United International University, MBA Faculty: Rashed Mohammad Saadullah Assistant Professor School of Business and Economics

1. Business Statistics (BUS 505) Assignment 6
Answer to the question no -34
1
Let,
Ai = Frequency of Unit [i = 1, 2, 3….]
Bj = Purchase of generic products [j = 1, 2, 3….]
a. P(A1 ) 1B  = P(A1) P(B1)
= .19 .79
= .12
b. ( A1│B3) =
P A  B
( 1 3)
P ( B
3)
.19
=
.27
=.7037
c. ( A1│B3) = P(A1) P(B3)
.19 = .79 .27
.19 ≠ .2133
So, they are not statistical independent.
d. (B1│A2) =
P B  A
( 1 1)
P ( A
2)
=
07.
21 .
=.333
e. P(B1 ) 2A  =P(B1) P(A1)
.07 = .19 .21
.07 ≠ .0399
So, they are not statistical independent.
f. P(A1) = P(B1)+P(B2)+P(B3)
=.12+.48+.19
=.79
g. P (B3) = P (A1) +P (A2)
= .19+.08
=.27
h. P (A1B3) =P(A1)+P(B3)- P(A1 B3)
= .79+.27-.19
=.87
FREQUENCY
OF VISIT
PURCHASE OF GENERIC PRODUCT
OFTEN(B1) SOMETIMES(B2) NEVER(B3)
TOTAL
Frequent(A1) .12 .48 .19 .79
Infrequent(A2) .07 .06 .08 .21
TOTAL .19 .54 .27 1.00

2. Business Statistics (BUS 505) Assignment 6
Answer to the question no -36
2
Let,
Ai =Traded Stock [i = 1, 2, 3….]
Bj =Read Business section [j = 1, 2, 3….]
a. P (B3) =P (A1) +P (A2)
=.04+.21
=.25
b. P (A1) =.18+.10+.04
=.32
c. P (A1│B3) =
P A  B
( 1 2)
P ( B
2)
=
04.
25.
=.16
d. P (B3│A1) =
P B  A
( 3 1)
P ( A
1)
=
04.
32.
=.125
e. Does not read regularly= P ( 1B ) = 1- P (B1)
=1-.34
= .66
P (A1│ 1B ) =
P A  B
( 1 1)
P ( B
1)
We Know, P (A1) =P (A1  B1) +P (A1  1B )
.32 =.18 + P (A1 1 B )
P (A1  1B ) = .32-.18 =.14
P (A1 1 B ) =
14.
66.
=.212
Traded
Stock
READ BUSSINESS SECTION
Regular (B1) Occasional(B2) Never (B3) Total
Yes (A1) .18 .10 .04 .32
No (A2) .16 .31 .21 .68
Total .34 .41 .25 1.00

3. Business Statistics (BUS 505) Assignment 6
Answer to the question no -37
PART SUBCONTRACTOR
A(B1) B(B2) C(B3) Total
Good(A1) .27 .30 .33 .90
Defective
3
(A2)
.02 .05 .03 .10
Total .29 .35 .36 1.00
Let, Ai=Particular Sensitive part [i = 1, 2, 3….]
Bj =Subcontractor [j = 1, 2, 3….]
a. P(A2)= .02+.05+.03
=.10
b. P (B2) = .30+.05
= .35
c. P (A2│B2) =
B A P 
) 2 2 (
B P
) 2 (
05.
=
35.
=.1428
d. P(B2│A2)=
A B P 
) 2 2 (
A P
) 2 (
=
05.
10.
=.5
e. P(A1  B2)=P(A1)P(B1)
= .90 .29
.27≠ .261
So, they are not Statistical independent.
f. P(A│G)=
P AG
( )
P ( G
)
=
27 .
9.
=.30
P (B│G) =
P B G
( )
P ( G
)
=
30.
9.
=.33
P (C│G) =
P C G
( )
P ( G
)
=
33.
.9
=.37
In terms of quality subcontractor C is most reliable.
Answer to the question no -38
Worked
Problem
EXPECTED GRADE
A(B1) B(B2) C(B3) Below C(B4) Total
YES (A1) .12 .06 .12 .02 .32
NO (A2) .13 .21 .26 .08 .68
Total .25 .27 .38 .10 1.00
Let, Ai = Workers related problem [i = 1, 2, 3….]
Bj = Expected grade [j = 1, 2, 3….]

4. Business Statistics (BUS 505) Assignment 6
4
a. P (A1) =.12+.06+.12+.02
= .32
b. P (B1) =.12+.13=.25
c. P (B1│A1) =
A B P 
) 1 1 (
A P
) 1 (
=
12.
32.
=.375
d. P (A1│B1) =
B A P 
) 1 1 (
B P
) 1 (
12.
=
25.
=.48
e. Let, R Denotes the Event “Students who worked additional hours expected a Grade
below A”
 P (R│Y) =
Y R P 
( )
Y P
( )
[P(R) =P(C  Y) +P(X<C Y)
14.
=
32.
=.12+.02
=.14 P(R  Y) =.14
f. P (A1  B1) =P (A1) P (B1)
.12 = .32 .25
.12 ≠ .08
So, they are not statistically independent.
Answer to the question no -41
Let,
A denotes the event “A campus student club member of men who receive the material.”
B denotes the event “A campus student club member of women who receive the
material.”
C denotes the event “A campus student club member of student who joins the club.”
So, P (A) = .40 P (B) = .60
P(C  A) = .07 and P(C B) = .09
a) P(C) = P(C  A) + P(C B) = .07 + .09 = .16
b)
P B C
( )
( )
( / )
P C
P B C

 =
09.
16.
= .5625

5. Business Statistics (BUS 505) Assignment 6
Answer to the question no -42
A P
) ( ) P(B/A 2 2
5
Let,
1A denotes the event, “Interest rate in next year will be more than 1% than this year.”
2A denotes the event, “Interest rate in next year will be more than 1% lower than this
year.”
3 A denotes the event, “Interest rates next year within 1% than this year.”
B denotes the event, “Significant earnings growth in the next year.”
So, P ( 1A ) = .25 P ( 2A ) = .15 P( 3A ) = 1-[ P ( 1A )+P ( 2A )] = .60 
P (B/ 1A ) = .10 P(B/ 2A ) = .80 P(B/ 3A ) = .50
a) P ( 1A  B) = P(B/ 1A ) P ( 1A ) = .10  .25 = .025
b) P(B) = P ( 1A  B) + P( 2A  B) + ( 3A  B)
= P(B/ 1A ) P ( 1A ) + P(B/ 2 A ) P ( 2 A ) + P(B/ 3 A ) P( 3 A )
= .1  .25 + .8  .15 + .5  .60 = .445
c) P ( 2A /B) =
B A P 
( ) 2
B P
( )
=
B P
( )
=
12.
445.
= .270
Answer to the question no -43
Let,
A denotes the event, “Blue-collar employees modified health care plan.”
B denotes the event, “Blue-collar employees a proposal to change the work schedule.”
P (A) = .42 P(B) = .22 and P(B/A) = .34
a) P (A  B) = P(B/A) P (A) = .34  .42 = .1428
b) P(A ) B  = P(A) + P(B) - P (A B)
= .42 + .22 - .1428 = .4972
c) P(A/B) =
P A B
( )
P ( B
)
=
.1428
.22
= .649