Week 5 Lecture Math 221 Mar 2012

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

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

 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

 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.

 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).

 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).

 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).

 Minitab

 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).


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.

 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

 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}

 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

 Conditional Probability Examples
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)
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)

 Probability Distributions
 All Probabilities must be between 0 and 1and 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)

 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)

 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

 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

 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

 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.)

 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.)

 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.

 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.

 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”


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

 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)

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

 P(X≥5)

P(X≥5) = P(X=5) + P(X=6) =
0.001536 + 0.000064 =
0.0016

 P(X<3)

P(X<3) = P(X=0) + P(X=1) +
P(X=2) = 0.262144 +
0.393216 + 0.245760 =
0.90112

 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”

 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

 Pivot Tables

Chocolate Vanilla Total

Girls 13 6 19

Boys 17 5 22

Total 30 11 41

 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

 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

 Pleasenote that I posted a presentation in
the Statcave last week that shows how to do
factorials, combinations and permutations in
Excel

 AlsoI posted a presentation on doing
binomial, geometric and Poisson distribution
calculations in Excel.

 Both of these may be helpful to you!

 Comesee me and download the charts at
www.facebook.com/statcave

 You don’t have to be a Facebook person….

Week 5 Lecture Math 221 Mar 2012

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Week 5 Lecture Math 221 Mar 2012