Download Logarithm, Exponent Rules, Counting Principles, Permutations, Combinations, and Binomial and more Study notes Computer Science in PDF only on Docsity!
Date: 08/28/
Course: COT 5405
Semester: Fall 2003
Instructor: Arup Guha
Logarithm/Exponent Rules
log (^) a ( xy )log ax log ay
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 2 log max( , )
log ( ) log ( 2 max( , ))
x y
x y x y
a a
a a
x y x a
y
a
log ( ) log
a
x
x
b
b
a
log
log
log
Note : This rule allows us to change the base of a log if we don't like the one it is currently in.
b x^ by
y x
log log
; to prove this,
Let
by c x
log .
Then, by definition of logarithm and exponent,
log (^) x c log (^) b y.
Using logarithm rule three then to change the base, we get:
b
y
x
c
2
2
2
2
log
log
log
log
Rewriting,
b
x
y
c
2
2
2
2
log
log
log
log
Then, by using the common base rule backwards,
log (^) y c log (^) b x.
Use the definition of a log to yield
y c
b x
log .
Now, substitute for c:
b x^ by
y x
log log
Exponents
(^) a b ab x x x
a b ab ( x ) x
(^) a ab ab x x x
b
Note: This is a very common error. Avoid it!!
x
a y
a = (xy)
a
Basic Principles of Counting (necessary for probability)
- 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.
- 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.
- 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:
Example: 3, 2, 1, 4, 5, 6, …, k
2, 3, 1, 4, 5, 6, …, k
In general,
n k
n P nn n n k n k
Binomial Theorem
Given, ( x + y )
n , in general we have:
n n n n n x y
n
n x y
n
n x y
n x y
n x y
n
x y x y x y x y
0 1 1 2 2 1 1 0
1
...
0 1 2
( )( )( )...( )
When we set x=1 and y=1, we find:
n
k
n
k
n
x y
0
Also, since any choice of k objects out of n corresponds to the remaining n-k objects, we have:
n k
n
k
n
With respect to the binomial coefficient, when bounding combinations,
1
1 ...
1
1 n k
k
n
k
n
k
n
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
k
n is the minimum term, the lower bound is:
k
k
n
k
n
for large n and k.
To get the upper bound,
!
( 1 )...( 1 )
k
nn n k
k
n (^)
.
Since each value in the numerator is less than or equal to n, we have
(i)
k!
n
k
n
k
Using Stirling’s approximation, we get
n
e
n n (^)
2 = n!
or simply: n!
e
n
n
^
for most values.
So, by substituting
n
e
n
for k! in equation (i) above, we obtain
k
k
k
k
en
e
k
n
k
n
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!
There are 36 possibilities, each one occurring with equal probability.
P(2) =
P(6) =
P(9)=
Assumptions of probabilities
(where P ( e ) is the probability of an event e and s is the sample)
- 0 ^ P ( e )^1
-
e S
P ( e ) 1
- 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 )
- Conditional probability
P B
P A B
P A 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.
In general,
P ( H )[ P ( R )* P ( H | R )][ P ( R )* P ( H | R )].
Submitted by: Fredrick Okumu and Christina Spradlin
