School

University of Texas, Arlington **We aren't endorsed by this school

Course

CSE 5301

Subject

Statistics

Date

Sep 4, 2023

Type

Other

Pages

4

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Data Analysis and Modeling Techniques (Assignment-1)
2.2, 2.4, 2.6, 2.8, 2.9, 2.10, 2.15, 2.17, 2.19
2.2.
Suppose that after 10 years of service, 40% of computers have problems
with motherboards (MB), 30% have problems with hard drives (HD), and 15%
have problems with both MB and HD. What is the probability that a 10-year old
computer still has fully functioning MB and HD?
Ans :-
P(MB)=0.4
P(HD)=0.3
P(MB ∩ HD)=0.15
P(MB U HD) = P(MB)+P(HD)- P(MB ∩ HD)
P(MB U HD) = 0.4 + 0.3 - 0.15
= 0.55
1-P(MB U HD) = 1-0.55
Answer = 0.45 (By taking out all the problem probability from complete
set )
2.4.
Among employees of a certain firm, 70% know C/C++, 60% know Fortran,
and 50% know both languages. What portion of programmers
(a) does not know Fortran?
(b) does not know Fortran and does not know C/C++?
(c) knows C/C++ but not Fortran? (d) knows Fortran but not C/C++?
(e) If someone knows Fortran, what is the probability that he/she knows C/C++
too? (f) If someone knows C/C++, what is the probability that he/she knows
Fortran too?
known values are
P(C) = 0.7
P(F)=0.6
P(C∩F)=0.5
Ans 2.4.a ->
Answer = 1- P(F) = 1-0.6 = 0.4
Ans 2.4.b -> Answer = 1-P(F)-P(C)+P(C∩F) = 1- 0.6 - 0.7 + 0.5 = 0.2
Ans 2.4.c -> Answer = P(C) - P(C∩F) = 0.7 - 0.5 = 0.2
Ans 2.4.d -> Answer = P(F) - P(C∩F) = 0.6 - 0.5 = 0.1
Ans 2.4.e -> Answer = P(C/F) = P(F ∩ C)/P(F) = 0.5/0.6 = 5/6

Ans 2.4.f -> Answer = P(F/C) = P(F ∩ C)/P(C) = 0.5/0.7 = 5/7
2.6.
Under good weather conditions, 80% of flights arrive on time. During bad weather,
only 30% of flights arrive on time. Tomorrow, the chance of good weather is 60%. What is
the probability that your flight will arrive on time?
Given P(F/GW) = 0.8
P(F/~GW) = 0.3
P(GW) = 0.6
P(~GW) = 1-0.6 = 0.4
P(F) = P(F
∩ GW
) + P(F
∩ ~GW
)
P(F) = P(F/GW)*P(GW) + P(F/~GW)*P(~GW)
P(F) = (0.8 * 0.6) + (0.3*0.4) = 0.48 + 0.12 = 0.6
2.8.
A shuttle's launch depends on three key devices that may fail independently of
each other with probabilities 0.01, 0.02, and 0.02, respectively. If any of the key devices
fails, the launch will be postponed. Compute the probability for the shuttle to be
launched on time, according to its schedule.
P(A) = 0.01
P(B) = 0.02
P(C) = 0.02
P(~A ∩~B∩~C) = (1-P(A))*(1-P(B))*(1-P(C)) =0.99 * 0.98 * 0.98 = 0.950796
2.9.
Successful implementation of a new system is based on three independent
modules. Module 1 works properly with probability 0.96. For modules 2 and 3, these
probabilities equal 0.95 and 0.90. Compute the probability that at least one of these
three modules fails to work properly.
P(A) = 0.96
P(B) = 0.95
P(C) = 0.90
Answer -> 1-P(A ∩ B ∩ C)
= 1-(0.96*0.95*0.90) = 1- 0.8208 = 0.1792

2.10.
Three computer viruses arrived as an e-mail attachment. Virus A damages the
system with probability 0.4. Independently of it, virus B damages the system with
probability 0.5. Independently of A and B, virus C damages the system with probability
0.2. What is the probability that the system gets damaged?
P(A)=0.4
P(B)=0.5
P(C)=0.2
Answer -> 1-P(~A ∩ ~B ∩ ~C) = 1 -((1-0.4)*(1-0.5)*(1-0.2)) = 1-(0.6*0.5*0.8) =
1-0.24 = 0.76
2.15.
A computer program consists of two blocks written independently by two different
pro- grammers. The first block has an error with probability 0.2. The second block has an
error with probability 0.3. If the program returns an error, what is the probability that
there is an error in both blocks?
P(A)=0.2
P(B)=0.3
P(A
∩ B
) = 0.06
P(A U B) = P(A) + P(B) - P(A
∩ B
) = 0.2+0.3-0.06 = 0.44
Answer -> P(A ∩ B)/P(A U B)
= 0.06/0.44 = 0.136363636363636
2.17.
A computer assembling company receives 24% of parts from supplier X, 36% of
parts from supplier Y, and the remaining 40% of parts from supplier Z. Five percent of
parts supplied by X, ten percent of parts supplied by Y, and six percent of parts supplied
by Z are defective. If an assembled computer has a defective part in it, what is the
probability that this part was received from supplier Z?
P(X)=0.24 , P(D/X)=0.05
P(Y)=0.36 , P(D/Y)=0.1
P(Z)=0.4 , P(D/Z)=0.06
P(Z/D)=P(Z)*P(D/Z)/P(D) -> Answer
P(D)=P(D
∩X
)+P(D
∩Y
)+P(D
∩Z
) = (0.24*0.05)+(0.36*0.1)+(0.4*0.06)=0.072
P(Z/D) = 0.4 * 0.06/0.072
P(Z/D)
= 0.024/0.072 =1/3
P.T.O