Nov 2012 Week 5 Lecture for Math 221

1. B. Heard
These charts are not to be posted or used without
my written permission. Students can download a
copy for their personal use.

2.  Preparing for the Week 5 Quiz
 Factorials
 Combinations/Permutations
 Probability
 Probability Distributions
 Discrete/Continuous Distributions
 Binomial Distribution
 Poisson Distribution
 Pivot Tables

3.  Factorials
 n! simply means n x (n-1) x (n-2) x …. 3 x 2 x 1
 For example 5! = 5x4x3x2x1 = 120, so 5!=120
 Always remember that 0! = 1 (NOT ZERO)
 Additional examples
 3! + 4! = (3x2x1) + (4x3x2x1) = 6 + 24 = 30
 5(5-3)! = 5 x 2! = 5 x (2x1) = 5x2 = 10
 5(5! – 3!) = 5(120 – 6) = 5(114) = 570
 4!(0!) = 24(1) = 24
 3!/0! = 6/1 = 6

4.  Combinations/Permutations
 Combinations – The number of ways of something
happening when order DOES NOT matter.
 Permutations – The number of ways of something
happening when order DOES matter.
 The number of ways to pick 5 out of 10 players to
start on a basketball team would be a combination.
 The number of ways to pick 5 out of 10 players to
play 5 different positions would be a permutation.

5.  Combinations/Permutations (continued)
 The number of ways to pick 3 committee
members out of 11 people would be a
combination because there are not distinct
positions mentioned (order doesn’t matter).
 The number of ways to pick 3 committee
members out of 11 people where 1 would be the
President, Vice-President and Treasurer would be
a permutation because there are distinct
positions mentioned (order does matter, you not
only could be picked, but you could be one of
three different positions).

6.  Combinations/Permutations (continued)
 Let’s Do these in Minitab
 The number of ways to pick 3 committee members out of 11
people would be a combination because there are not distinct
positions mentioned (order doesn’t matter).
 This is a combination where we are picking 3 from 11 (Sometimes noted
11C3).
 The number of ways to pick 3 committee members out of 11
people where 1 would be the President, Vice-President and
Treasurer would be a permutation because there are distinct
positions mentioned (order does matter, you not only could be
picked, but you could be one of three different positions).
 This is a permutation where we are picking 3 from 11 (Sometimes noted
11P3).

7.  Combinations/Permutations (continued)
 Minitab
 Let’s look at the combination where we are picking 3 from 11
(Sometimes noted 11C3).
 In Minitab, Choose “Calc”, then “Calculator”
 Type C1, or any other blank column in the “Store result in variable” box
 In the Expression box, type “Combinations(11,3)”
 In the cell you chose, you will see your answer of 165
 This means there are 165 ways to pick a committee of 3
from 11 people (remember order didn’t matter).

8.  Combinations/Permutations (continued)
 Minitab

9.  Combinations/Permutations (continued)
 Minitab
 Let’s look at the permutation where we are picking 3 from 11
(Sometimes noted 11P3).
 In Minitab, Choose “Calc”, then “Calculator”
 Type C1, or any other blank column in the “Store result in variable” box
 In the Expression box, type “Permutations(11,3)”
 In the cell you chose, you will see your answer of 990
 This means there are 990 ways to pick a committee of 3
from 11 people where there are THREE DISTINCT
POSITIONS (order DID matter).

10.  Combinations/Permutations (continued)
 Minitab

11.  Combinations/Permutations (continued)
For the same numbers, the number of permutations will
always be larger! This is because there are distinct
positions.
As an example
3C3 = 1 (there is only one way to pick three from three)
3P3 = 6 (there are 6 ways to pick 3 people from 3 to serve
in 3 distinct positions, think about it 3 could be
President, 2 are left to serve as Vice-President, one is
left to serve as Treasurer, 3x2x1 = 6
In our previous example we had 11P3 = 990 which could be
looked at as 11 x 10 x 9 = 990. Why? 11 could be Pres.,
10 VP, and 9 Treasurer.

12.  Probability
 Probability is simply the number of desired
events over the total number of events that can
happen.
 Sound complicated? It’s not
 What is the probability of rolling a 5 on six-sided die?
 One side with a five/six sides = 1/6

13.  Probability (continued)
 Sample Spaces (What can happen?)
 For a regular light switch, the sample space would be
{on, off}
 For a six-sided die, the sample space would be
{1,2,3,4,5,6}
 For a new baby, the sample space would be {girl, boy}
 For suits in a card deck, the sample space would be
{hearts, diamonds, clubs, spades}

14.  Probability (continued)
 Examples
 What is the probability of drawing a Jack from a
standard deck of cards? (There are 4 cards out of the
52 that are Jacks) The probability would be 4/52 or
1/13 simplified
 What is the probability of drawing a red Jack from a
standard deck of cards? (There are 2 cards out of the
52 that are red Jacks) The probability would be 2/52
or 1/26 simplified
 What is the probability of drawing a Jack of Hearts
from a standard deck of cards? (There is only 1 Jack of
Hearts out of the 52) The probability would be 1/52

15.  Probability (continued)
 Conditional Probability Examples
 What is the probability of drawing a Jack from a
standard deck of cards, if you first drew a 3 of clubs
and didn’t replace it? (There are 4 cards out of the 51
that are Jacks) The probability would be 4/51
(Remember you didn’t replace the 3 of clubs)
 What is the probability of drawing a Jack from a
standard deck of cards, if you first drew a Jack of
clubs and didn’t replace it? (There are 3 Jacks left out
of the 51) The probability would be 3/51 or 1/17
simplified (Remember you didn’t replace the Jack of
clubs)

16.  Probability Distributions
 All Probabilities must be between 0 and 1 and
the sum of the probabilities must be equal to 1.
 Examples
 If X = {5, 10, 15, 20} and P(5) = 0.10, P(10) =
0.20, P(15) = 0.30, and P(20) = 0.40, can the
distribution of the random variable X be
considered a probability distribution?
 YES, because all probabilities (0.10,0.20,0.30,0.40)
are between 0 and 1 and they add up to 1
(0.10+0.20+0.30+0.40 = 1)

17.  Probability Distributions (continued)
 Examples
 If X = {5, 10, 15, 20} and P(5) = 0.20, P(10) =
0.20, P(15) = 0.20, and P(20) = 0.20, can the
distribution of the random variable X be
considered a probability distribution?
 No, the probabilities (0.20,0.20,0.20,0.20) are
between 0 and 1 BUT they DO NOT add up to 1
(0.20+0.20+0.20+0.20 = 0.80)

18.  Probability Distributions (continued)
 Examples
 If X = {-5, A, 7, 0} and P(-5) = 0.50, P(A) = 0.10,
P(7) = 0.10, and P(0) = 0.30, can the distribution
of the random variable X be considered a
probability distribution?
 YES, the probabilities (0.50,0.10,0.10,0.30) are
between 0 and 1 and they add up to 1
(0.50+0.10+0.10+0.30 = 1)
 {-5, A, 7, 0} Does Not matter, you can have negative
numbers, letters, colors, names, etc. in the sample
space but you couldn’t have negative values as
PROBABILITIES

19.  Probability Distributions (continued)
 Examples
 Given the random variable X = {100, 200} with
P(100) = 0.7 and P(200) = 0.3. Find E(X).
 Simple
 E(X) = Sum of X(P(X) (add the values times their
probabilities)
 E(X) = 100(0.7) + 200(0.3) = 70 + 60 = 130

20.  Probability Distributions (continued)
 Examples of Probability values
 Which of these can be probability values?
 3/5 YES
 0.004 YES
 1.32 NO
 43% YES
 -0.67 NO
 1 YES
 5 NO
 0 YES

21.  Discrete/Continuous Distributions
 Simple explanation
A discrete probability distribution has a
finite number of possible outcomes. (People,
items, distinct things that can not be
measured infinitesimally)
A continuous probability is based on a
continuous random variable such as a persons
height or weight. (GOOD EXAMPLES ARE
UNITS OF MEASURE LIKE HEIGHTS, WEIGHTS,
VOLUME, ETC.)

22.  Discrete/Continuous Distributions
 Simple Example
 The number of cans of soda in your refrigerator is
discrete (0,1,2,3 etc.)
 The amount of soda IN the can is continuous
(ounces can be split and split and split, etc.)

23.  Binomial
and Poisson Distributions
Know the difference between the two!
A good hint is that a Poisson usually give you
an average number of something per time
period and a Binomial gives you a probability
and a number of times/trials/etc.

24.  Binomial Distribution using Minitab
 The way is to show you an example.
 Let’s
say we have a binomial experiment
with p = 0.2 and n = 6 and you are asked to
set up the distribution and show all x values
and the mean, variance and standard
deviation.

25.  Binomial Distribution using Minitab
 Open a new Minitab Project
 Since n= 6, put 0,1,2,3,4,5,6 in column C1
 (Don’t forget the zero)
 Go to Calc>>Probability Distributions>>Binomial
 Change Radial Button to “Probability”
 Put 6 in for number of trials and 0.2 in for event
probability
 Put “C1” in for Input Column
 Click “OK”

26.  Binomial Distribution using Minitab
You would write
X = {0,1,2,3,4,5,6}
P(x=0) = 0.2621
P(x=1) = 0.3932
P(x=2) = 0.2458
This is your
Etc. distribution, we P(x=3) = 0.0819
now have to
calculate the P(x=4) = 0.0154
mean, variance
and standard P(x=5) = 0.0015
deviation.
P(x=6) = 0.0001

27.  Binomial Distribution using Minitab
 Remember we had p = 0.2 and n = 6
 Mean?
 E(X) = np so E(X) = 6(0.2) = 1.2 (the mean)
 Why “E(X)”? Because we would “expect” the outcome
1.2 times out of the 6 times we did it.
 Variance?
 V(X) = npq (q is just “1-p”), so
 V(X) = 6(0.2)(1-0.2) = 6(0.2)(0.8) = 0.96 (the variance)
 Standard Deviation?
 Std Dev. = Square Root of the Variance = √0.96 =
0.9216 (the standard deviation)

28.  Binomial Distribution using Minitab
 Same Example, what if you were asked
 P(X≥5)
 P(X<3)
 Etc.
 Use your probabilities (next page)

29.  Binomial Distribution using Minitab
 P(X≥5)
P(X≥5) = P(X=5) + P(X=6) =
0.001536 + 0.000064 =
0.0016

30.  Binomial Distribution using Minitab
 P(X<3)
P(X<3) = P(X=0) + P(X=1) +
P(X=2) = 0.262144 +
0.393216 + 0.245760 =
0.90112

31.  Poisson Distribution with Minitab
Find P(x=3)
For a Poisson Distribution with
mean = 5
Probability Density Function
Poisson with mean = 5
x P( X = x )
3 0.140374
Poisson Distribution using Minitab
Go to Calc>>Probability
Distributions>>Poisson
Change Radial Button to “Probability”
Put 5 in for number of trials and 3 in
for “Input Constant”
Click “OK”

32.  Poisson Distribution with Minitab
What is the probability that X≤3?
Use Cumulative Distribution Function
Poisson with mean = 5
x P( X <= x )
3 0.265026

33.  Pivot Tables
Chocolate Vanilla Total
Girls 13 6 19
Boys 17 5 22
Total 30 11 41

34.  Pivot Tables
Find P(Girl) Chocolate Vanilla Total
P(Girl) = 19/41
Girls 13 6 19
Find P(Vanilla)
P(Vanilla) = 11/41
Boys 17 5 22
Find P(Girl who likes
Chocolate)
Total 30 11 41
13 out of the 41 so it is
13/41

35.  Pivot Tables
Find P(Girl given they Chocolate Vanilla Total
like chocolate)
P(Girl|Choc) = 13/30 Girls 13 6 19
Find P(Like Vanilla
Boys 17 5 22
given they are a boy)
P(Vanilla|Boy) = 5/22
Total 30 11 41