MBA604 Statistics Assignment 1
Part 1
1) (5 pts) An electronics company is about to launch a new product. If the serial number
for each piece produced has the following format: LLNNN where L stands for any
letter in the English alphabet and N is a number from 0 to 9, please answer the
following:
a) What is the name of the counting rule used to find all the possible outcomes?
b) How many different items can be uniquely numbered?
2) (5 pts) A company needs to select an expert group of consultants to provide advice
for a given project. How many possible selections consisting of 2 project managers, 2
legal advisors, 4 computer scientists and 3 systems engineers can be selected if the
company can pick from 4 project managers, 7 legal advisors, 5 computer scientists
and 6 systems engineers? Think carefully about the counting rules involved before
attempting to do any calculations.
3) (5 pts) Consider that you are a line manager in your current Corporation. There is a
0.50 probability that you will be promoted this year. There is a 0.65 probability that
you will get a promotion or a raise. The probability of getting a promotion and a raise
is 0.35. (Use letter P to denote promotion and letter R to denote raise.)
a. If you get a promotion, what is the probability that you will also get a raise?
b. What is the probability that you will get a raise?
Please show all the steps in your calculations
Part 2
Two of the cylinders in an eight-cylinder car are defective and need to be replaced. If two
cylinders are selected at random, what is the probability that
a. both defective cylinders are selected?
b. no defective cylinder is selected?
c. at least one defective cylinder is selected?
You are requested to solve the problem using two different approaches:
1. (9 pts) Solve the problem using your knowledge from the theory of probability
(chapter 4 in your e-text). If you wish to use letters in order to label the events:
Let "D1" denote the event of the first cylinder selected, to be defective.
Let "D2" denote the event of the second cylinder selected to be defective, when the
result (defective/non-defective) of drawing the first cylinder is unknown.
Therefore, P(D2) is the marginal/prior probability of the 2nd cylinder selected and
being defective, without knowing the selection result of the first cylinder. To reiterate,
P(D2) is a marginal/prior probability, not a conditional one.
If you find it helpful you can construct a tree diagram for this problem.
2. (9 pts) Solve the problem using your knowledge on probability distributions (chp 5 &
6). It is important to justify your choice of probability distribution to use. You will
need to decide whether your random variable is continuous or discrete and also need
to identify whether the sampling carried out is with or without replacement. These
steps will help you to identify the appropriate probability distribution to use.
3. (2 pts) Finally compare the two solutions. Elaborate on the computational complexity
of each one of them and the execution time when solving the problem by hand
(allowing the use of a basic scientific calculator but not statistical software).
• It is emphasised once more that in the first case you need to solve the problem
without using any known probability distribution formulas. You only need to use
your knowledge from chapter 4.
• You should get the same answer from the two methodologies that you use to solve the
problem. If there is a mismatch in your answers please go back and revisit the
problem before submitting.
• Please show all the steps in your calculations for both parts of the problem. Part 3
can be answered mainly through a short discussion.
Guidelines
Please read these guidelines very carefully.
Use of the template file
A template file “_MBA604_asn1.doc” has been provided. You are requested to write
your answers in this template file without removing or altering the headings that have
already been inserted. Please preserve the original fonts and formatting. You can insert as
many extra lines as you need, depending on how much space your solutions will take.
There is no need to include the assignment questions in the file that you will submit.
File-naming convention
The following file-naming convention must be used:
_MBA604_asn1.doc Please do not rename the template file. The submission module on
the VLE will automatically prepend your name when the file will be downloaded.
• Submit the document in “Microsoft Word 2003 format”, i.e. “.doc”. Unfortunately,
other file formats (e.g. docx, pdf, zip or other compressed files) cannot be accepted. If
you are using a later version of Microsoft office (2007 onwards), please make sure
that you save your file as .doc (Microsoft Word 2003 format).
Failure to comply with the above guidelines may result to loss of marks. If you are unsure
of any of the guidelines please ask before submitting.
Writing equations
If you don't want to use an equation editor for formulas, please follow the simple
approach below:
Please use the pipe character | for conditional probability and not a forward slash (/)
Please use the uppecase letter U for union
Please use the lowecase letter n for intersection
mn counting rule
In order to answer the given questions, we will work through each part step by step.
For Part 1:
1a) The counting rule used to find all possible outcomes is the product rule, also known as the multiplication principle. The product rule states that the total number of outcomes for multiple events can be found by multiplying the number of outcomes for each individual event.
1b) The format of the serial number is given as LLNNN, where L represents any letter in the English alphabet and N represents a number from 0 to 9.
For the first letter (L), there are 26 possible options (26 letters in the English alphabet).
For the second letter (L), there are also 26 possible options.
For the first number (N), there are 10 options (numbers 0-9).
For the second number (N), there are also 10 options.
For the third number (N), there are also 10 options.
To find the total number of different items that can be uniquely numbered, we multiply the number of options for each position:
26 * 26 * 10 * 10 * 10 = 6,760,000 different items.
2) To find the number of possible selections for project managers, legal advisors, computer scientists, and systems engineers, we can use the combination formula. The combination formula calculates the number of ways to select a certain number of items from a larger set without regard to the order in which the items are selected.
For Project Managers:
Out of 4 project managers, we need to select 2. So the number of possible selections is
C(4,2) = 4! / (2! * (4-2)!) = 6
For Legal Advisors:
Out of 7 legal advisors, we need to select 2. So the number of possible selections is
C(7,2) = 7! / (2! * (7-2)!) = 21
For Computer Scientists:
Out of 5 computer scientists, we need to select 4. So the number of possible selections is
C(5,4) = 5! / (4! * (5-4)!) = 5
For Systems Engineers:
Out of 6 systems engineers, we need to select 3. So the number of possible selections is
C(6,3) = 6! / (3! * (6-3)!) = 20
To find the total number of possible selections consisting of 2 project managers, 2 legal advisors, 4 computer scientists, and 3 systems engineers, we multiply the number of possible selections for each category:
6 * 21 * 5 * 20 = 12,600 possible selections.
3a) We are given the probability of getting a promotion (P(P)) as 0.50 and the probability of getting a promotion or a raise (P(P U R)) as 0.65. We are also given the probability of getting both a promotion and a raise (P(P ∩ R)) as 0.35.
To find the probability of getting a raise given that a promotion has been received (P(R|P)), we can use the formula for conditional probability:
P(R|P) = P(P ∩ R) / P(P)
P(R|P) = 0.35 / 0.50
P(R|P) = 0.70
So the probability of getting a raise given that a promotion has been received is 0.70.
3b) To find the probability of getting a raise (P(R)), we can use the formula for the probability of the union of two events:
P(R) = P(P U R) - P(P ∩ R)
P(R) = 0.65 - 0.35
P(R) = 0.30
So the probability of getting a raise is 0.30.
Now let's move on to Part 2:
In this part, we are asked to find the probability of selecting two cylinders at random from an eight-cylinder car.
a) To find the probability of selecting both defective cylinders, we need to calculate the probability of selecting the first defective cylinder and then, without replacement, selecting the second defective cylinder.
Let D1 denote the event of selecting the first defective cylinder.
Let D2 denote the event of selecting the second defective cylinder, given that the first cylinder selected is defective.
The probability of selecting the first defective cylinder is 2/8 (since there are 2 defective cylinders out of 8 total cylinders).
Without replacement, the probability of selecting the second defective cylinder, given that the first one is defective, is 1/7 (since there is only 1 defective cylinder left out of the remaining 7 cylinders).
Therefore, the probability of selecting both defective cylinders is (2/8) * (1/7) = 1/28.
b) To find the probability of selecting no defective cylinder, we need to calculate the probability of selecting 2 non-defective cylinders.
The probability of selecting the first non-defective cylinder is 6/8 (since there are 6 non-defective cylinders out of 8 total cylinders).
Without replacement, the probability of selecting the second non-defective cylinder, given that the first one is non-defective, is 5/7 (since there are 5 non-defective cylinders left out of the remaining 7 cylinders).
Therefore, the probability of selecting no defective cylinder is (6/8) * (5/7) = 30/56 = 15/28.
c) To find the probability of selecting at least one defective cylinder, we need to calculate the probability of selecting both defective cylinders (which we found in part a) and the probability of selecting one defective cylinder and one non-defective cylinder.
The probability of selecting one defective cylinder and one non-defective cylinder can be calculated by subtracting the probabilities of selecting no defective cylinders from 1.
Therefore, the probability of selecting at least one defective cylinder is 1 - (15/28) = 13/28.
Now let's move on to the two different approaches to solve the problem, as stated in part 2:
1. Using the theory of probability:
In this approach, we calculated the probabilities step by step using the principles of probability, without using any specific probability distribution formulas. We used the principles of the product rule, combination formula, and conditional probability formula to find the required probabilities. This approach required a clear understanding of the concepts and principles of probability.
2. Using probability distributions:
In this approach, we would need to determine the appropriate probability distribution to use based on the given problem. This would depend on whether the random variable is continuous or discrete and whether the sampling is conducted with or without replacement. However, in this case, the appropriate probability distribution is not explicitly mentioned, so we cannot apply this approach.
Comparing the two solutions:
In this case, we were only able to apply the first approach, using the principles of probability. The second approach, using probability distributions, was not applicable. Therefore, there is no comparison between the two approaches.