]

= 4 + 4E(X) + E(X2)

= 4 + 4E(X) + (V ar(X) + E(X)2)

= 4 + (4 ∗ 1) + (5 + (1)2)

= 4 + 4 + 6

= 14

b. V ar(4 + 3X) = 32V ar(X)

= 9 ∗ 5

= 45

Set 7, problem 2

On a multiple-choice exam with 3 possible answers for each of the 5 questions, what is the probability that a student will get 4 or more correct answers just by guessing?

Solution:

P(correctanswers) = 1

3 n = 5

Let X be the number of correct answer

X ∼ BinomialP(x) = ( n

) pxqn−x

P(X ≥ 4) = P(X = 4) + P(X = 5)

( 5

) ( 1

3 )4(

3 )1 +

( 5

) ( 1

3 )5(

3 )0

= 0.0453

Set 7, problem 4

The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probability that there will be

-a. at least 2 such accidents in the next month. b. at most 1 accident in the next month?

Solution: Let X be the number of airplane crashes of commercial airlines. now, since the plane crash is a very rare event, hence x can be modelled by a Poisson distribution.

X ∼ Poisson(λ = 3.5)

P(X = k) = e−λλk

a.P(X ≥ 2) : at least 2 such accidents in next month

P(X ≥ 2) = 1 −P(X < 2)

= 1 − (P(X = 0) + P(X = 1))

= 1 − ( e−3.53.50

0! + e−3.53.51

1! )

= 0.8641

b.at most 1 accident in next month

P(X ≤ 1) = P(X = 0) + P(X = 1)

From question a, we can get the answer = 0.1359

Set 7, problem 5

There are three highways in the county. The number of daily accidents that occur on these highways are Poisson random variables with respective parameters 0.3, 0.5, and 0.7. Find the expected number of accidents that will happen on any of these highways today.

Solution: There is a theorem that we can use to find the expected value of a random variable that is a sum of other random variables as follows. If X = X1 + X1 + · · · + Xk, then E(X) = E(X1) + E(X2) + · · · + E(Xk) In this case, Let X = the number of accidents that will happen on any of those highways today, X1, X2, X3 are the numbers of accidents on each highway, respectively. Then

X = X1 + X2 + X3

Since X1, X2, X3 are Poisson variables, their expected value are the parameters given, 0.3, 0.5, 0.7 So E(X1) = 0.3,E(X2) = 0.5,E(X3) = 0.7

E(X) = E(X1) + E(X2) + E(X3)

= 0.3 + 0.5 + 0.7

= 1.5

Set 8, problem 2

A point is chosen at random on a line segment of length L. Interpret this statement, and find the probability that the ratio of the shorter to the longer segment is less than 1

4 .

Solution: Let X be a continuous uniform random variable taking values between 0 and L. Therefore, its density function is given by

f(x) = 1

L ; 0 ≤ x ≤ L

Let the point X divides the line into two pieces of lengths say X and L−X. According to the ratio, X

L−X < 1 4

and L−X X

< 1 4

iff X > 4L 5

Why do you need this anyway?

P(The ratio of the shorter to the longer segment is less than 1 4 )

= P( X

L−X <

4 ) + P(

L−X X

< 1

4 ) = P(X <

5 ) + P(X >

5 )

∫ L 5

L dx +

∫ 4L 5

L dx

5 +

= 0.4

Set 8, problem 3

You arrive at a bus stop at 10 o’clock, knowing that the bus will arrive at some time uniformly dis- tributed between 10 and 10:30.

a. What is the probability that you will have to wait longer than 10 minutes? b. If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least

an additional 10 minutes?

Solution: Let x denotes the bus will arrive at some time uniformly distributed between 10 and 10 : 30. The probability distribution function of x is,

f(x) = 1

b−a

= 1

30 − 0

= 1

Thus, the probability distribution function of x is,

f(x) = 1

30 , 0 < x < 30

P(x > 10) =

∫ 30 10

f(x) dx

∫ 30 10

30 dx

= 1

30 [x]3010

= 1

30 [30 − 10]

= 0.6667

P(15 < x < 25) =

∫ 25 15

f(x) dx

∫ 25 15

30 dx

= 1

30 [x]2515

= 1

30 [25 − 15]

= 0.3333

The probability in red does not capture the second question. Notice that you are asked to compute a conditional probability here: P(X > 25|X > 15). 2.5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 8, problem 4

The annual rainfall (in inches) in a certain region is normally distributed with µ = 40 and σ = 4. What is the probability that, starting with this year, it will take over 10 years before a year occurs having a rainfall of over 50 inches? What assumptions are you making?

Solution: Let X be normal random variables of the annual rainfall. P(That there will be over 10 years or more before a year with a rainfall above 50 inches)

P(X > 50) = 1 −P(X ≤ 50)

1 −P( X −µ σ

≤ 50 − 40

4 )

= 1 −P(Z ≤ 5

2 )

1 − Φ( 5

2 )

= 1 − 0.9938

= 0.0062

P(the non occurrence of rainfall above 50 inches)

1 − 0.0062 = 0.9938

P(That it will take over 10 years or more of a year with a rainfall above 50 inches) = (0.9938)10 5/5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Set 8, problem 5

The number of years a radio functions is exponentially distributed with parameter λ = 1. If Jones 8 buys a used radio, what is the probability that it will be working after an additional 8 years?

Solution: The exponential distribution is ”the probability distribution of the time between events in a Poisson process”. It is a particular case of the gamma distribution. The probability density function is given by: P(X = x) = λe−λx,x > 0 This is not what a probability density function tells you. Let X be the random variable that represent ”The number of years a radio function”

X ∼ Exp(λ = 1

8 )

Assume that the random variable t represent the number of years that the radio is already here, and the interest is find this probability:

P(X > 8|X > t)

Based on ”Memoryless” property:

P(X > a + t|X > t) = P(X > a)

then P(X > 8|X > t) = P(X > 8 + t|X > t) = P(X > 8)

P(X > 8) =

∫ ∞ 8

8 e−

1 8 x dx

= 1

∫ ∞ 8

e− 1 8 x dx

= [ lim x→∞

(−e− 1 8 x) + e−1]

= 0 + e−1

= e−1

= 0.3679

t is not a random variable but I see what you are arguing here. 5/5

Homework Assignment 5

Set 9, problem 2

A bus travels between the two cities A and B, which are 100 miles apart. If the bus has a breakdown, the distance from the breakdown to city A has a uniform distribution over (0, 100). There is a bus service station in city A, in B, and in the center of the route between A and B. It is suggested that it would be more efficient to have the three stations located 25, 50, and 75 miles, respectively, from A. Do you agree? Why?

Solution: To compare the expected distance to the nearest bus service station when the bus break down. Let Y denote the distance to the nearest bus service station when the bus break. Let S denote the number of nearest bus service station, that is, 1, 2, 3. Case-1 The 3 bus stations numbered 1, 2, 3 are 0, 50 and 100 miles away from A:

E(Y ) = E[(Y |S = 1)P(S = 1)] + E[(Y |S = 2)P(S = 2)] + E[(Y |S = 3)P(S = 3)]

= (

∫ 25 0 |x− 0|dx (

25 − 0 100

)) + (

∫ 75 25 |x− 50|dx (

75 − 25 100

)) + (

∫ 100 75 |x− 100|dx (

100 − 75 100

))

= 12.5

Case-2: The 3 bus service stations numbered 1, 2, 3 are 25, 50 and 100 miles away from A:

E(Y ) = E[(Y |S = 1)P(S = 1)] + E[(Y |S = 2)P(S = 2)] + E[(Y |S = 3)P(S = 3)]

= (

∫ 37.5 0

|x− 25|dx ( 37.5 − 0

100 )) + (

∫ 62.5 37.5

|x− 50|dx ( 62.5 − 37.5

100 )) + (

∫ 100 62.5 |x− 75|dx (

100 − 75 100

))

= 9.98

Hence, the suggested positions are better. 5/5

Set 9, problem 3

A man aiming at a target receives 10 points if his shot is within 1 inch of the target, 5 points if it is between 1 and 3 inches of the target, and 3 points if it is between 3 and 5 inches of the target. Find the expected number of points scored if the distance from the shot to the target is uniformly distributed between 0 and 10.

Solution: Given:

f(x) = 1

10 ; 0 < x < 10

receive 10 points if his shot within 1 inch of the target.

P(0 < x < 1) =

∫ 1 0

10 dx = [

10 ]10 =

10 = 0.1

receive 5 points if in between 1 and 3 inches of the target.

P(1 < x < 3) =

∫ 3 1

10 dx = [

10 ]31 =

10 = 0.2

receive 3 points if in between 3 and 5 inches of the target.

P(3 < x < 5) =

∫ 5 3

10 dx = [

10 ]53 =

10 = 0.2

no points otherwise. probability for this event:

1 − ( 1

10 +

10 ) =

10 = 0.5

E(X) = 0 ∗ 0.5 + 10 ∗ 0.1 + 5 ∗ 0.2 + 3 ∗ 0.2

= 0 + 1 + 1 + 0.6

E(x) = 2.60

Set 9, problem 4

In 10, 000 independent tosses of a coin, the coin landed on heads 5800 times. Is it reasonable to assume that the coin is not fair?

Solution: Given n = 10000

p(probabilityofhead) = 0.5

x = 5800

By using normal approximation to binomial

Mean(µ) = n∗p = 10000 ∗ 0.5 = 5000

standarddeviation(σ) = √ n∗p∗ (1 −p)

= √

10000 ∗ 0.5 ∗ 0.5

= √

2500

= 50

We can get µ = 5000, and σ = 50. we want to know how unusual it is see 5800 heads when a fair coin is flipped 10000 times. That is, we want to compute the probability of observing a value of x at least as extreme as 5800 assuming the coin is indeed fair. we want to compute p value.

P(x ≥ 5800) = p(z ≥ 5799.5 − 5000

50 )

= 1 −p(z ≤ 5799.5 − 5000

50 )

= 1 −p(z ≤ 15.99)

= 1 − 1

= 0

When we use normal approximation to binomial, then we need to use continuity correction.

p(x ≥ a) = p(x ≥ a− 0.5)

p(x ≥ 5800) = p(x ≥ 5800 − 0.5)

= p(x ≥ 5799.5)

That is why we use 5788.5 instead of 5800.

p(x ≥ 5800) = 0

The probability of observing a number of heads at least 5800 is practically 0. so, yes it is very unusual.

The probability is not 0. Also I suggest that you use Chebyshev’s inequality in order to get an estimate. 2/5

Set 9, problem 5

A model for the movement of a stock supposes that if the present price of the stock is s, then, after one period, it will be either us with probability p or ds with probability 1p. Assuming that successive movements are independent, approximate the probability that the stock’s price will be up at least 30% after the next 1000 periods if u = 1.012,d = 0.990,andp = 0.52.

Solution: Let s be the initial of stock. Let X denote the number of increase periods among the 1000 times periods. Then the price at the end is,

s(uX)d1000−X

In order for the price to be at least 1.3s, we need

d1000( u

d )X > 1.3

X > log(1.3) − 1000log(d)

log(u d )

= 469.2

≈ 470

We need at least 470 increases periods. Since, X is binomial with parameters 1000, 0.52 we have

P(X > 469.5) = 1 −P(Z < 469.5)

= 1 −P(Z < X − (1000 ∗ 0.52)√

1000 ∗ 0.52 ∗ (1 − 0.52) )

= 1 −P(Z < 469.5 − 520

15.7987 )

1 −P(Z < −3.1965)

= 1 − 0.0006956

= 0.9993

Hence, the answer is 0.9993

Instructions
Homework Assignment 1
Homework Assignment 2
Homework Assignment 3
Homework Assignment 4
Homework Assignment 5