sta_5325_hw_1_ramin_shamshiri

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Ramin Shamshiri STA 5325, HW #1 Due 05/18/09

STA 5325, Homework #1

Due May 18, 2009

Ramin Shamshiri

UFID # 90213353 Note: Problem numbers are according to the 6

th text edition. Selected problem are highlighted.

Assignment for Monday, May 11th

, 2009

1.4. Acquired immunodeficiency syndrome (AIDS) has become one of the most devastating diseases in modern

society. The number of cases of AIDS (in thousands) reported in 25 major cities in the United States during 1992 are as follows:

38.3 6.2 3.7 2.6 2.1

14.6 5.6 3.7 2.3 2 11.9 5.5 3.4 2.2 2

6.6 4.6 3.1 2.2 1.9 6.3 4.5 2.7 2.1 1.8

a. Construct a relative frequency histogram to describe these data. b. What portion of these cities reported more than 10,000 cases of aids in 1992

c. If one of the cities is selected at random from the 25 for which preceding data were taken, what is the

probability that it will have reported fewer than 3000 cases of AIDS in 1992?

Solution a:

Solution b: 3 out of 25 cities which is 12%

Solution c: Total number of cities less than 3000 cases = 11, so probability = 11/25=0.44 or 44%

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2.2. Suppose that A and B are two events. Write expressions involving unions, intersections, and complements

that describe the following:

a. Both events occur.

b. At least one occurs. c. Neither occurs

d. Exactly one occurs.

Solution a: Both events occur means A AND B, which is intersection, A∩B

Solution b: At least one event occurs means either A OR B, which is union, A∪B

Solution c: Neither occurs means NOT A OR B, which is 𝐴 ∪ 𝐵

Solution d: Exactly one occurs means A AND NOT B , or B AND NOT A, which is A∩𝐵 or B∩𝐴 2.12. A survey classified a large number of adults according to whether they were diagnosed as needing eyeglasses to correct their reading vision and whether they use eyeglasses when reading. The proportions falling

into the four resulting categories are given in the table.

Uses Eyeglasses for Reading

Needs Glasses Yes No

Yes 0.44 0.14

No 0.02 0.40

If a single adult is selected from the large group, find the probabilities of the events defined below:

a. The adult needs glasses.

b. The adult needs glasses but does not use them. c. The adult uses glasses whether the glasses are needed or not.

Solution a: P(needs glasses)= 0.44+0.14=0.58 Solution b: P (adult needs glasses BUT not use them)= 0.14

Solution c: P(use glasses)= 0.44+0.02=0.46

2.15. Hydraulic landing assemblies coming from an aircraft rework facility are each inspected for defects.

Historical records indicate that 8% have defects in shaft only, 6% have defects in bushings only, and 2% have

defects in both shafts and bushings. One of the hydraulic assemblies is selected randomly. What is the probability that the assembly has

a. A bushing defect? b. A shaft or bushing defect?

c. Exactly one of the two types of defects?

d. Neither type of defect?

Solution a: P(bushing defect)=0.06+0.02=0.08

Solution b: P(shaft or bushing)=0.08+0.06+0.02=0.16

Solution c: P(exactly one types of defects)=0.08+0.06=0.14 (both type of defect should not be included) Solution d: P(Neither type)=1-P(defect)=1-0.16=0.84

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2.16. Suppose two balanced coins are tossed and the upper faces are observed.

a. List the sample points for this experiment.

b. Assign a reasonable probability to each sample point. ( Are the sample points equally likely?) c. Let A denote the event that exactly one head is observed and B the even that at least one head is

observed. List the sampling points in both A and B.

d. From your answer to (c) find P(A), P(B), P(A∩B), P(A∪B), and P(A ∪B)

H=Head, B=Back Solution a: Set of all samples points are: S={HB,BH,HH,BB}

Solution b: if the coins are normal, we can assign equal probability to each sample point which is 0.25.

Solution c: A={HB,BH } and B={HB,BH,BB}.

Solution d:

P(A)=2/4

P(B)=3/4

P(A∩B)=1/2

P(A∪B)=3/4

P(A ∪B)=1

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Assignment for Tuesday, May 12th

, 2009

2.21. Two additional jurors are needed to complete a jury for a criminal trial. There are six prospective jurors,

two women and four men. Two jurors are randomly selected from the six available.

a. Define the experiment and describe one sample point. Assume that you need describe only the two

jurors chosen and not the order in which they were selected.

b. List the sample space associated with this experiment. c. What is the probability that both of the jurors selected are women?

Solution a: The experiment is to randomly select two jurors out of a group of 6 jurors that includes two women and 4 men.

Solution b: Let w1 and w2 denotes women and m1, m2 and… denote men.

S={(w1,m1),(w1,m2),(w1,m3),(w1,m4),(w2,m1),(w2,m2),(w2,m3),(w2,m4),(m1,m2),(m1,m3),(m1,m4),

(m2,m3),( m2,m4),( m3,m4),( w1,w2)}

Solution c:

Probability that the first juror is a woman = 2/6

Probability that the second juror is a woman=1/5 Probability that both jurors selected are woman = (2/6)*( 1/5)=1/15

2.27. An airline has six flights from New York to California and seven flights from California to Hawaii per

day. If the flights are to be made on separate days, how many different flight arrangements can the airline offer

from New York to Hawaii?

Solution:

Using m.n rule, there would be 6*7=42 possible ways to fly from NY to CA.

2.33. How many different seven-digit telephone numbers can be formed if the first digit cannot be zero?

Solution:

1 2 3 4 5 6 7

9 10 10 10 10 10 10

There would be 9*106 possible phone numbers with seven digits.

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2.44. A group of three undergraduate and five graduate students are available to fill certain student government

posts. If four students are to be randomly selected from this group, find the probability that exactly two undergraduates will be among the four chosen.

Solution:

Combination of 4 out of 8,

𝑛

𝑟 =

8

4 =

8!

4! 4!= 70

Number of ways that exactly two undergrad students can be selected:

3

2 =

3!

2!= 3

5

2 =

5!

2! 3!= 10

P(exactly two undergrad students):

3 × 10

70=

3

7

2.50. A balanced die is tossed six times and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1,2,3,4,5 and 6 in any order?

Solution: 1 2 3 4 5 6

1 5/6 4/6 3/6 2/6 1/6

Probability that the numbers recorded are 1,2,3,4,5 or 6 in any order is

(1)(5/6)(4/6)(3/6)(2/6)(1/6)=5!

65 =120

7776= 0.01543209

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Assignment for Wednesday, May 13th

, 2009

2.57. If two events, A and B, are such that P(A)=0.5, P(B)=0.3 and P(A∩B)-0.1, find the following:

a. P(A׀B) b. P(B׀A)

c. P(A׀A∪B)

d. P(A׀A∩B)

e. P(A∩B׀A∪B)

Solution a:

𝑃 𝐴׀𝐵 =𝑃 𝐴 ∩ 𝐵

𝑃 𝐵 =

0.1

0.3=

1

3

Solution b:

𝑃 𝐵׀𝐴 =𝑃 𝐵 ∩ 𝐴

𝑃 𝐴 =

0.1

0.5=

1

5

Solution c:

𝑃 A׀A ∪ B =𝑃 𝐴 ∩ A ∪ B

𝑃 A ∪ B

=𝑃 𝐴 ∩ 𝐴 ∪ 𝐴 ∩ 𝐵

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

=𝑃 𝐴 ∪ 𝐴 ∩ 𝐵

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

=𝑃 𝐴 + 𝑃 𝐴 ∩ 𝐵 − 𝑃(𝐴 ∩ (𝐴 ∩ 𝐵))

𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

=𝑃 𝐴

𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)=

0.5

0.7=

5

7

Solution d:

𝑃 A׀A ∩ B =𝑃 𝐴 ∩ A ∩ B

𝑃 A ∩ B

=𝑃(A ∩ B)

𝑃(A ∩ B)= 1

Solution e:

𝑃 A ∩ B׀A ∪ B =𝑃 A ∩ B ∩ (A ∪ B)

𝑃 A ∪ B

=𝑃[( A ∩ B ∩ A ) ∪ ( A ∩ B ∩ B)]

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B

=𝑃[(A ∩ B) ∪ (A ∩ B)]

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B

=𝑃[(A ∩ B)

𝑃 𝐴 + 𝑃 𝐵 − 𝑃 A ∩ B =

0.1

0.7=

1

7

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2.60. A survey of consumers in a particular community shows that 10% were dissatisfied with plumbing jobs done in their home. Half the complaints dealt with plumber A, who does 40% of the plumbing jobs in the town.

a. Find the probability that a consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.

b. Find the probability that a consumer will obtain a satisfactory plumbing job, given that the plumber was

A.

Solution a:

P(A)= Probability that plumber A does the job is 40%

P(B)=Probability that consumers are dissatisfied is 10%

Half the complains dealt with plumber A, means probability that plumber A does the job given the probability

that consumers are dissatisfied, So P(A׀B)=50%

Finding the probability that a consumer will obtain an unsatisfactory plumbing job, given that the plumber was

A, is

𝑃 𝐵׀𝐴 =𝑃 𝐵 ∩ 𝐴

𝑃 𝐴 =

𝑃 P(A׀B) .𝑃(𝐵)

0.4=

0.5(0.1)

0.4=

0.5

4= 0.125

Solution b:

𝑃 𝐵 ׀𝐴 = 1 − 𝑃 𝐵׀𝐴 = 1 − 0.125 = 0.875

2.65. If P(A)>0, P(B)>0, and P(A)<P(A׀B), show that P(B)<P(B׀A).

Solution:

If P(A)<P(A׀B) then we have: 𝑃 𝐴 <𝑃 𝐴∩𝐵

𝑃 𝐵

Or 𝑃 𝐵 <𝑃 𝐴∩𝐵

𝑃 𝐴

This is equivalent as: 𝑃 𝐵 <𝑃 𝐵∩𝐴

𝑃 𝐴 , or 𝑃 𝐵 < P(B׀A)

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2.69. In s game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, H or miss,

M. The game requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand

for the third. Her chance of scoring a hit with her right hand is 0.7 and with her left hand is 0.4. Assume that the

results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?

Solution:

RIGHT LEFT RIGHT

H H M (0.7)(0.4)(0.3)=0.084

M H H (0.3)(0.4)(0.7)=0.084 H H H (0.7)(0.4)(0.7)=0.196

0.084+0.084+0.196=0.364

2.73. Consider the following portion of an electric circuit with three relays. Current will flow from point a to

point b if there is at least one closed path when the relays are activated. The relays may malfunction and not

close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of 0.9

a. What is the probability that current will flow when the relays are activated? b. Given that current flowed when the relays were activated, what is the probability that relay 1

functioned?

Solution a:

Probability of closing properly for each relay is 90%, therefore probability of not closing is 1-0.9=0.1

So the probability that current will flow when relays are activated is 1-P(not closing 3 relays) which is:

1-(0.1*0.1*0.1)=1-(0.001)=0.999

Solution b:

Probability of current flowed=0.999 Probability of relay 1 functioned if current flowed is:

𝑃 𝐴׀𝐵 =𝑃 𝐴 ∩ 𝐵

0.999=

0.9

0.999= 0.9009

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Assignment for Thursday, May 14th

, 2009

2.71. Two events A and B are such that P(A)=0.2, P(B)=0.3 and 𝑃 𝐴 ∪ 𝐵 = 0.4. Find the following:

a. 𝑃 𝐴 ∩ 𝐵 b. 𝑃 𝐴 ∪ 𝐵 c. 𝑃 𝐴 ∩ 𝐵

d. 𝑃 𝐴 ׀𝐵

Solution a: 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵 𝑃 𝐴 ∩ 𝐵 = 0.2 + 0.3 − 0.4 = 0.1

Solution b: 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵 = 1 − 𝑃 𝐴 + 1 − 𝑃 𝐵 − (1 − 𝑃 𝐴 ∪ 𝐵 )

= 0.8 + 0.7 − 0.6 = 0.9

Solution c: 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∪ 𝐵 = 1 − 𝑃 𝐴 ∪ 𝐵 = 1 − 0.4 = 0.6

Solution d:

𝑃 𝐴 ׀𝐵 =𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

0.3 − 0.1

0.3=

2

3

2.87. An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 5 sees a corresponding ad on television. One in 100 sees both. One in 3 actually

purchases the product after seeing the ad, 1 in 10 without seeing it. What is the probability that a randomly

selected potential customer will purchase the product?

Solution:

P(magazine ad)=P(A)=1/50

P(TV add)=P(B)=1/5

P(A∩B)=1/100

P(ad)=P(A∪B)=1/50+1/5-1/100=21/100

P(NOT ad)=P(NOT A AND NOT B)=P(𝐴 ∩ 𝐵 )=1- P(A∪B)=79/100

P(buy after ad)=P(buy׀ ad)=1/3=𝑃(buy ∩𝑎𝑑 )

𝑃(𝑎𝑑 )

So, 𝑃 buy ∩ 𝑎𝑑 =1

3.

21

100=

7

100

P(buy׀ NOT ad)=1/10=𝑃(buy ∩𝑁𝑂𝑇 𝑎𝑑 )

𝑃(𝑁𝑂𝑇 𝑎𝑑 )

So, 𝑃 buy ∩𝑁𝑂𝑇 𝑎𝑑 =79

100.

1

10=

79

1000

P(buy)= 𝑃 buy ∩ 𝑎𝑑 + 𝑃 buy ∩𝑁𝑂𝑇 𝑎𝑑 =7/100+79/1000=0.149

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2.88. Three radar sets, operating independently, are set to detect any aircraft flying through a certain area. Each set has a probability of 0.02 of failing to detect a plane in its area.

a. If an aircraft enters the area, what is the probability that it goes undetected? b. If an aircraft enters the area, what is the probability that it is detected by all three radar sets?

Solution a: (0.02)( 0.02)( 0.02)=8*10-6

Solution b: (0.98)( 0.98)( 0.98)=0.941192

2.90. A lie detector will shows a positive reading (indicate a lie) 10% of the time when a person is telling the

truth and 95% of the time when the person is lying. Suppose two people are suspected in a one-person crime

and (for certain) one is guilty and will lie. Assume further that the lie detector operates independently for the truthful person and the liar.

a. What is the probability that the detector shows a positive reading for both suspects?

b. What is the probability that the detector shows a positive reading for the guilty suspect and a negative reading for the innocent suspect?

c. What is the probability that the detector is completely wrong-that is, that it gives a positive reading for

the innocent suspect and a negative reading for the guilty? d. What is the probability that it gives a positive reading for either or both of the two suspects?

Solution:

From the problem, 10% of times that the person is really saying the truth, the lie detector shows lie which is

Positive. So 90% of times that a person is really saying truth, the lie detector gives Negative.

Also, 95% of the time that a person is saying lying, the lie detector shows lie which is positive and 5% of times that a person is lying, the lie detector shows Negative.

For each person, we have four cases: 1- He is saying the truth and lie detector shows it as a lie (Positive) : 10%

2- He is saying the truth and lie detector shows it as a truth (Negtive): 90%

3- He is lying and lie detector shows it as a lie (Positive) :95%

4- He is lying and lie detector shows it as a truth (Negative): 5%

Bases on this, we can answer the questions:

Solution a: (0.1)(0.95)=0.095

Solution b: (0.9)(0.95)=0.855

Solution c: (0.1)(0.05)=0.005 Solution d: (0.1)+(0.95)-(0.1)(0.95)=0.955

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2.104. A study of Georgia residents suggests that those who worked in shipyards during World War II were subjected to a significantly higher risk of lung cancer. It was found that approximately 22% of those persons

who had lung cancer worked at some prior time in a shipyard. In contrast, only 14% of those who had no lung

cancer worked at some prior time in a shipyard. Suppose that the proportion of all Georgians living during World War II who have or will have contracted lung cancer is 0.04%. Find the percentage of Georgians living

during the same period who will contract (or have contracted) lung cancer, given that they have at some prior

time worked in shipyard.

Solution:

Pr(Cancer)=P(C)=0.0004

P(𝐶 )=0.9996 Pr(work given cancer)=P(B׀C)=0.22

Pr(work given NOT cancer)=P(B׀𝐶 )=0.14 What is the probability of Cancer given work, P(C ׀B)=?

𝑃 C׀B =𝑃 𝐶 ∩ 𝐵

𝑃 𝐵 =

𝑃 𝐶 .𝑃 B׀C

𝑃 𝐶 ∩ 𝐵 + 𝐶 ∩ 𝐵

=𝑃 𝐶 .𝑃 B׀C

𝑃 𝐶 . B׀C + 𝐶 . B׀C =

0.0004 0.22

0.0004 0.22 + 0.9996 0.14 = 0.0006

2.111. A student answers a multiple-choice examination question that offers four possible answers. Suppose

that the probability that the student knows the answer to the question is 0.8 and the probability that the student will guess is 0.2. Assume that if the student guesses, the probability of selecting the correct answer is 0.25. If

the student correctly answers a question, what is the probability that the student really knew the correct answer?

Solution:

Pr(guess)=P(A)=0.2

P(know)=P(B)=0.8 P(correct)=P(C)

P(C׀A)=0.25

P(C׀B)=1

P(C)= (0.8)(1)+(0.2)(0.25)=0.85

P(B׀C)=P(NOT guess ׀ C)

=P(𝐴 ׀C)

=𝑃 𝐴 ∩ 𝐶

𝑃 𝐶 =

𝑃 𝐶 ׀𝐴 𝑃 𝐴

0.85=

1 0.8

0.85= 0.9412

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Assignment for Friday, May 15th

, 2009

3.1. When the health department tested private wells in a county for two impurities commonly found in drinking

water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B.

(Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well.

Solution:

𝑃 𝐴 ∩ 𝐵 = 0.2

𝑃 𝐴 = 0.4

𝑃 𝐵 = 0.5

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∪ 𝐵 => 0.2 = 1 − 𝑃 𝐴 ∪ 𝐵 Or 𝑃 𝐴 ∪ 𝐵 = 0.8

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∪ 𝐵 = 0.5 + 0.4 − 0.8 = 0.1 P(Y=0)=0.2

P(Y=2)=0.1

P(Y=1)=0.7, (since P(Y) should sum to 1 over all Ys)

3.4. Consider a system of water flowing through valves from A to B. Valves 1,2, and 3 operate independently, and each correctly opens on signal with probability 0.8. Find the probability distribution for Y, the number of

open paths from A to B after the signal is given. (Note that Y can take on the values 0,1, and 2.)

Solution:

P(each valve open)=0.8

P(all valve open correctly)==> there would be two paths from A to B, or P(Y=2)= (0.8)(0.8)(0.8)=0.512 P(valve 1 does not open AND either valve 2 OR 3 does not open)

=P(Not openV1∩ (𝑁𝑜𝑡 𝑜𝑝𝑒𝑛𝑉2 ∪𝑁𝑜𝑡 𝑂𝑝𝑒𝑛𝑉3))

=P(Not open V1)*P(𝑁𝑜𝑡 𝑂𝑝𝑒𝑛 𝑉2 ∪ 𝑁𝑜𝑡 𝑂𝑝𝑒𝑛 𝑉3)=0.2*(0.2+0.2-0.04)=0.072 So, P(Y=0)=0.072

Consequently P(Y=1)=1-(0.512+0.072)=0.416

3.10. Let Y be a random variable with p(y) given in the accompanying table. Find E(Y), E(1/Y), E(Y2-1), and

V(Y).

y 1 2 3 4

P(y) 0.4 0.3 0.2 0.1

Solution:

𝐸 𝑌 = 𝑦𝑝 𝑦 = 0.4 + 0.6 + 0.6 + 0.4 = 2

𝐸 1

𝑌 =

1

𝑦𝑝 𝑦 = 0.4 + 0.15 + 0.2/3 + 0.1/4 = 0.6417

𝐸 Y2 − 1 = 𝐸 𝑌2 − 1 = 0.4 + 4 × 0.3 + 9 × 0.2 + 16 × 0.1 − 1 = 4

𝑉 𝑌 = 𝐸 𝑌2 − 𝐸 𝑌 2

= 5 − 22 = 1

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3.21. A potential customer for an $85,000 fire insurance policy possesses a home in an area that according to

experience, may sustain a total loss in a given year with probability of 0.001 and 50% loss with probability 0.01. Ignoring all other partial losses, what premium should the insurance company charge for a year policy in

order to break even on all $85,000 policies in this area?

Solution:

P(total loss)=0.001

P(50% loss)=0.01

P(0% loss)=1-(0.001-0.01)=0.989

𝐸 𝑌 = 𝑦𝑝 𝑦 = 85000 0.001 + 0.5 ∗ 85000 0.01 + 0 ∗ 85000 0.989 = 510

3.23. Let Y be a discrete random variable with mean 𝜇 and variance 𝜎2. If a and b are constants, use Theorem

3.3 through 3.6 to prove that:

a) E(aY+b)=aE(Y)+b=a 𝜇+b

b) V(aY+b)=a2V(Y)=a

2𝜎2

Solution a:

Let g1(Y)=aY and g2(Y)=b

E(aY+b)=E[g1(Y)+ g2(Y)]=E(g1(Y))+E(g2(Y))=E(aY)+E(b)=aE(Y)+E(b)= a 𝜇+b

Solution b:

V(aY+b)=E[Ay+b-( a 𝜇+b)]2=E[aY-a 𝜇+b-b]

2=E[a(Y- 𝜇)]2

=E[a2(Y- 𝜇)

2]

=a2E(Y- 𝜇)

2

=a2V(Y)

=a2𝜎2

sta_5325_hw_1_ramin_shamshiri

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