HUDM4122 Probability and Statistical Inference February 16, 2015.

HUDM4122Probability 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)

= 0.40951

Solution B (class)

= 0.32805

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 5 days in a row

• 10*10*10*10*10 = 100,000




• 10*10*10*10*10 = 100,000• A.K.A. too many to list out




• 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


• It turns out sleep deprivation is bad for cognition


• 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

• CXXXX• XCXXX• XXCXX• XXXCX• XXXXC

5 days

In each of 5 days,1 answer where right

5 daysCorrect answer

And 4 more days where wrong; and there are 9 wrong answers

5 casesCorrect answer

Wrong answers

Wrong days

Each day is independent from other days – that’s why this is the correct math

5 casesCorrect 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


• 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


• Let’s say you have k stages of data collection• And there are n1 ways to accomplish the first

stage• And n2 ways to accomplish the 2nd stage

• And n3 ways to accomplish the 3rd stage

• And nk ways to accomplish the kth stage

• Then the sample space = n1 * n2 * n3 … * nk

Any questions about theExtended mn rule?

• Let’s say you have k stages of data collection• And there are n1 ways to accomplish the first

stage• And n2 ways to accomplish the 2nd stage

• And n3 ways to accomplish the 3rd stage

• And nk ways to accomplish the kth stage

• Then the sample space = n1 * n2 * n3 … * nk

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?A B

Flipping a fair coin Flipping same fair coin again


Flipping a fair coin Flipping same fair coin againFlipping a biased coin Flipping same biased coin again



Bob parties late night before midterm

Bob falls asleep during midterm





Bob’s grade on midterm Bob’s grade on final





Bob’s grade on midterm Bob’s grade on finalBob disrupts class Bob gets expelled





Bob’s grade on midterm Bob’s grade on finalBob disrupts class Bob gets expelledBob gets expelled Bob takes job at McDonald’s






Bob takes job at McDonald’s Bob wins lottery






Bob takes job at McDonald’s Bob wins lotteryBob wins lottery Bob becomes ill due to

congenital heart problem

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



• But if friend 1 comes out OK• The probability of a bad outcome for friend 2

is 1/5, not 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?



• 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


• If friend 2 comes out OK• The probability of a bad outcome for friend 3

is 1/4



is 1/4


is 1/3



is 1/2



is 1/2

• If friend 5 comes out OK



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


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


• Let’s say I’m a roadie for the band Van Halen– Professors need to moonlight to make ends meet

in this city


• 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




• 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?


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?



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?


More general case

• If we only want to pick r objects out of the n total objects, the equation becomes

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

• The number of combinations of r objects out of n total objects is

Example

• I have five friends, and three tickets to see Ferrari Truck

• How many combinations of friends could I potentially bring?

Example

• I have five friends, and three tickets to see Ferrari Truck

• How many combinations of friends could I potentially bring?

• = = = = = 10

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

• 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?

• = = = =

• 35.8 million

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

• Number of playdates that involve frenemy = 5– Frenemy plus each of 5 friends

• Total number of combinations (2 of 6)

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

• Probability of A and B equals • Probability of A• Multiplied by • Probability of B, given A

• Formally

Example

• P(Q) = 0.2• P(R|Q) = 0.7

• P(Q R) = ?

Example

• P(Q) = 0.2• P(R|Q) = 0.7

• P(Q R) = 0.14

You try it

• P(F) = 0.7• P(G|F) = 0.5

• P(F G) = ?

You try it

• P(F) = 0.7• P(G|F) = 0.5

• P(F G) = 0.35

You try it

• P(X) = 0.3• P(X|Y) = 0.9

• P(X Y) = ?

You try it

• P(X) = 0.3• P(X|Y) = 0.9

• P(X Y) = ?

• Impossible to calculate from information given

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

• A mathematical transformation of the General Multiplication Rule

• That rule was


• If you divide both sides by P(A)

• Which resolves to


• P(B|A) =


• P(B|A) =

• Note that P(A) can’t equal 0, or you’re dividing by 0…

Example

• P(B|A) =

• P(Bob parties late, night before exam) = 0.5• P(Bob does badly on exam AND parties late) =

0.35

• What is the probability that does badly, given that he parties late?

Example

• P(B|A) =

• P(Bob parties late, night before exam) = 0.5• P(Bob does badly on exam AND parties late) =

0.35

• What is the probability that does badly, given that he parties late? – 0.7

You try it

• P(B|A) =

• P(Student is from a wealthy family) = 0.25• P(Student goes to college AND comes from

wealthy family) = 0.21

• What is the probability that a student goes to college, given that he/she comes from a wealthy family?

You try it

• P(B|A) =

• P(Student is from a wealthy family) = 0.25• P(Student goes to college AND comes from wealthy

family) = 0.21

• What is the probability that a student goes to college, given that he/she comes from a wealthy family?– 0.84

What if we want to determine

• P(B|A) =

• P(Student is from a wealthy family) = 0.25• P(Student goes to college AND comes from

wealthy family) = 0.21

• What is the probability that a student comes from a wealthy family, given that he/she went to college?

What if we want to determine

• P(B|A) =

• P(Student is from a wealthy family) = 0.25• P(Student goes to college AND comes from wealthy

family) = 0.21

• What is the probability that a student comes from a wealthy family, given that he/she went to college?– We can’t tell, using this formula…

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


HUDM4122 Probability and Statistical Inference February 16, 2015.

Documents

wrong slide

beaver slide

professor slide

cognition slide

excited slide

midterm slide

data collection slide

days correct answer