Some Initial Ideas of Probability:
The Fundamental Counting Principle: Let E
1
and E
2
be two events. The first event E
1
can occur in m different ways. After E
1
occurs, E ways that the two events can occur is m
2
⋅
can occur in n different ways. The number of n
These are the ideas that have been discussed before in previous classes. Example: If you have
3 choices of sandwich, 2 choices for soup, and 3 choices for drink, how many different lunches can you have? This was previously approached by creating tree diagrams. chicken lemonade
Pepsi ham potato tea lemonade
Pepsi tea lemonade
So for the ham sandwich we have
6 different lunches (3*2). The same will be true for all sandwich choices. Therefore, each sandwich has 6 choices and there are 3 sandwiches, there are 18 different meals. 3*2*3 chicken Pepsi tea turkey tuna potato chicken lemonade
Pepsi tea lemonade
Pepsi tea potato lemonade
Pepsi tea
This is a simple counting principle to determine the number of possible outcomes.
You try one: A local phone number is seven digits. The first digit can not be a 0 or a 1. How many local phone numbers are possible for one area code? Any other special cases to omit?
Page 1
Kevin Thompson revised 11/20/2006

The first answer is 8 ⋅ 10 6
The second answer is yes. Can a local number be 911-5555? How many phone numbers do we need to omit?
Ex/ 1 Suppose we have a group of 30 people. We want to choose a steering committee to be made up of three people from the group. How many ways can this be done?
Solution: This problem is a standard combinations problem. Previously someone might have told you that since order doesn’t matter (committee ABC is essentially the same as committee
BCA, CBA, ACB, BAC, and, CAB—notice there are 6 or 3! repeats) Therefore, we know it is a combination problem. Therefore we want
30
C
3
and it is given by the formula
30 !
3 !
⋅ 27 !
.
Therefore, there are 4060 ways to choose a committee of three from 30.
If you understood all of that, fine, great. If you understood some, or none read on. I am going to propose a new (or first) way to think about this type of problem using an ordered partition.
The notation that I am going to use is
30
27 , 3
=
30 !
27 !
⋅ 3 !
and is read “30 choose 27 choose 3”.
Notice that by writing it in this fashion it looks more like the algorithmic approach to the problem given above. In addition though, this way of thinking as more advantages.
Given the previous problem, here is how we would set it up. We know we have 30 members to choose from so we will be choosing members to form the committee from the 30. Thus
30
?
, but know I have to decide what to choose. Think about from the thirty, the groups that will be formed. Well, there is a group of three people that will form the committee, and there is a group of 27 people that remain that will NOT serve on the committee. So, you’re
PARTITIONING the initial group of 30 into a group of 27 and a group of 3.
Page 2
Kevin Thompson revised 11/20/2006

Notice the partitioned groups must add together to be the whole amount
So we have
30
3 , 27
=
30
27 , 3
=
30 !
3 !
⋅ 27 !
= 4060 ways to form the committee. This is the number you would get if you counted up all the possible comities and avoid repetitions (again the reason for the 3!)
This problem is now a standard permutation problem where order is said to be important . (It matters if Johnny is President or Secretary-Treasurer.) Previously, we may have done this problem by doing the following:
President V. President Sec/ Treas
There are 30 ways to fill the president slot. Leaving 29 people to fill the V.P. slot and 28 people to fill the Sec/ Treas slot since you wouldn’t want one person to have two of those jobs.
Therefore, the solution is 30 ⋅ 29 ⋅ 28 or 24,360 ways. (notice this is 6 times (3!) more than if there was a committee of 3. 6*4060)
This method is still valuable. Notice this is effectively how the telephone number problem was solved.
Using an ordered partition works as well here too.
We have a group of 30 to choose from. So,
30
?
, but now I have to decide what to choose.
Again, we are partitioning this group of 30 into smaller groups. The sum of the smaller
Page 3
Kevin Thompson revised 11/20/2006

groups needs to equal the sum of the whole . We will have a group of 1 (for the president), a group of 1 (for the vice president), a group of 1 for the Sec/Treas, and a group of 27 of the remaining people holding no office. Therefore, we have:
1 ,
30
1 , 1 , 27
=
1 !
⋅ 1 !
30
⋅ 1 !
!
⋅ 27 !
=
30
27
!
!
=
30 ⋅ 29 ⋅ 28 = 24,360 ways.
Say we want to multiply out ( x + y ) 4 =
( x + y
) ( x + y
) ( x + y
) ( x + y
)
. It is pretty easy to see that there will be only 1 x 4 term. A harder question is how many x 3 y terms will be present?
What is its coefficient? (There is a connection to Pascal’s Triangle for this question as well)
This too can be thought of as a group in which you are partitioning. We have 4 factors of
( x + y
)
. The binomial theorem can also be thought of using the idea of an ordered partition.
So our group to be partitioned is of size 4. Three of the factors need to contribute an x and 1 of the factors must contribute a y . So, we have
3
4
, 1
ways to get an x 3 y term. So, 4 is the coefficient.
Here is the connection to Pascal’s Triangle:
Row(n) Columns (m)
1 2 3 4 5 6 7 8
0
1
2
3
4
5
6
7
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
You try: For the expansion of connection above. y 30
For the expansion of ( x + y ) n
Look at the nth row.
The first entry is the coefficient of the x n . Then continue to reduce the power of x by 1 and increase the powers of y by 1. Each time moving one space horizontally in order to obtain the coefficient
( x + y ) 5 Write out the complete expansion. Note the
Ex/ Find the coefficient of the term partitions. x 10 of the expansion of ( x + y ) 40 . Use order
Page 4
Kevin Thompson revised 11/20/2006

Homework
1.
How many 10 digit phone numbers are available if the first digit of area code cannot be
1 or zero and the first digit of the prefix cannot be a one or a zero?
2.
In how many ways can a six question T/F test be answered?
3.
A student may answer any 10 questions from a total of 12 questions on an exam. In how many ways can this be done? (Think ordered partition)
4.
A combination lock has three numbers needed to open it. The combination goes from 1 to 49 inclusive. How many combinations are there? (Isn’t our language wonderfully accurate—this is actually a permutation—so it should be called a permutation lock)
(ordered partition)
5.
On a particular lottery there are 50 numbers. In how many ways can a player select six of the numbers? (Will a player still win if the select the numbers in different orders?
What groups are you forming with the numbers?)
6.
Expand ( x + y ) 6 .
Page 5
Kevin Thompson revised 11/20/2006

7.
What is the coefficient of the x 3 y 5 in the expansion of
(
2 x + y
)
8 ? (Think of 2 x as a whole.)
8.
Find the coefficient of x 3 y 9 z in the expansion of ( x + y + z ) 13 ?
9.
A company has established an incentive program to identify and reward their best workers. Among those nominated, 5 are awarded 8% bonuses and all other nominees receive 4% bonuses. In how many ways can the bonuses be assigned if 16 employees have been nominated?
10.
Someone suggests it might just be easier to award 1 st , 2 nd , and 3 rd place bonuses of
15%, 12%, and 10% to the top three employees while all other nominees receive a 4% bonus. If 16 employees are nominated, in how many ways can this be done?
11.
Which decision making process seems easier for the selection committee (in 9 or 10)?
Which process would be easier if there are only 14 nominations the next year?
12.
If the United States Supreme Court voted randomly on a certain case, what is the probability that it would be a 5 to 4 decision? (could be 5 to 4 either way)
Page 6
Kevin Thompson revised 11/20/2006