Answer all questions. All work must be shown to qualify for partial credit Also check to make sure all
your work is legible and clear before sending it.
1. The average age for a person getting married for the first time is 26 years (U.S. News and World
Report, June 6, 1994). Assume the ages for first marriages have a normal distribution with a
standard deviation of four years.
a) What is the probability that a person getting married for the first time is younger than 23 years
of age?
b) What is the probability that a person getting married for the first time is in his or her twenties?
c) 90% of people getting married for the first time get married before what age?
2. A management company must first decide whether to undertake a market research survey. If
the market research study is conducted, the outcome will either be favorable (F) or unfavorable
(U). Assume there are only two decision alternatives, D1 and D2, and two states of nature, S1
and S2. The payoff table showing profit is as follows:
State of Nature
Decision Alternative
S1
S2
D1
100
300
D2
400
200
a) Using the following probabilities, what is the optimal decision strategy?
P(F) = 0.56
P(S1/F) = 0.57
P(S1/U) = 0.18
P(S1) = 0.40
P(U) = 0.44
P(S2/F) = 0.43
P(S2/U) = 0.82
P(S2) = 0.60
b) Find the Efficiency of this market research. What would be your recommendation?
c) How much will this company be willing to pay for this market research?
3. The demand for Carolina Industries” product varies greatly from month to month. Based on the
past two years of data, the following probability distribution shows the company’s monthly
demand:
Unit Demand
Probability
300
0.20
400
0.30
500
0.35
600
0.15
a. If the company places monthly orders equal to the expected value of the monthly
demand, what should Carolina’s monthly order quantity be for this product?
b. Assume that each unit demanded generates $70 in revenue and that each unit ordered
costs $50. How much will the company gain or lose in a month if it places an order
based on your answer in part (a) and the actual demand for the item is 300 units?
c. What are the variance and standard deviation for the number of units demanded?
4. Lawson’s Department Store faces a buying decision for a seasonal product for which
demand can be high or low. The purchaser for Lawson’s can order 1, 2, or 3 lots of the
product before the season begins but cannot reorder later. Profit projections (in thousands
of dollars) are shown.
State of Nature
Decision Alternative High Demand (S1)
Low Demand (S2)
Order 1 lot, D1
60
50
Order 2 lots, D2
80
30
Order 3 lots, D3
100
10
P(Sj)
0.4
0.6
a) Find the optimal decision using the expected value approach.
b) Use the expected loss opportunity method to find the Expected Value of Perfect
Information.
c) Conduct a graphical sensitivity analysis identifying the decisions for the different P(S1).
d) For the optimal solution, given the current probabilities of the state of nature, what would
be the range of optimality (that is the range of values for its payoff where that decision
alternative will still remain optimal) for its payoffs compared to the closest Expected
Value.
5. A survey on British Social Attitudes asked respondents if they had ever boycotted goods for
ethical reasons (Statesman, January 28, 2008). The survey found that 23% of the respondents
have boycotted goods for ethical reasons.
a) In a sample of six British citizens, what is the probability that two have ever boycotted goods for
ethical reasons?
b) In a sample of six British citizens, what is the probability that at least two respondents have
boycotted goods for ethical reasons?
c) In a sample of ten British citizens, what is the probability that between 3 and 6 have boycotted
goods for ethical reasons?
d) In a sample of ten British citizens, what is the expected number of people that have boycotted
goods for ethical reasons? Also find the standard deviation.
Introduction to Probability by Christian N. Madu
Probability is a numerical measure of chance or the likelihood that an event will occur. In everyday life,
we make use of probabilities. As we prepare to go to work or school, we make an assessment of the
weather condition. We ask ourselves questions such as, what is the likelihood of rainfall today. What is
the likelihood of snow today? Based on our previous knowledge of similar weather conditions, we may
subjectively come up with a probability assignment of the likelihood that the particular event (rain or
snow) may occur. The probability we assign is in a scale of 0 to 1 or in terms of percentages, from 0 to
100%. The probability assigned shows our degree of belief that the particular event will take place. If we
assign a value that is less than 0.5, it means we believe that the event is more likely not to occur.
However, a higher probability that is greater than 0.5 shows we have a higher degree of belief that the
event will take place. The extreme values like 0 and 1 show that we believe the event will either not
occur or that we are certain the event will take place respectively. A probability assignment of 0.5 shows
equal likelihood of occurrence.
There are three major ways to assign probabilities namely the classical method, relative frequency
method, and the subjective method.
With the classical method approach, we assume equal likelihood of occurrence. Typical
examples of this will be when we toss a fair coin or roll a balanced die. In tossing a fair coin, we
know that the coin has two faces. These are the only possibilities or outcomes we can expect
when such an experiment as tossing a coin is performed. However, because it is a fair coin,
either outcome has equal likelihood of occurrence. So the possible number of outcomes is 2
which are Head and Tail. These make up what we call a Sample Space or Universal Set. If we use
S to denote a Sample Space and H for Head and T for Tail, we can express S in a set form as
S = {H, T}
Since any one of them must occur in a single toss of a coin, we say the probability of the sample
space is 1 and since there are two possible outcomes, they can equally share that probability
thus
P(S) = P(H) + P(T) = 1
Where P(H) = P(T) = ½
Similarly, when we roll a die, we can identify S and express it as follows:
S = {1, 2, 3, 4, 5, 6}
Since P(S) =1, we can share that probability to the six elements or members of the sample set. In
other words, the probability of obtaining any one face = 1/6.
The classical method of assigning probabilities is easy and fast but we know that real life does
not operate on the assumption of equal likelihood of occurrence. There are certain events that
may have higher frequency of occurrence. There are for example, some periods when we may
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expect higher amounts of rainfall or higher temperatures. We cannot possibly assume equal
likelihood of occurrence when we have historical data.
Relative Frequency Method – With the relative frequency method, we are assigning
probabilities based on historical data we have. Let us take the experiment for rolling of a die as
an example. Suppose that we have conducted this experiment several times and each time we
record the face that is observed, we may find out that certain faces may be more likely to
appear than others. However, we must have done this so many times for the probabilities
obtained to be stable in other for us to rely on the probabilities we are obtaining. Continuing
with this example, suppose we are able to roll this die 100,000 times and each time, we record
the face that is observed. The frequencies of the faces that are observed are given in the Table
below:
Outcome
1
2
3
4
5
6
Total
Frequency
10,000
20,000
40,000
15,000
10,000
5,000
100,000
Relative Frequency
0.10
0.20
0.40
0.15
0.10
0.05
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Notice from the frequencies that are presented here, that the face of 3 seems to appear more
often than any other face. Conversely, the face of 6 is the least occurring face. Armed with this
information, we cannot continue to claim equal likelihood of occurrence. So, we need to divide
each frequency by the sum of the frequencies which is the number of times we repeated the
experiment. In other words, the probability or likelihood that the face of 1 will occur is
10000/100,000 = 0.1 and the likelihood that the face of 3 will occur is 40000/100000 = 0.4.
These probability values we are obtaining are what we call relative frequencies. Thus, the
probabilities are relative to how often they do occur. We can therefore, conclude that another
name for probabilities is relative frequency.
We have constructed a frequency distribution and we see that all the laws of probabilities are
satisfied. Namely, we observe that the probability of any outcome is in the interval of 0 and 1.
We also notice that the sum of the probabilities is equal 1 which is the probability of the sample
space.
Subjective Method – We briefly discussed this when we were trying to process information on
the possibility of rainfall or snow. This method of assigning probabilities is subjective because it
does not rely completely on data. We can assign probabilities based on our intuition or
experiences about previous events. Suppose for example, a sales forecast estimates that the
probability of demand being 5000 units in the next period is 0.45. However, the manager
believes that there are some market trends that are not captured in the historical data and is
more optimistic of what demand may be and assigns a probability of 0.55. This new probability
is subjective because it is based partly on data and partly on the manager’s intuition. Because it
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is subjective, it cannot be replicated. Different managers may have different assessment of the
same data and situation, and may come up with different probabilities.
We shall be focusing primarily on the use of classical method and relative frequency method for
assigning probabilities but all these methods have roles to play as we assign probabilities to the
outcome of experiments. There are many situations where we may not have enough data or data may
not even exist so expert judgment may be needed to subjectively estimate probabilities.
Why do we need to know probabilities?
We go through all these efforts to assign probabilities and we may want to ask, why do we need to know
probabilities? The answer is simple. We need to know probabilities to be able to make a better decision.
If we go back to the example we presented above about the likelihood of a rainfall. If we assign a high
probability say 0.7 to the likelihood of rainfall in a given day, obviously it means we need to prepare
ourselves before leaving the house to school or to work. No one will like to be rained on on the way to
school or work. The high probability means you should dress up properly in anticipation of rain, take
your raincoat and umbrella and protect your useful materials that may be rained on or damaged by
water.
Since future events are not know with certainty, probabilities help us to estimate their likelihood of
occurrence and such information can help us to make informed decisions.
Counting Rules, Combinations, and Permutations
To be able to assign probabilities, you must be able to know the sample space. Therefore, you should be
able to count all the possible experimental outcomes. To identify these experimental outcomes, we
conduct an experiment. The experiment here is different from the ones in physical or natural sciences.
In each repetition of the experiment, a different outcome may be observed. Thus, we refer to this as a
random experiment. Notice that in most sciences, the experiments are replicable but in our case, the
outcomes are due to chance.
Take for example, the experiment to toss a fair coin three times and the question is to find the
probability of getting exactly 3 heads from the three tosses of a coin.
To answer this question, we need to know the number of possible outcomes from this experiment. One
way we can do this is to draw a tree diagram that shows the multiple sequences for this experiment. We
start by doing the first toss then the second and third tosses.
Figure 1 shows the tree diagram for this experiment. From the last branches after the third toss, we can
count eight possible outcomes from the experiment.
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Suppose we extend this problem to a case where we toss a coin two times and roll a die once, we also
have a sequence of experiments to conduct. The first step will be to do the first toss, the second step is
the second toss, and the third step will be to roll a die. Again, a tree diagram is used to represent the
sequence of this experiment. This is shown in Figure 2 about here.
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From these two examples, we can count 8 and 24 possible experimental outcomes respectively. If we
were to expand these problems by adding more steps, we may come to a situation where we may not
even have enough space to list all the possible sample points on a paper. This process of using tree
diagrams to identify all the possible outcomes from an experiment may be expedient but it is very
cumbersome. Rather than using the tree diagram to identify the sample points, we use the counting
rule whenever we conduct multiple-step.
Counting Rule
5
If an experiment involves a sequence of steps, the total number of experimental outcomes is given as
the product of the possible outcomes from each step.
Starting from our first example, we can identify the number of outcomes from each step of the
experiment. For example, in the three tosses of a fair coin, we have three steps, each with two
outcomes as follows
n1 = 2, n2 = 2, n3 = 2.
Applying the counting rule, we have n1 x n2 x n3 = 2 x 2 x 2 = 8
Thus, there will be 8 experimental outcomes from tossing a coin three times. Similarly, we can find the
number of possible outcomes for the second experiment as follows:
n1 = 2, n2 = 2, n3 = 6.
Applying the counting rule, we have n1 x n2 x n3 = 2 x 2 x 6 = 24
The counting rule is therefore a more concise and efficient way of finding the number of experimental
outcomes from an experiment.
Combination
Another counting rule allows to select x objects from n distinct objects. This is what we refer to as
combination. For example, suppose we want to select two people from a group of 5 people, how many
possible ways can this selection be made?
If we label the five people as A, B, C, D, E, we can do the selection as follows:
AB, AC, AD, AE
BC, BD, BE
CD, CE
DE
So, there are 10 possible ways we can do the selection of 2 objects from 5 distinct objects. Thus, the
selection of x objects from n distinct objects can be written in a simple mathematical form as
Insert equation 1
Permutation
Permutation is another counting rule. In this case, the order of selection is of importance. Notice that in
combination, AB is the same as BA. We did not care who is selected first. However, in permutation, the
order of selection is important. If we use the same example of five people labeled A, B, C, D, E, we may
be interested in how we can select the 1st and 2nd positions. In this case, the possible selections will be
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AB, AC, AD, AE
BC, BD, BE
CD, CE
DE
BA, CA, DA, EA
CB, DB, EB
DC, EC
ED
Because of the order, there are more possibilities. In fact, in this case, it is double. We can alternatively
identify the number of possible selections as follows:
The first step of the experiment will be to select the 1st position from the five people, and the 2nd step
will be to select the 2nd position from the remaining four people. Thus, applying the counting rule, we
have that n1 = 5 and n2 = 4.
Thus, n1 x n2 = 5 x 4 =20.
These twenty possibilities are listed in the table above. The model for permutation is presented below:
Insert equation 2.
Defining a random variable
If we go back to the first example, we can express the sample space S as a set consisting of eight
elements which are given as
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
This list will continue to increase exponentially as we increase the number of tosses. For example, a toss
of the coin four times will generate 16 possible outcomes while a toss of the coin five times will
generate 32 possible outcomes. If we go one more step to six tosses, we end up with 64 possible
outcomes. This becomes difficult to represent and will lead to both ink and paper waste. It is not
efficient. In statistics, we prefer to work with numbers than text. The use of numbers makes it easier to
represent information concisely. We shall therefore, define a random variable as a numerical
assignment to the outcome of an experiment and this assignment is due to chance occurrence.
Going back to our initial problem, we may wish to define the random variable x = number of heads or
number of tails from tossing a coin three times. From this definition, since the experiment is repeated
thrice, we know that the following possibilities exist: We can get 0 head, 1 head, 2 heads, or 3 heads. In
other words,
X = 0, 1, 2, 3
We can use this to develop the probability distribution for x by counting from the set of S, the number of
times we observe 0,1,2,3 heads.
x
0
1
2
3
Total
f(x)
1/8
3/8
3/8
1/8
1
F(x)
1/8
4/8
7/8
8/8
7
Notice that there are two sets of probabilities here namely f(x) and F(x). The small f(x) is known as the
probability density (mass) function and represents the probability of an exact value. For example, f(0)
represents the probability of obtaining 0 head and f(1) is the probability of obtaining 1 head. However,
the big F(x) is the cumulative probability or the sum of probabilities up to and including x. Therefore,
F(2) is the sum of probabilities up to and including 2. Thus,
F(2) = f(0) + f(1) + f(2) = 1/8 + 3/8 + 3/8 = 7/8.
Because this is the sum of probabilities, if x = 0, 1, 2,…, n, then F(n) = 1.
Frequency Distribution of Tossing a Coin Thrice
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
Series1
The figure above shows the distribution of the number of heads from tossing a coin three times. As you
can see, the observation of heads once and twice have the higher frequency of occurrence. We also
have a symmetric distribution meaning that we can divide this distribution into two equal parts.
Discrete random variable
In this case, the random variable is described to be discrete. It is discrete because we are able to count
for example, the number of heads from the experiment. We can definitely state the probabilities of 0, 1,
2, or 3 heads. Any time we are able to count the number of objects in an experiment, we can describe it
with a discrete random variable. The objects we are counting can be either finite or infinite. What
matters is the ability to be able to count. We shall apply this knowledge as we review discrete
probability distributions.
We shall focus on three discrete probability distributions namely
Binomial
Hypergeometric
Poisson
Binomial Distribution
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The Binomial distribution is applicable when we sample with replacement. In other words, we replace
the sampled item before taking the next sample. As a result, there is a slim possibility of sampling the
same object more than once. Thus, the samples are independent and it will not be the most appropriate
for sampling human population. However, because most times we may sample from a large population,
it will not make much difference whether we sample with or without replacement.
To use the binomial distribution, we check to make sure that the following conditions are satisfied:
1.
2.
3.
4.
The sequence of trials in an experiment is independent.
There are only two mutually exclusive outcomes namely success or failure.
The probability of an outcome is constant from one trial to the next.
The outcome from each trial is independent and it is not influenced or affected by the outcome
of previous trials.
Our experiment of tossing a coin three times is a typical application of the binomial distribution. We can
see that all these conditions are met.
First of all, each toss of the coin is ...

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