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EQT 272 PROBABILITY AND STATISTICS
UNIVERSITI MALAYSIA PERLIS
TEST 1 ANSWER SCHEME
EQT 272 – PROBABILITY AND STATISTICS
SEMESTER 1
2012/2013
EQT 272 PROBABILITY AND STATISTICS
SEMESTER 1
2012/2013
Question 1:
A sample of three calculators is selected from a manufacturing line, and each calculator is
classified as either defective or acceptable. Let A, B, and C denote the events that the first,
second, and third calculators respectively, are defective.
a) Describe the sample space for this experiment with a tree diagram.
(4 Marks)
b) Find the probabilities of the following events;
(8 Marks)
i) P ( A)
ii) P (B )
iii) P( A
B)
iv) P( B
C)
a)
1st calculator
2nd calculator
3rd calculator
D3
D2
A3
D3
D1
A2
A3
A2
D1D2D3
D1D2A3
 1M (branch)
D1A2D3
 1M(label)
D1A2A3
D3
A1D2D3
A3
A1D2A3
D2
A1
Result
D3
A3
A1A2D3
A1A2A3
S {D1 D2 D3 , D1 D2 A3 , D1 A2 D3 , D1 A2 A3 , A1 D2 D3 , A1 D2 A3 , A1 A2 D3 , A1 A2 A3 }
 1M
 1A(correct sample
outcomes)
b) A = defective calculator 1
= { D1D2D3, D1D2A3, D1A2D3 ,D1A2A3}
 1M
 1A
EQT 272 PROBABILITY AND STATISTICS
P( A)
4
8
SEMESTER 1
2012/2013
1
2
B = defective calculator 2,
 1M
= { D1D2D3, D1D2A3 , A1D2D3 , A1D2A3 }
 1A
P(B)
4
8
1
2
A ∩ B = defective calculator 1 and 2
= { D1D2D3, D1D2A3 }
P( A
B
B)
2
8
 1A
1
4
C = all defective calculator 2, all defective calculator 3
= { D1D2D3, D1D2A3, A1D2D3 , A1D2A3 , D1A2D3 , A1A2D3}
P( B
 1M
C)
6
8
3
4
 1M
 1A
8Marks
EQT 272 PROBABILITY AND STATISTICS
SEMESTER 1
2012/2013
Question 2:
In a genetics experiment, the researcher mated two Drosophila fruit flies and observed the traits
of 300 offspring. The results are shown in the table.
Wing Size
Eye color
Normal
Miniature
Normal
140
6
Vermillion
3
151
One of these offspring is randomly selected.
a) What is the probability that the fly has normal eye color and normal wing size?
(2 Marks)
b) What is the probability that the fly has vermillion eyes?
(2 Marks)
c) What is the probability that the fly has either vermillion eyes or miniature wings?
(4 Marks)
a) Let A = fly has normal eye color, B = fly has normal wing size
Probability that the fly has normal eye color and normal wing size
= P( A
=
140
300
 1M
B)
7
15
0.4667
b) Let C = fly that has vermillion eyes
3 151 77
P(C )
0.5133
300
150
c) Let D = fly that has miniature wings
6 151 157
P(D)
0.5233
300
300
P(C D) P(C ) P( D) P(C D)
77 157 151 8
0.5333
=
150 300 300 15
 1A
 1M, 1A
 1M, 1A
 1M, 1A
EQT 272 PROBABILITY AND STATISTICS
SEMESTER 1
2012/2013
Question 3:
a) A wireless garage door opener has a code determined by the up or down setting of 12
switches. How many outcomes are in the sample space of possible codes?
(3 Marks)
b) A home security was designed to have password consists of five characters.
i. How many different passwords are possible if the first two characters are any
lowercase letter and the remaining characters are numbers?
ii)
(3 Marks)
How many different passwords are possible if the first two characters are any
lowercase letter and the remaining characters are numbers, and the first character must be
‘r’ letter?
a)
( 2C1 )12 4096 possible codes
(or 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 4096)
(3 Marks)
 1M(choose 1 from 2)
 1M(power of 12)
 1A(correct answer)
b) 26 26 10 10 10 676 000 different password
 1M(multiply 26 and 10)
 1M(using multiplication)
 1A(correct answer)
c) 1 26 10 10 10 26000 different password
 1M(multiply 1)
 1M(using multiplication)
 1A(correct answer)
EQT 272 PROBABILITY AND STATISTICS
SEMESTER 1
2012/2013
Question 4:
Let x represent the number of times a customer visits a grocery store in a 1-week period as
shown in the table.
x
P (X = x)
a)
b)
c)
d)
0
0.1
1
0.4
2
0.4
3
0.1
Find the expected value and variance of X.
Find the cumulative distribution function of X.
Find the probability that X is greater than 1.
Find P (1≤ X ≤3).
a) E(X ) 0(0.1) 1(0.4) 2(0.4) 3(0.1) 1.5
E ( X 2 ) 0 2 (0.1) 12 (0.4) 2 2 (0.4) 32 (0.1)
Variance = E(X2) – (E(X))2
= 2.9 – (1.5)2
=0.65
2.9
(5 Marks)
(2 Marks)
(2 Marks)
(2 Marks)
 1M, 1A
 1M, 1A
 1A
b)
x
P(X=x)
F(X)
0
0.1
0.1
1
0.4
0.5
2
0.4
0.9
3
0.1
1
 1M
 1A
c) P (X > 1) = P (X =2) + P (X = 3)
= 0.4 + 0.1
= 0.5
 1M
 1A
d) P (1≤ X ≤3) = P(X=1) + P(X=2) + P(X=3)
= 0.4+0.4+0.1 = 0.9
 1M
 1A