Sampling for Tickets Probability Problem

HUDM4122

Probability and Statistical Inference

February 16, 2015

In the last class

•

We started Ch. 4.4  in Mendenhall, Beaver, &

Beaver

Today

•

Ch. 4.4-4.6  in Mendenhall, Beaver, & Beaver

Today

•

Sampling without Replacement

•

Permutations

•

Combinations

•

Independence

•

Conditional Probability

We ended last class with this problem

•

I call a radio station, where they make me pick a

number between 1 and 10

•

If I get today’s winning number, I get free tickets

to hear Justin Bieber (oh lucky me)

•

Let’s say I call 5 days in a row

•

What is the probability I get tickets on exactly one

day? (my daughter will still be excited)

Solution A (professor)

Solution B (class)

These represent two different

problems

•

One (my solution) is sampling without

replacement

•

The other is the sum of five independent

events

•

So which one is right?

Let’s first compute the sample space

•

I call a radio station, where they make me pick

a number between 1 and 10

•

I call 5 days in a row

•

10*10*10*10*10 = 100,000

Let’s first compute the sample space

•

I call a radio station, where they make me pick

a number between 1 and 10

•

I call 5 days in a row

•

10*10*10*10*10 = 100,000

•

A.K.A. too many to list out

Let’s first compute the sample space

•

I call a radio station, where they make me pick

a number between 1 and 10

•

I call 5 days in a row

•

10*10*10*10*10 = 100,000

•

A.K.A. too many to list out

–

Or is it?

bieber-tix-example.xlsx

bieber-tix-example.xlsx

•

0.32805

So, the professor was wrong

So, the professor was wrong

•

It turns out sleep deprivation 

is

 bad for

cognition

So, the professor was wrong

•

It turns out sleep deprivation 

is

 bad for

cognition

•

Don’t try this on your midterm

So why was this correct?

5 days to win 1 ticket

5 days

In each of 5 days,

1 answer where right

5 days

Correct answer

And 4 more days where wrong;

and there are 9 wrong answers

5 cases

Correct answer

Wrong answers

Wrong days

Each day is independent from other days –

that’s why this is the correct math

5 cases

Correct answer

Wrong answers

Wrong days

Questions? Comments?

The sample space for multi-stage data

collection can be calculated using

•

 The Extended 

mn 

rule

The Extended 

mn 

rule

•

Let’s say you have k stages of your data

collection

The Extended 

mn 

rule

•

Let’s say you have k stages of data collection

–

Unlike the book, I don’t call it an 

experiment

,

because that term is usually given a more specific

meaning by researchers

The Extended 

mn 

rule

•

Let’s say you have k stages of data collection

•

And there are n

1

 ways to accomplish the first

stage

•

And n

2

 ways to accomplish the 2

nd

 stage

•

And n

3

 ways to accomplish the 3

rd

 stage

•

And n

k

 ways to accomplish the k

th

 stage

•

Then the sample space = n

1 

* n

2 

* n

3 … 

* n

k

Any questions about the

Extended 

mn 

rule?

•

Let’s say you have k stages of data collection

•

And there are n

1

 ways to accomplish the first

stage

•

And n

2

 ways to accomplish the 2

nd

 stage

•

And n

3

 ways to accomplish the 3

rd

 stage

•

And n

k

 ways to accomplish the k

th

 stage

•

Then the sample space = n

1 

* n

2 

* n

3 … 

* n

k

Note that there doesn’t have to be the

same probability in each stage!

This can come in useful in cases that

are not truly independent

•

Unlike 

the Justin Bieber example

Independence

•

Two events A and B are 

independent 

if A does

not affect B and B does not affect A

Which of these are independent?

Which of these are independent?

Which of these are independent?

Which of these are independent?

Which of these are independent?

Which of these are independent?

Which of these are independent?

Which of these are independent?

Example of probability calculation with

non-independent events

•

Let’s say that I invite 6 friends over to play

Beer Hunter

This can come in useful in cases that

are not truly independent

•

Let’s say that 6 friends decide to play Beer

Hunter

•

Rules are

–

6 cans of beer

–

1 violently shaken before playing and then

shuffled

–

Each person chooses a can and opens it

Initial Sample Space = 6

•

1 Bad outcome

•

5 Perfectly fine outcomes

–

If you don’t like beer, imagine it’s root beer

Probability of bad outcome

•

The probability of a bad outcome for friend 1

is 1/6

Probability of bad outcome

•

The probability of a bad outcome for friend 1

is 1/6

•

But if friend 1 comes out OK

•

The probability of a bad outcome for friend 2

is 1/5, not 1/6!

Probability of bad outcome

•

The probability of a bad outcome for friend 1

is 1/6

•

But if friend 1 comes out OK

•

The probability of a bad outcome for friend 2

is 1/5, not 1/6!

•

Does everyone see why?

Probability of bad outcome

•

The probability of a bad outcome for friend 1

is 1/6

•

But if friend 1 comes out OK

•

The probability of a bad outcome for friend 2

is 1/5, not 1/6!

•

Does everyone see why?

–

Friend 1 already opened a beer can, and it went

ok

–

This only leaves 5 closed beer cans

Probability of bad outcome

•

If friend 2 comes out OK

•

The probability of a bad outcome for friend 3

is 1/4

Probability of bad outcome

•

If friend 2 comes out OK

•

The probability of a bad outcome for friend 3

is 1/4

•

If friend 3 comes out OK

•

The probability of a bad outcome for friend 4

is 1/3

Probability of bad outcome

•

If friend 4 comes out OK

•

The probability of a bad outcome for friend 5

is 1/2

Probability of bad outcome

•

If friend 4 comes out OK

•

The probability of a bad outcome for friend 5

is 1/2

•

If friend 5 comes out OK

Probability of bad outcome

•

If friend 4 comes out OK

•

The probability of a bad outcome for friend 5

is 1/2

•

If friend 5 comes out OK

•

Friend 6 will need to get a clean shirt

True sample space

•

6 *5 * 4 * 3 * 2 * 1

•

This is called 

sampling without replacement

Any questions?

Now you do an example

•

In pairs

Now you do an example

•

Let’s say I’m a roadie for the band Van Halen

Now you do an example

•

Let’s say I’m a roadie for the band Van Halen

–

Professors need to moonlight to make ends meet

in this city

Now you do an example

•

Let’s say I’m a roadie for the band Van Halen

•

They have a “no brown M&M’s” clause in their

contract, and if any of the four members get a

brown M&M, I’m fired

Now you do an example

•

Let’s say I’m a roadie for the band Van Halen

•

They have a “no brown M&M’s” clause in their

contract, and if any of the four members get a

brown M&M, I’m fired

•

I’ve just handed them a bowl with 20 M&Ms,

including one brown M&M

•

Each band member takes 1 M&M without

looking

Now you do an example

•

Let’s say I’m a roadie for the band Van Halen

•

They have a “no brown M&M’s” clause in their

contract, and if any of the four members get a

brown M&M, I’m fired

•

I’ve just handed them a bowl with 20 M&Ms,

including one brown M&M

•

Each band member takes 1 M&M without looking

•

What is the sample space?

•

What is the probability I get fired?

Questions? Comments?

Another example: permutations

•

An application of sampling without

replacement

•

How many orderings can you have between a

certain number of objects?

Example

•

Let’s say that I’m redecorating my office in

preparation for a visit from a funder from the US

army (“Bob”), a funder from the National Science

Foundation (“Janet”), and a funder from the US

Department of Education (“Ed”)

•

I want to place Bob’s book, Janet’s book, and Ed’s

book in a place of honor next to my desk

•

How many different orders can I put their books

in?

Example

•

The first book could be Bob’s, Janet’s, or Ed’s

•

If the first book is Bob’s, the second book can

only be Janet’s or Ed’s

•

If the first book is Bob’s, and the second book

is Janet’s, then the third book can only be Ed’s

•

We’re back to the same math of 3*2*1

Any questions?

Formal equations

•

The sample space for n stages, sampling

without replacement, is

•

n * (n-1) * (n-2) * (n-3) * (n-4) until (n-k)=1

•

This is written n!

–

Pronounced “n factorial”

Formal equations

•

The number of permutations for n objects,

taking all of them together, is

•

Still n!

•

n * (n-1) * (n-2) * (n-3) * (n-4) until (n-k)=1

Do it yourself

•

What is the number of permutations for 4

objects?

•

What is the number of permutations for 6

objects?

•

What is the number of permutations for 10

objects?

Do it yourself

•

What is the number of permutations for 4

objects?

–

4*3*2*1=24

•

What is the number of permutations for 6

objects?

–

6*5*4*3*2*1=720

•

What is the number of permutations for 10

objects?

–

10*9*8*7*6*5*4*3*2*1= 3,628,800

Ryan’s daughter suggests

•

“Daddy, why don’t we try organizing all the

books on your bookshelves in every possible

way?”

•

I own approximately 600 books

•

Is this a good idea?

Questions? Comments?

More general case

Using general case equation

•

If I want to find out how many orderings of 2

books I can get from 6 total books

–

n!/(n-r)!

–

6!/(6-2)!

–

6!/4!

–

720/24

–

30 possible orderings

Using general case equation

•

If I want to find out how many orderings of 4

books I can get from 6 total books

–

n!/(n-r)!

–

6!/(6-4)!

–

6!/2!

–

720/2

–

360 possible orderings

Any questions?

Related problem: Combinations

•

If we don’t care about order, but only want to

know how many combinations of items we

can get

Combination formula

Example

•

I have five friends, and three tickets to see

Ferrari Truck

•

How many combinations of friends could I

potentially bring?

Example

Example

•

I have 600 books, and want to take 3 books on

a ridiculously long flight to the First

Uzbekistani Conference on Educational Data

Mining

•

How many combinations of books could I

potentially bring?

Example

Your turn

•

Peter’s Pizzeria has 6 toppings, and a 2-

topping special

•

How many combinations of toppings could

you get, and have the special?

Questions?

An application

•

In my son’s play group, he has 6 playmates

–

5 friends and 1 frenemy

•

What is the probability that on a specific

playdate with 2 friends, it will involve the

frenemy?

Can be written as

Can be written as

•

Number of playdates that involve frenemy = 5

–

Frenemy plus each of 5 friends

•

Total number of combinations (2 of 6) = 15

•

5/15 = 1/3

Any questions?

Conditional Probability

•

Let’s take two non-independent events, A and

B

•

P(A | B) =

•

Probability of A,

•

Given that we know that B occurred

Conditional Probability

•

Let’s take two non-independent events, A and B

•

P(A | B) =

•

Probability of A,

•

Given that we know that B occurred

•

Note that this tells us 

nothing 

about

P(A | ~B)

P(A) overall

General Multiplication Rule

•

Probability of A and B equals

•

Probability of A

•

Multiplied by

•

Probability of B, given A

General Multiplication Rule

Example

Example

You try it

You try it

You try it

You try it

A Concrete Example

•

P(Bob parties late, night before exam) = 0.5

•

P(Bob does badly on exam | parties late) = 0.7

•

P(Bob does badly on exam | ~parties late) = 0.2

•

What is the probability that Bob parties late and

does badly?

A Concrete Example

•

P(Bob parties late, night before exam) = 0.5

•

P(Bob does badly on exam | parties late) = 0.7

•

P(Bob does badly on exam | ~parties late) = 0.2

•

What is the probability that Bob parties late and

does badly?

•

0.35

Conditional Probability Formula

Conditional Probability Formula

Conditional Probability Formula

Conditional Probability Formula

Example

Example

You try it

You try it

What if we want to determine

What if we want to determine

Upcoming Classes

•

2/18 Bayes Theorem

–

Ch. 4-7

–

HW3 due

•

2/23 Discrete Random Variables and Their

Probability Distributions

–

Ch. 4-8

Homework 3

•

Due in 2 days

•

In the ASSISTments system

Questions? Comments?