QMB 3600 Final Exam Review 2022 Detailed Solutions

This document is for review purposes and does not represent every type of problem that may be on the 40
question QMB3600 cumulative final exam.

1. This problem is in reference to students who may or may not take advantage of the opportunities
provided in QMB such as homework. Some of the students pass the course, and some of them do not
pass. Research indicates that 40% of the students do the assigned homework. Of the students who do
homework, there is an 80% chance they will pass the course. The probability of not passing if the student
does not do the home work is 90%. What is the probability of a student not doing homework or passing?
A. .32
B. .52
C. .94
D. .92
E. .38
Solutions:
P(Hwk) =0.40 P(Pass | Hwk) = 0.80 P(Not Pass | Not Hwk) = 0.90
P(Not Hwk or Pass) = ?
PASS NOT PASS
HWK 0.32 0.08 0.40
NOT HWK 0.06 0.54 0.60
0.38 0.62 1
P(Pass & Hwk) = P(Pass | Hwk) * P(Hwk)
0.32 = 0.80 * 0.40

P(Not Pass & Not Hwk) = P(Not Pass | Not Hwk) * P(Not Hwk)
0.54 = 0.90 * 0.60

P(Not Hwk or Pass) = P(Not Hwk) + P(Pass) – P(Not Hwk & Pass)
0.92 = 0.60 + 0.38 - 0.06

2. Suppose that for a certain football game the probability that the home team will be ahead at half time
is 0.60 and the probability that the home team will be ahead at half time as well as at the final gun is 0.45.
What is the probability that the home team will win this game given that it is ahead at the half?
A. .45
B. .75
C. .15
D. .27
E. None of the above.

Solutions:

P(HalfTime ) = 0.60 P(HalfTime & Final) = 0.45 P(Final | HalfTime) = ?

P(Final | HalfTime) = P(Final & Half) / P(Half)
0.75 = 0.45 + 0.60




1

,3. A company markets two products (Product A and Product B) through mail order. The company will
market them in sequence with the first mail order offer for product A. It f feels that there is a 30%
chance that any customer will purchase product A. Product B is offered some months later. It is felt, for
product B, that there is a 30% chance of selling product B to a customer if the customer purchased
product A and a 5% chance of selling product B to a customer who did not purchase product A.

What is the probability of not selling product B to a particular customer?
A. .875
B. .665
C. .125
D. .210
E. None of the above
Solutions:
P(A) =0.30 P(B | A) = 0.30 P(B | Not A) = 0.05 P(Not B) = ?
B NOT B
A 0.09 0.21 0.30
NOT A 0.035 0.665 0.70
0.125 0.875 1
P(B & A) = P(B | A) * P(A)
0.09 = 0.30 * 0.30
P(B & Not A) = P(B | Not A) * P(Not A)
0.035 = 0.05 * 0.70
P(Not B) = 1 – P(B)
0.875 = 1 - 0.125

4. A quality control department finds that it accepted only 5% of all bad items and it rejected only 1% of
good items. A supplier has just delivered a shipment of a certain item. Past records show that only
90% of the parts of that supplier are good. If the department accepts an item, what is the probability that
the item is bad? Round your answer to five decimal places.
A. 0.05000
B. 0.00558
C. 0.35714
D. 0.09635
E. None of the above
Solutions:
P(Accept | Bad) =0.05 P(Reject | Good) = 0.01 P(Good) = 0.90 P(Bad | Accept) = ?
Accept Reject
Good 0.891 0.009 0.90
Bad 0.005 0.665 0.10
0.896 0.875 1

P(Accept & Bad) = P(Accept | Bad) * P(Bad)
0.005 = 0.05 * 0.10
P(Reject & Good) = P(Reject | Good) * P(Good)
0.009 = 0.01 * 0.90
P(Bad | Accept) = P(Bad & Accept) / P(Accept)
0.00558 = 0..896


2

, 5. The probability that house sales will increase over the next six months is estimated at 0.25. It is also
estimated that the probability is 0.74 that 30 year fixed-loan mortgage rates will increase over this period.
Economists estimate that the probability is 0.89 that either housing sales or interest rates will increase.
The probability that both house sales and interest rates will increase is estimated at:
A. .100
B. .185
C. .705
D. .900
E. .500
Solutions:

P(Sales Inc ) = 0.25 P(Rates Inc) = 0.74 P(Sales Inc or Rates inc) = 0.89 P(Sales Inc & Rates Inc) = ?
P(Sales Inc or Rates inc)= P(Sales Inc) + P(Rates Inc) - P(Sales Inc & Rates Inc)
0.89 = 0.25 + 0.74 - X
P(Sales Inc & Rates Inc) = 0.100

6. Over the last 100 business days, Harry had 20 customers on 30 of those days, 25 customers on 20 days,
35 customers on 30 days, 40 customers on 10 days, and 45 customers on 10 days. What is the variance of
the number of Harry’s customers?
A. 30
B. 38
C. 59
D. 75
E. 83
Solutions:
For the best results, do the calculation in Excels or using the TI-83/84 graphing calculators

Customers
(x)
Days
F(x)
EV
=
Sum
(x
*
F(x))
Variance
=
Sum[
(x-­‐
EV)^2
*
F(x)]

20
30
0.3
6
30

25
20
0.2
5
5

35
30
0.3
10.5
7.5

40
10
0.1
4
10

45
10
0.1
4.5
22.5



100
1
30
75







EV
Variance


7. What is the probability that exactly 1 out of 10 cars experience a breakdown if the probability of a
breakdown is 30%?
A. .5121
B. .1211
C. .0282
D. .3828
E. .4276
Solutions:

N = 10 P = 0.30 X =1
Excel: =BINOM.DIST(x,n,p,TRUE/FALSE)->>>>>>>=BINOM.DIST(1,10,0.3,FALSE)->>0.1221
TI-83: 2nd->>>VARS->>> 0: binompdf(n,p,x)->>>>>>> binompdf(10,.30,1,p,x)->>0.1221
3