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For a Copy of this Review Lecture, go to
http://www.people.virginia.edu/~ttb/EIT/EITREVIEW1.PDF
Review Problems:
In examining sample problems that have appeared in past,
several concepts appear quite commonly.
1. Basic probability theory concepts:
basic set theory (probability space) concepts
mutually exclusive, collectively exhaustive events
Probabilities of Unions and Intersections
Independent events
Conditional Probability
Theorem of Total Probability
Bayes Theorem
2. Combinatorial Rules
Equally likely outcomes (card and dice problems)
Permutations
Combinations
3. Random Variable Concepts
Probability density, probability mass functions
Cumulative Distribution function
Mean, standard deviation
4. Discrete Distributions
Binomial Distribution
Poisson Process and Poisson Distribution
5. Continuous Distributions
Normal Distribution
Students t- Distribution
Central limit theorem
Set Theory Concepts:
You could encounter a problem involving set theory.
Here's a simple one.
P.0. Consider all of the people in the U.S. Draw a Venn
diagram showing all the adults as A and all the children
(under 18) as C. Draw the set showing all students
(including professional students) as S Show the set of all
people who are studying or teaching history at some level.
Shade in the portion of the diagram that corresponds to
adults who are history students.
Permutations:
When the order of an arrangement matters, we use
permutations to calculate the number of possibilities:
1. For the number of arrangements of n objects (or people),
the total is just n!
2. For k objects out of n, the correct number, when the
order matters, is
n!
(n − k )!
Let's apply these rules to some problems.
P.1. How many ways can a party of 7 people arrange
themselves in a row of 7 chairs?
ANS: The first person has a choice of 7 chairs, the second
a choice of 6, the third a choice of 5, etc. for a total of
(7)(6)(5)(4)(3)(2)(1)=7! arrangements.
P.2. What if there are more than 7 chairs. (Say there are 15
chairs.)
ANS: Then the first person has a choice of 15 chairs, the
second 14 chairs, and so on. The last person has a choice of
9 chairs. So the number of possibilities is
(15)(14)(13)(12)(11)(10)(9)=15!/(15-7)!
(In general, k items out of n items may be chosen in
n!/(n-k)! ways, if the order matters, as indicated above.
P.3. How many ways can a party of 7 people arrange
themselves around a circular table that seats 7?
ANS: In this case, we only distinguish between different
relative arrangements. Therefore, after the first person is
seated, there remain (6)(5)(4)(3)(2)(1)=6! arrangements for
the rest of the party.
P.4. An urn contains 8 balls. How many ordered samples
of size 3 can be chosen (i) with replacement (ii) without
replacement.
ANS:
(i) For sampling with replacement, there are 8 possible
choices for the first ball, 8 for the second, and 8 for the
third. Therefore, there are
8 ×8 ×8 = 83 = 512 different ordered pairs with replacement
(ii) For sampling without replacement, there are 8 possible
choices for the first ball, 7 for the second, and 6 for the
third. Therefore there are
8 ×7 ×6 = 336 different ordered pairs without replacement.
Combinations:
If the order doesn't matter, we divide the total number of
ways that k objects can be taken out of n objects by the k!
ways that the k objects chosen can be arranged. The total
number is
n!
(n − k )! k!
P.5. In how many ways can a committee of 3 men and 2
women be chosen from 7 men and 5 women?
ANS: For questions like this, we don’t care what the order
of picking is, we only care who is on the committee at the
end. How many ways can we pick 3 men from 7 men if
order doesn’t matter? Here, we use the combinatoric
formula
n!
7!
=
= 35 ways of picking 3 men out of 7
(n − k )! k! (7 − 3)!3!
and
5!
= 10 ways of picking 2 women out of 5 women.
(5 − 2)!2!
The total number of different committees that could be
picked is thus
35 ×10 = 350 different committee makeups.
P.6. In how many ways can 7 toys be divided among 3
children if the youngest gets 3 toys, and each of the others
gets two?
ANS: If order mattered, we could give the youngest 3 toys
first in 7 ×6 ×5 ways. Then we could give the next child
toys in 4 ×3 ways. Finally we could give the last child toys
in 2 ×1 ways, for a total of 7! ways. However, once the
first child has three toys, it doesn’t matter what order they
were received. Each combination of 3 toys could be
obtained in 3! ways. Likewise, the second child could
obtain 2 toys in 2! ways, and the third child could obtain 2
toys in 2! ways. Thus, the number of distinguishable
combinations is
7!
= 210 ways.
3!2!2!
P.7. Three light bulbs are chosen from 15 bulbs of which 5
are defective. Find the probability that (i) none is defective
(ii) exactly one is defective (iii) at least one is defective.
ANS: The best way to handle this problem is to recognize
the Hypergeometric distribution. However, assuming we
don’t know that, let’s reason it out.
(i) If none are defective, then we pick 3 from the 10 good
ones, and none from the 5 bad ones. The number of ways in
which this could happen is
10 5  10! 5!
= 120 ways.
   =
3
0
7
!
3
!
0
!
5
!
  
The total number of ways we could pick 3 light bulbs from
15 is
15 15!
= 455 ways. Therefore the probability of no
 =
3  12!3!
defectives is 120/455=0.264
(ii) Following the above reasoning, the probability of one
defective is
10 5 
  
2 1  = 0.495
455
(iii) The probability of at least one defective is
P( N ≥ 0) = 1 − P( N = 0) = 1 − 0.264 = 0.736
Binomial Distribution:
If an experiment is conducted n times, and the probability
of success is p on each trial, then, the probability of k
successes out of n trials is (Binomial distribution)
n  k
pk =  p (1 − p ) n − k for 0 ≤ k ≤ n
k 
P.8. A fair coin is tossed 6 times, or equivalently, six fair
coins are tossed. Call heads a success. Then,
n = 6, p = 0.5 , q = 0.5
(i) What is the probability that exactly two heads will
occur. By the binomial distribution
n  k n − k
P( K = k ) =  p q
k 
P ( K = 2) =
6!
(0.5) 2 (0.5) 4 = 0.234
2!4!
(ii) What is the probability of at least 4 heads?
6!
6!
6!
(0.5) 4 (0.5) 2 +
(0.5) 5 (0.5)1 +
(0.5) 6
4!2!
5!1!
6!0!
= 0.344
P ( k ≥ 4) =
(ii) What is the probability of no heads?
6 
P( K = 0) =  (0.5) 0 (0.5) 6 = 0.0156
0 
(Note: this problem is somewhat simpler than most, since
in this case p=q, while usually that isn’t the case.
P.9. Over a long period of time, a particular manufacturing
plant has been found to produce 7 defective toaster ovens
for every 1000 toaster ovens manufactured. An appliance
dealer receives a shipment of 100 toasters. What is the
probability that there are one or fewer defective toasters in
the shipment?
This is also a binomial problem. Let
p=0.007=Probability of a toaster oven being defective
q=1-p=0.993=Probability it's not defective
n=100=size of the shipment
100 
P( k = 0) =  (0.007) 0 (0.993)100 = 0.495
0 
100 
P( k = 1) =  (0.007)1 (0.993) 99 = 0.349
1 
P( k ≤1) = P( k = 0) + P( k = 1) = 0.884
Poisson Process:
If events can occur at any time t, and the probability of
occurrence in any short time interval dt is λdt, independent
of occurrence in any other time interval, then the probability
of k occurrences at time t is given by the Poisson process
(λt ) k e − λt

P( n(t ) = k ) = p(k ; t , λ) =  k!

0
k = 0,1,2,...
k<0
P.10. Patrons arrive at a service counter at a rate of 5 per
minute. Assume that, during the interval of interest the rate of
arrivals is constant and follows a Poisson process. Then
λ= 5 patrons / minute
(a) What is the expected number of patrons in 30 minutes?
E [ X (30)] = λt
= (5 patrons / minute)(30 minutes)
= 150 patrons
(b) What is the probability that no customers arrive during
a 2 minute period?
(5 × 2) 0 − 5×2
P[ X ( 2) = 0] = p( 0;5,2) =
e
= 0.0000454
0!
(c) What is the probability that exactly 10 customers arrive
in a 2 minute period?
(5 × 2)10 − 5×2
P[ X ( 2) = 10] = p(10;51
, )=
e
= 0125
.
10!
Although the Poisson process was derived from an arrival
time perspective, it can also be used to consider spatial
occurrences,
P.11. In a particular region of the country, interstate
highway pavements of a certain age are found to have 15
cracks per mile that need repair, on average. Here the
arrival rate is in cracks per mile.
(a) What is the expected number of repairs to cracks in a
ten mile section of interstate?
E[X(10)]=(15)(10)=150 cracks
(b) What is the probability of 10 or fewer cracks that need
repair in a 1 mile section? For this problem,
λx = 15 ×1 = 15
P[ X (1) ≤10] = e
− 15×1
15 (15) 2
(15)10
[1 +
+
+L +
]
1
2!
10!
= 0118
.
Later, we’ll show an easier way to evaluate this probability.
(c) What is the probability of no cracks in ½ mile? Here
P[ X ( 12 ) = 0] = e − 15×0.5 = 0.000553
So, its pretty likely that we will need to repair some
cracks in a ½ mile stretch.
Normal Distribution:
Perhaps the most commonly used continuous distribution.
If X is a random variable, with mean µ X and standard
deviation σX , the Normal probability density function is
 (x − µ X )2 
1

f ( x; µ X , σX ) =
exp
−
2


2πσX
2σX


This can't be integrated in closed form, but the CDF of the
closely related standard Normal R.V. is widely tabulated.
To find the probability that X ≤ x , calculate the standard
variate value
z=
x − µX
σX
and enter the tables with that value. The relationship is
x − µ X
P( X ≤ x ) = Φ ( z ) = Φ 
 σ
 X




Alternately, given a probability p, enter the tables with that
value to find
x = σX Φ − 1 ( p ) + µ X
P.12. Suppose the temperature T during June is normally
distributed with a mean of 68°F and a standard deviation of
6°F.
(i) What is the probability that the temperature is between
70°F and 80°F?
80 − 68
70 − 68
70 ≤T ≤80]]= P 
P[
≤Z ≤
6 
 6

= P[
.3333 ≤ Z ≤2]
= Φ (2) − Φ (0.3333) = 0.9772 − 0.6293
= 0.348
(ii) 70°F lies how many standard deviations above the
mean? From above, we obtained
70 − 68
= 0.3333
6
(iii) What is the probability that the temperature is above
80°F? From the previous calcs, P(T ≤80) = .9772 so
P(T > 80) = 1 − 0.9772 = 0.023
t -Distribution:
Basic result: A sample of size n is drawn from a normal
population. The mean and standard deviation are both
unknown. Then the sample mean X allows the definition
of a random variable
T=
X− µ
S/ n
T has a t distribution with n-1 degrees of freedom.
P.13. Suppose that 15 measurements were made during the
month of June with a sample mean of 68°F and a sample
standard deviation of 8°F. What is the 95% (two sided)
confidence interval for the sample mean?
ANS:
With 15 measurements, the number of degrees of freedom
of the t statistic is 14. The one-sided 97.5% t value is 2.145
(corresponds to a two-sided value of 95%) Therefore,
2.145 =
68 − µ
⇒ 63.569 ≤µ ≤72.431
8 / 15
with probability 0.95
Central Limit Theorem: Basic Result
Given a sample of size n from a population, the sample
1 n
mean X = ∑ X i approaches a normally distributed r.v.
n i =1
as n becomes large. If the underlying population has
standard deviation σx the sample mean has standard
deviation σx / n .
P.14. A one gallon can of a certain paint will cover (on the
average) 513.3 square feet of surface area with a standard
deviation of 31.5 square feet. What is the probability that
the mean area covered by a sample of 40 of these cans will
be between 510 and 520 square feet?
For the 40 cans, µ = 513.3, σ = 31.5 / 40 = 4.981
Hence,
520 − 513.3
510 − 513.3
P(510 ≤ X ≤520) = P 
≤Z ≤
4.981 
 4.981

= P( − 0.663 ≤ Z ≤1.345)
= Φ (1.345) − Φ ( − 0.663)
= Φ (1.345 + Φ (0.663) − 1
= .911 + .746 − 1
= 0.657
Total Probability; Bayes Theorem
Suppose the probability space is subdivided into a group of
n
.
mutually exclusive, collectively exhaustive sets {E i }
i =1
For any event A in the space,
1. Theorem of Total Probability
n
P( A) = ∑ P( A / E i ) P ( E i )
i =1
2. Bayes Theorem:
Since P( A ∩ B ) = P( A | B ) P ( B ) = P( B | A) P ( A) , it follows
that
` P( B | A) =
P ( A | B ) P( B )
P( A)
3. Combining 1. and 2.,
P( E i | A) =
P( A | E i ) P( E i )
n
∑ P( A | E ) P( E )
i =1
i
i
A typical problem is given below.
P.15. A TV manufacturer has three plants (A, B, C)
Typically, 4 out of 600 TV's from plant A will be defective
under warranty, 10 out of 900 from Plant B, and 5 out of
1000 from Plant C. A dealer gets 30% of his stock from
Plant A, 45% from Plant B, and 25% from Plant C. A
customer buys a TV from the dealer, and it is defective
under warranty. What is the probability that it was
manufactured in Plant A?
Let A=The TV is manufactured in plant A
B=The TV is manufactured in plant B
C=The TV is manufactured in plant C
D=The TV is defective under warranty
From the above
P( D | A) = 4 / 600 = 0.00667
P( D | B ) = 10 / 900 = 0.0111
P( D | C ) = 5 / 1000 = 0.005
P( A) = 0.30
P( B ) = 0.45
P(C ) = 0.25
Hence, by the Theorem of Total Probability,
P( D ) = P( D | A) P( A) + P( D | B ) P( B ) + P( D | C ) P (C )
= (0.00667)(0.3) + (0.0111)((0.45) + (0.005)(0.25)
= 0.00825
It follows from Baye's Theorem that
P( A | D ) = P( D | A) P( A) / P( D )
= (0.00667)(0.3) / 0.00825
= 0.243