# data[].attributes.body
Group leader: Nur Sara Syuhada
Team members:
1. Afifah
2. Aisyah Sumayyah
3. Nur Sabrina

(a) The PMF utilized is the same as random experiment 5 because the question states that the transmission continues until the receiving modem makes its first fault. X can be regarded as a negative binomial random variable because it has two possible outcomes: 0 or 1.

(b) We need to find the probability for at least 10 bits, hence P(X≥10) = 1-P(X<10) for X=1,2,3,4,5,6,7,8,9

# data[].attributes.body
Team members:
1.Norsaidatulain binti Asral
2.Farah Amira
3. Noramalia

a. Let X be the number of bits transmitted until the random makes its first error.First error is include at the number of bits transmitted. The experiment will stop when the number of error occur is meet the requirement.X is a geometric random variable.

b. The probability that at least 10 bits will be transmitted, P(X>=10). p=10
P(X>=10)=1-P(X<10)=0.3874

# data[].attributes.body
a)       Since the modem transmit is an independent event then P (a and b) = P(a)P(b). let P(a) equal to probability to make error, y = number of events to occurs than a= p^y, while for condition only make one error therefor y=1, a = p. Let P(b) equal to probability to not make error than a= (1-p) ^y. Assume x is number of bits transmitted and reject the situation modem make error in first transmit, we able to conclude y= x-1.  Therefor f(x) = p(1-p) ^(x-1) for x >=1, otherwise 0.

b)      Let p =0.1, P(x>=10) = 1- P (x < 10) = 1 – 0.6126 = 0.3874

c)       Let p = probability to make error, then similar concept in a , P(a)= p ^y ,P(b)= (1-p)^(100-y) , however we need to consider the combination situation , therefor f(y)= 100Cy(p^y)( (1-p)^(100-y)), otherwise 0 .

d)      By apply the pmf in c, P(y=2) = 0.1849

e)      Similar concept in a but take note there is 2 error in first 11 number of bits transmitted, f (12) = (11C2) *(0.25^3) (1-0.25) ^ (12-3) = 0.0645

# data[].attributes.body
1. Nurul Aqilah Bt Saiful Bahril
2. Ummi Fakhriah Bt Mohd Noor
3. Nur Diana Shafiah Bt Jasin

a) Since X is the number of bits transmitted where it is detected from its first error, and the transmitted bit is either 0 or 1,  X is a geometric random variable (negative binomial). Hence, the PMF is similar to random experiment 5.

c) The PMF of X, the number of errors observed from a random sample of 100 transmitted bits  is similar to random experiment 4.

d) Suppose Y be the random variables that counts the error at the receiver with 2 errors (y=2), p=0.01 and modem transmit on  100 bits. Thus, the probability of Y is similar to random experiment 2.

# data[].attributes.body
Team members:
1. Nur Syaurah Binti Daud
2. Nurfarhaniza Binti Ramlee
3. Nurul Ika Mirlia Binti Zarimi
4. Salwa Alhana Binti Mohd Sofian

# data[].attributes.body
1. Nurjannatul Najwa (leader)
2. Nurul Afrina
3.Nurhazira
4. Nurin Hidayah
5. Nursyafiqah Nabilah

a) Let X be the number of bit transmitted until modem receive the first error. The number of bits transmitted is including the first error.
X can be described as negative binomial random variables as number of outcomes is 0 and 1. The success event here is 1 and the experiment is now stops when the transmission modem make its first error.

b) p=0.1, probability that >=10 transmitted

P(X >= 10) = 1-P(X < 10)
= 1- P(X=9) - P(X=8) - ... - P(X=0)
= 0.3874

c) modem transmits = 100

Solve the problem using pmf formula random experiment 4

# data[].attributes.body
Team members:
1. Jasmine Wang Thye Wei
2. Oh Sing Yee
3. Lew Poh Chyi
4. Lwi Wen Ling

a. X can be described as negative binomial random variable as number of outcomes is only 2 which is either 0 or 1. The success event here is 1 and the experiment now stops dependent on the number of success events required. The probability of success is constant throughout the trials. The number of trials is a random variable.

b. Discrete
X~G(0.1)
P(X>= 10)= 1 - P(X<10)
= 1 - 0.1 (0.9^0 + 0.9^1 + 0.9^2
+...+ 0.9^8)
= 0.3874

c. The number of modem transmit is 100 bits which is the number of trials. Hence, if Y is the random variable that counts the number of error, Y can be describe as binomial random variable as number of outcomes is either 0 or 1. The success event here is 0 and it is constant throughout the 100 trials.

d. Y~b(100, 0.01)
P(Y=2) = 100C2 (0.01)^2 (0.99)^98
= 0.1849

e. X~NB(3, 0.25)
f(12)=11C2 (0.25)^3 (0.75)^9
= 0.0645

# data[].attributes.body
Team members:
1. Ilyana Najihah (Leader)
2. Nurfarah Hazira
3. Nurin Ainun Miza
4. Siti Aisyah

a) Since the question mentions that the transmission continues until receiving modem makes its first error, the PMF used is the same as random experiment 5. X can be described as geometric since it gives two outcomes which are 0 or 1.

b) We want to find the probability that at least 10 bits will be transmitted, where P(X>=10) with the given value p=0.1. We can find the probability by using 1-P(X<10). Thus, 1-[P(X=1)+P(X=2)+…+P(X=9)].

# data[].attributes.body
Team members :
Muhammad syazwan (Leader)
Nur izati
Nur hazreena

a) For question (a) the transmission continue until the modem receive the first error. X be a random variable that count the number of transmission.  X can be described as negative binomial random variable as number of outcomes is only two which is either 0 or 1. PMF of the question same as random experiment 5

b)  P(x≥10) = 1 - P(x<10)
P(X=x) = nCx (1-p)^(n-x) p^x
= 10C1 (1-0.1)^(10-1) (0.1)^(1)
= 0.3874

# data[].attributes.headline
 Khairunnas, Muhammad Fikri, Badrul Amin.

# data[].attributes.subject
 Khairunnas, Muhammad Fikri, Badrul Amin.

# data[].attributes.body
Question (a):
Let X be the number of transmitted, p is the probability of the error receiving modem, x counts the number of transmitted. X is a geometric random variable, the probability will be in a range [0,1]. The possibility will be two, which is if the x is lower than 1, the value will be 0 and more than 1 the value shown.

Question (c):
Given X is the number of errors in the random sample of 100 transmitted bits. The success event is Y where Y is the pmf . The probability is only 2 possibilities which are 0 or range 0<y<1.

# data[].attributes.body
TEAM MEMBERS:
1) SIVANES CHANDRAN (LEADER)
2) THARISHINI RAJA

Question (a)

Let X be the random variable that counts the number of bits transmitted counted the first error made which also described as negative binomial. k is equal to 1 since the transmission stop when the first error made. Thus, the probability mass function is same as random experiment 5.

Question (b)

P(X ≥ 10) = 1-P(X<10)

= 1-0.1(1+0.9+

= 0.3874

Question (d)

p=0.01, y=2, n=100

P(Y=y) = nCy

P(Y=2) = 100C2

= 0.1849

Question (e)

X =12, p=0.25 and k=3.

f(12) = (x-1)C(k-1)(

= (11C2)(

= 0.0645

# data[].attributes.body
team members :
1. juwariah (leader)
2. nabilah farhana
3. nurul anis suraya
4. nurul hadirah

a) Let X be the number of bits transmitted. X can be described as negative binomial random variable as number of outcome is only two which the transmitted bit is either 0 or 1. The pmf is similar  as the random experiment 5.

b)  the question mention that at least 10 bits will be transmitted, so we can use the formula
P[x>=10]=1-p[x<10]

# data[].attributes.body
Team members:

1.Noor Hafizah binti Mohd Yusoof

2.Nur Farah Alia binti Zahrullail

3.Nur Irdina binti Salim

Discussion 2.3 (b)

From the discussion question b, p=0.1 and value of probability at least 10 bits.

Solution:

From that we can find the probability whereas
P[X>= 10]
1-P[X<10]

= 1-(P (X=0) + P (X=1) + P(X=2)  + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)  + P(X=8) +  P(X=9)).

Hence, we will get the answer 0.3874.

# data[].attributes.body
Team Members :
Chong Ee Xuan
Khoo Xu Yi
Tan Theng Joe
Tan Zhi Qing

a) Let X be there number of bits transmitted until the receiving modem makes its first error. The number of bits transmitted counted including the 1st error. The situation is same as random experiment 5 where the experiment will stop when the number of error occur is meet the requirement.

b) P(X >=10) = 1- P(X <10)
=1-0.1 (1+0.9+0.9^2+0.9^3+0.9^4+0.9^5+0.9^6+0.9^7+0.9^8）
=0.3874

# data[].attributes.body
Members:
Vini Bong Ee Woon
Vivian Sylvia Ting
Yeoh Yen Ping

Discussion:
a) Let X be the random variable that counts the number of bits transmitted until the first error is made. The number of bits transmitted counted including the first error made. X can be described as negative binomial random variable. The experiment will stop when the first error is made. The probability mass function is the same as in the random experiment 5.

b)Since p=0.1 and the number of bits transmitted, x >= 10, we can substitute the values into the function from a) to get the probability.

P(x>=10) =1-P(x<10)

=1-[f(1)+f(2)+...+f(9)]

=1-[0.1+0.09+0.081+0.0729+0.0656+0.0590+0.0531+0.0478+0.043]

=0.3874

# data[].attributes.body
Group leader: Nurul Aisyah
Team Members:
1. Nur A'ina Farahah
2. Nor Farzana
3. Nur Alya Aisyah

d) We want to know the probability of two errors occur at the receiver in 100 transmission. Given p=0.01 and we can use the pmf in experiment 2 to find the probability of two errors occur throughout 100 transmission. The pmf is denoted as:
P(Y=y)=nCy(1-p)^(n-y)p^y
Let n=100, y=2 and p=0.01. Hence,
P(Y=2)=100C2(1-0.01)^(100-2)•(0.01)^2
=0.1849

e) The question ask the probability of 12 bits transmitted if the transmission continues until the receiving modem makes 3 errors? In that case, X is a negative binomial distribution as  p=0.25 and k=3. The number of bits transmitted is 12.
Solution:
11C2 x (.25)^3 x (1-0.25)^9
=0.0645

# data[].attributes.body
Team members :

1) Nurul Ain Najihah (Leader)

2) Nurul Hanisah

3) Farah Hanis

4) Farrah Qhairunniza

e) If the transmission continues until the receiving modem makes three errors, what is the probability of 12 bits transmitted? P=0.25

Answer :

From the formula given, (x-1)C(k-1)((p)^k)((1-p)^(x-k)) where

x = the number of trials

k = number of errors

p = the probability given

The value of x=12, k=3, p=0.25.

By using the formula (12-1)C(3-1)((0.25)^3)((1-0.25)^(12-3)). The answer is 0.0645.

# data[].attributes.headline
Lui Dickson(L), Lee Wei Cong, Wong Ying Beng

# data[].attributes.subject
Lui Dickson(L), Lee Wei Cong, Wong Ying Beng

# data[].attributes.body
Group leader: Lui Dickson

Team members:
1.Lee Wei Cong
2.Wong Ying Beng

Discussion question (e):
From the question e, we can know that the solution is 11C2 x 0.25^3 x (1-0.25)^9 = 0.0645. It is matched with random experiment.

Let X be the random variable that counts the number of bits transmitted until the receiving modem makes 3 errors.

X can be described as negative binomial random variable as number of outcomes is only two which is either 0 or 1 (signal that arrives). The experiment stops dependent on the number of errors requires which is 3 errors. The number of trials is now a random variable.

# data[].attributes.body
If the modem transmits 100 bits, what is the PMF of Y, the numbers of errors?

Let Y as the number of error then PMF of Y if the X=100 is

Group leader- Fazlin
Team members-
1. Nur Afifah
2. Nur Aliazahra
3. Maryam Abidah Hanim

# data[].attributes.body
Group leader: Nabil
Members:
1. Naqiuddin
2. Firdaus
3. Rasman

In question (a), let X be the number of bits transmitted, p is the probability of error that was made by modem, x counts the number of bits transmitted. X is a geometric random variable since X since it only give us 2 possible answers, which are 0 and 1.

# data[].attributes.body
Discuss in Group Quiz 2.3 (Yates and Goodman). Write down the process of understanding this exercise. You can select an expert in your group or you can have a brainstorming session to understand the answer and guide that has been given in the lecture presentation slides.

# data[].attributes.body
Team Members :
1) Aiman Syamimi
2) Nur Batrisyia
3) Siti Hajar
4) Sharifah Azira Nazrah

a)  let X be the number of bits transmitted, p is the probability of error that was made by modem, x counts the number of bits transmitted. Since X is a geometric random variable, the probability is in the range of [0,1] if the x is larger than 1 if else it will return a value 0.

b) From the question, it state that the probability that at least 10 bits will be transmitted where P(X>=10). From that, we can find the probability where 1 - P(X<10). where P(X<10)= P(X=0) + P(X=1) + P(X=2) + P(X=3) +P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9).

c)  Let X be the number of errors in random sample of 100 transmitted bits. X can be described as binomial random variable as number of outcomes is only two which is either 0 or 1. The success event here is error.

# data[].attributes.body
Team members:
1. Muhammad Asri(ketua)
2. Muhammad Rafiq
3. Siti Rokaiyah
4. Iffa Syazweena
Discussion for question 2.3(d):

From the question, given the value of p = 0.01 and the modem transmits 100 bits. The question ask what is the probability of Y=2 errors at the receiver. From the given important notes given, we can denote the question as the formula given:
nCr(p)^r(1-p)^n-r
where n is the number of trials, r is the number of error and p is the probability given.
Therefore we use information given to solve the question and the formula become
100C2(0.01)^2(1-0.01)^100-2
and the answer that we get is 0.1849

# data[].attributes.body
Group leader: Chan Chong Lee
Team members:
1.Gan Chun Hong
2.Lee Chi Poh
3.Lim Chau Jie
4.Ng Ching Kok

a) Since the question mentions the transmission continues until the receiving modem makes its first error， therefore we can catogory the question as random experiment 5. Thus the PMF is equal to random experiment 5.

b) The question mentions that at least 10 bits will be transmitted, thus x is greater or equal than 10. We will use 1 - P(X<10) to find the probabilty.

c. The question mentions that 100 bits have been transmitted, and assume x as be the random variable that counts the number of errors. The success event here is error. It is similar to random experiment 2.

d) p=0.01, and 100bits, y=2, therefore we substitute the value into the pmf of question c to get the answer.

e) The question is mention until 3 errors, it can be catogory in random experiment 4. Besides that, we can get x and k from the question where x=12, k=3 and p=0.25. We substitute the value insides the formula and get the answer.

# data[].attributes.headline
AYU FARHANA, FIRDA HARIRI, NUR'ATIFAH NADHIRAH

# data[].attributes.subject
AYU FARHANA, FIRDA HARIRI, NUR'ATIFAH NADHIRAH

# data[].attributes.body
Question (a) explains that the success event is 1 (until makes an error) and the experiment now stops dependent on the number of success events required. Plus, here we wanted to find the first bit so we use random experiment 5.

Question (b): We want to find the probability that at least 10 bits will be transmitted with given value p = 0.1. We can solve the question by using nCx(1-p)^(n-x)p^x where n=10, x=1 (success). It will become: P(X=1) = 10C1(1-0.1)^(10-1)(0.1) =0.3874

Question (c): Let X be the numbers of error in a random sample of 100 transmitted bits. X can be described as binomial random variable as number of outcomes is only two which is either 0 or 1 bit. The success event here is 0.

# data[].attributes.body
Team Members:
1. Nuraina Aqilah
2. Nur Auni Aisyah
3. Ayu Nur Athirah
4. Iffah Ufairah

Question (a)

Let X be the random variable that counts the number of bits transmitted until the first error is observed.

X can be described as geometric random variable as number of outcomes is only two which is either 0 or 1 . The success event here is 1 and the experiment now stops dependent on the number of success events required. The probability of success is constant throughout the trials. It makes an error with probability p, independent of whether any other is received correctly. The number of trials is now a random variable.  It's PMF as given.

Question (b)
p=0.1,
P( x >= 10) = 1 - P ( x <10)
Thus,
1 - (P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) + P ( x = 4 ) + P ( x = 5 ) + P ( x = 6 ) + P ( x = 7 ) + P ( x = 8 )
+ P ( x = 9 )
=0.3874
Since the probability that need to find is at least 10 bit transmitted so it can be calculate by finding the probability less than 10.

Question (c)

Let X be the number of errors in a random of 100 bits transmitted.

X can be described as binomial random variable as number of outcomes is only two which is either 0 or 1.

Its pmf
Y ~ b(n =100, p) f(y) = 100Cy p y (1− p) n−y ; x = 0,1,...,n 0 ; otherwise

Question (d)

substitute y=2 in c since it is mention that the error is 2.

# data[].attributes.body
Team Members:
1. Nur Najeeha Natasha binti Jefri (leader)
2. Nur Husna Amierah binti Mohd Zaperi
3. Che Engku Nur Syafiyyah binti Che Engku Dalam
4. Nabilla Huda binti Hairulnizam

a)
-value of random variable is 0 and 1 only
- error probability, p
a) transmission continue until modem makes first error
(hint: refer to the Random Experiment 5 in the slides)

Solution:
X~G(p)
f(x)= { p * (1-p) ^ (x-1); x>= 1}
{0          ; otherwise}

b) p=0.1 and value of probability at least 10 bits

Solution:
P( x >= 10) = 1 - P ( x <10)
Thus,
1 - (P ( x = 0 ) + P ( x = 1 ) + P ( x = 2 ) + P ( x = 3 ) + P ( x = 4 ) + P ( x = 5 ) + P ( x = 6 ) + P ( x = 7 ) + P ( x = 8 )
+ P ( x = 9 )

c) n = 100 ; pmf ?

Solution:
Since it transmits 100 bits of the bit whether 0 and 1, it is consider as binomial

So, the pmf of binomial is
P(X = x) = nCx (1-p)^ (n-x) * (p)^ (x)

d) p=0.01 ; n = 100 ;  P ( Y = 2)

Solution:
Since
The value of Y is given, we just substitute the value in the binomial random variable formula

e) makes 3 errors ; X = 12 ; p = 0.25

Solution:
We can use the negative binomial since the question ask for the probability when the error appear 3 times

f(12) = (11 C 2 )* ( 0.25)^ (3) * (1-0.25)^ (12-3) = 0.0645

# data[].attributes.body
Team members:
1. Khor Hui Jing
2. Kwek Yu Chen
3. Wong Xiang Yi

a)X is the geometric(negative binomial) random variable that counts the number of bits transmitted up to and including the 1st error made. The number of outcomes is either 0 or 1. Since the transmission of bit continues until the modem makes its 1st error, k is equal to 1 here. Thus, the pmf of X is as shown by Random Experiment 5.

b) Probability that at least 10 bits will be transmitted same with P(X≥10).
P(X≥10)
=1-P(X<10)
=1-0.1(1+0.9+0.9^2+...+0.9^8)
=0.3874

c)Given modem transmits=100bits
Y=binomial
So, using pmf formula as random experiment 4 to solve the problem.

d)Substitute p=0.01, y=2 and n=100 bits into the pmf of question c :  P(2 errors)=P(Y=2)=100C2*(0.01)^2*(1-0.01)^(100-2)=0.1849

e) Given k=3 (makes 3 errors); X=12; p=0.25
Case: Random experiment 4
f(12) = (11 C 2 )* (0.25^3) * (1-0.25)^ (12-3) = 0.0645

# data[].attributes.body
Team Members:
1. Muhammad Zuhairi bin Abu Nangim (Leader)
2. Nurul Haidah binti Saidon
3. Nurul Najiha binti Fadzil
4. Nurul 'Ain Nabila binti Abd Kadir

a)

Let X be the random variable that count the number of transmission until 1st error is observed. The number of transmission counted including the 1st error observed.

So X can be described as negative binomial random variable as the random outcomes is only two which is 0 or 1. The success event here is 0 and the experiment now stops depending on the number of success events required. The probability of success is constant throughout the trials. The number of trials is now random variable.

b)

We want to find the probability that at least 10 bits will be transmitted with given value p = 0.1. It is similar to the random experiment 2 where event here is 1. The probability of success is constant throughout the fixed trials. In this question, we can say that n=10, x=10 (success). It will become: P(X=1) = 10C1(1-0.1)^(10-1)(0.1)=0.3874

c)

It is said that the number of modem transmit is 100 bits which is the number of trials. Hence, if Y is the random variable that counts the number of error, Y can be describe as binomial random variable as number of outcomes is either 0 or 1. The success event here is 0 and it is constant throughout the 100 trials.

d)

Given that p=0.01 and number of modem transmitted is 100. The question ask the probability when there is 2 error. The question is similar to random experiment 2 where Y be the random variable that counts number of error. So, to calculate the probability mass function, we use P(X=x)=nCx(1-p)^(n-x)p^x. The x in this question is the y=2 which is number of event. n is the number of trials which is 100 and p is given which is 0.01.

e)

Let X be the random variable that counts the number of bits transmitted until the receiving modem makes 3 errors. The number of transmission counted including the receiving modem makes 3 errors is observed.

X can be described as negative binomial random variable as the number of outcomes (transmitted bit) is only two which is either 0 or 1. The success event here is 0 and the experiment now stops dependent on the number of success events required which is the receiving modem makes 3 errors. The probability of success is constant throughout the trials. The number of trials is now a random variable. So, from random experiment 4, the k for this question is 3. The probability of 12 bits transmitted is the x. Hence, from the explanation, we use the formula as in random experiment 4.

# data[].attributes.body
Group leader - Muhammad Shafie Bin Yusop

Team members :
1. Amirruliman
2. Carolyn Anak Junas
3. Intan Norhanim bt Abd Aziz

Discussion :

From the question, each time modem transmit one bit to receiving modem that analyses the bit whether 0 or 1. There will be an error with probability p.

From question a)
we will take X as the random number of bit until until the error received.

So we can use the pmf by random experiment 5 since our k=1.