Date: 08/28/2003
Course: COT 5405
Semester: Fall 2003
Instructor: Arup Guha
Logarithm/Exponent Rules

log a ( xy)  log a x  log a y
Note: The rule does not apply to
log a ( x  y) . A way to bound log a ( x  y) would be to use
the following relationship:
log a ( x  y )  log a (2 max( x, y ))
 log a 2  log a max( x, y )

log a ( x y )  y log a x

 log b x 

log a x  
log
a
b


Note: This rule allows us to change the base of a log if we don't like the one it is currently in.

y logb x  x logb y ; to prove this,
Let c  x b .
Then, by definition of logarithm and exponent,
log y
log x c  log b y .
Using logarithm rule three then to change the base, we get:
log 2 c log 2 y
.

log 2 x log 2 b
Rewriting,
log 2 c log 2 x
.

log 2 y log 2 b
Then, by using the common base rule backwards,
log y c  log b x .
Use the definition of a log to yield
y logb x  c .
Now, substitute for c:
y logb x  x logb y .
Exponents


x a x b  x a b
( x a ) b  x ab


x a  x ab  x a b
b
a a
x y = (xy)
Note: This is a very common error. Avoid it!!
a
Basic Principles of Counting (necessary for probability)
1. Addition Principle: if counting the members in two disjoint sets, X and Y, the total
number of members of the union is found by simple addition:
X Y  X  Y .
Where |X| is the cardinality of the set X.
2. Multiplication Principle: when counting the number of elements of ordered pairs in the
Cartesian Product of the sets X and Y, multiply the cardinalities of the sets as so:
|X*Y| = |X| * |Y|.
Example: Given (x, y) for
0  x  10 , 0  y  20 , the total number of elements is:
10*20 = 200.
3. Subtraction Principle: Counting elements from a universe, U, and you can count the
elements that you don’t want, then subtract this from |U| as so:
X U X .
This principle assumes that U is countable.
Permutations
Given n distinct objects (1, 2, 3, …, n), you want to order any k of them. Each order is distinct
and is counted separately. What we want is a k-tuple:
3, 2, 1, 4, 5, 6, …, k
2, 3, 1, 4, 5, 6, …, k
Example:
In general,
n
Pk  n(n  1)( n  2)...( n  k  1) 
n!
(n  k )!
Note: 0! = 1
The formula for permutations of objects with repetition is:
n!
n1! n 2 !...n k !
In other words:
n total objects: n1 of them are 1
n2 of them are 2
…
…
nk of them are k
Example:
Permutation of the word
Total permutations:
Some permutations would be:
DO1O2R
4! = 24
DO1O2R
DO2O1R
(In the case of permutations with repetition, we do not want to count both
permutations listed above, since they are in essence identical. For this
reason, we divide the total number of permutations by the factorial number
of repetitions of each letter.)
Actual number of distinct permutations:
totalPerms
4!
24


 12 .
D!*O!*R! 1!*2!*1! 2
Combinations
How many ways can we choose k objects out of n?
n
n!
    n C k
k!(n  k )!  k 
Order in combinations is insignificant.
Binomial Theorem
Given, (x+y)n , in general we have:
 ( x  y )( x  y )( x  y )...( x  y )
n
n
 n
 n  1 n 1  n  0 n
 x y    x y
   x n y 0    x n 1 y 1    x n 2 y 2  ...  
0
1
 2
 n  1
n
When we set x=1 and y=1, we find:
n
n
( x  y ) n    
k 0  k 
Also, since any choice of k objects out of n corresponds to the remaining n-k objects, we have:
n  n 
   

k  n  k 
With respect to the binomial coefficient, when bounding combinations,
 n   n  n  1   n  k  1 
    
...

1

 k   k  k  1  
where n, k are positive integers, and n  k .
When we subtract 1 from the numerator and denominator, successive terms are greater than or
equal to the previous term. Since
n
is the minimum term, the lower bound is:
k
k
 n  n 
    
k   k 
for large n and k.
To get the upper bound,
 n  n(n  1)...( n  k  1)
  
.
k!
k 
Since each value in the numerator is less than or equal to n, we have
(i)
 
 n nk
  
.
k!
k
Using Stirling’s approximation, we get
n
n
2n   = n!
e
n
or simply:
n
   n!
e
for most values.
n
n
So, by substituting   for k! in equation (i) above, we obtain
e
k
n
nk
 en 
  
  .
k
k 
k   k 
 
e
Stirling's Approximation and Factorial Function:
The factorial function is basic. By definition,
0! = 1.
Values may be computed recursively from given values by
(n + 1)! = (n + 1)n!.
For large n, the function is very large. A convenient approximation for n large is
the Stirling formula
n! ~
n
2n  
e
n
where ~ is read "asymptotically equal" and means that the ratio of the two sides of the equation
approaches 1 as n approaches infinity.
Source: Advanced Engineering Mathematics, 8th edition, Erwin Kreyszig, page 1067.
Probability
In most probability outcomes, we have a sample space or set of possible outcomes (e.g.
for a die throw, sample space = {1,2,3,4,5,6}).
Assuming each outcome in the sample space is equally likely,
P(success) = number of successes in the sample space
number of elements in the sample space
Note: To obtain a meaningful result, each outcome in your sample space should be
“equally likely”. Consider the rolling of 2 dice example to verify the above.
Example:
With the roll of two dice, what is the probability of obtaining a sum of 2, 6
and 8?
The key here is to be certain that the sample space is labeled properly. If
it is not, you won’t get the correct answer!
(1, 1)
(1, 2)
(1, 3)
…
…
(6, 5)
(6, 6)
There are 36 possibilities, each one occurring with equal probability.
P(2) =
1
36
P(6) =
5
36
P(9)=
4
36
Assumptions of probabilities
(where P(e) is the probability of an event e and s is the sample)
1.
0  P (e)  1
2.
 P ( e)  1
eS
3.
If events of A, B, C are disjoint,
P(A  B  C) = P(A) + P(B) + P(C)
In general,
P(A  B) = P(A) + P(B) - P(A  B)
4. Conditional probability
P( A | B) 
P( A  B)
P( B)
The probability of an event not occurring can be found by: 1  p( failure ) .
Example:
Given,
What is P(temp>90)?
Solution:
P(Temp > 90) = P(Rain  (Temp > 90)) + P((not rain)  (Temp>90))
P(Temp > 90) = (0.4)(0.3) + (0.6)(0.8)
P(Temp > 90) = 0.6
In general,
P( H )  [ P( R) * P( H | R)]  [ P( R) * P( H | R)] .
Submitted by: Fredrick Okumu and Christina Spradlin