Class 02Probability, Probability Distributions, Binomial

Distribution

What we learned last class…• We are not good at recognizing/dealing with randomness

– Our “random” coin flip results weren’t streaky enough.• If B/G results behave like independent coin flips, we know

how many families to EXPECT with 0,1,2,3,4 girls.– We expect 6/16 4-child families to have 2 each.– This is PROBABILITY

• We will compare the actual counts to the expected counts to judge whether the coin flip assumption is a good one.– To do this comparison, we will have to recognize that there will be

differences between actual and expected counts even if the coin flip assumption is a good one. • That is STATISITCS!

Probability is useful

• To make better (thoughtful) decisions.– Lend or reject.– Operate or wait and see.– Bunt or hit away.

• To help make sense of data– By comparing what happened to what can happen

by chance.

The First Probability Problem

Two men play chess. The first to win three games will receive two ducats. Play is interrupted with player A ahead 2 games to 1. How should the prize be divided between the two men? (circa 1400)

Flip a Fair Coin Draw a Card from a well shuffled Deck

Observe the weather tomorrow

P(Head)=0.5 P(Ace)=4/52 P(R)= ?

Probability Examples

Probability Fact: The Pr A will not happen is 1 minus the Pr it will happen (and vice versa).

Flip a Fair Coin Draw a Card from a well shuffled Deck

Observe the weather tomorrow

P(Head)=0.5 P(Ace)=4/52 P(R)= ?

P(Tail)=1-0.5 P(not an Ace) = 1-4/52 P(Rc)= 1-?

Not A is denoted Ac.

So if it is difficult to find P(A), try finding P(Ac) instead.

P(3 or fewer girls in 4) = 1 – P(4 boys)

P(some students here have the same birthday) = 1 – P(all have different birthdays)

(4.5)

Consider Two TrialsFlip a Fair Coin Draw a Card from a well

shuffled DeckObserve the weather

tomorrow

P(H)=0.5 P(Ace)=4/52 P(R)= ?

P (H,H)=(0.5)(0.5) P(Ace,Ace) = (4/52)(3/51) P(R1,R2)=P(R1)*P(R2│R1)

P(AandB) is written as P(A∩B) or P(A,B)

P(A∩B) = P(A) * P(B│A) always. THE MULTIPICATION LAW (4.12)

B and A are INDEPENDENT if Pr(B│A) = P(B) and vice versa. (4.9)

So Pr(A∩B) = P(A) * P(B) if A and B are independent. (4.13)

Prob of B given A

Conditional Probability

People who switched to ALLSTATE saved on average $348 per year.

http://www.couponsnapshot.com/merchant-Allstate-coupons-deals-5106.html

P(Amount of Saving│You swithed) does not equal P(Amount of Savings)

“Amount of Saving” and “Switching” are NOT independent.

Consider Two TrialsFlip a Fair Coin Draw a Card from a well

shuffled DeckObserve the weather

tomorrow

Pr(H)=0.5 Pr(Ace)=4/52 Pr(R)= ?

Pr(H,H)=(0.5)(0.5) Pr(Ace,Ace) = (4/52)(3/51) Pr(R1,R2)=Pr(R1)*Pr(R2│R1)

Pr(AandB) is written as Pr(A∩B)

Pr(A∩B) = P(A) * P(B│A) always.

B and A are INDEPENDENT if Pr(B│A) = P(B) and vice versa.

Pr(A∩B) = P(A) * P(B) if A and B are independent.

Coin Flips are independent Card draws are

not. (Unless we replace the first

card or the deck is HUGE)

Independence is often THE question

• Are boy/girl outcomes independent?– Does P(fourth child is a boy) change based on first

three outcomes?• Do players get “hot” or “in the zone”?• Does past fund performance predict future

performance?

The Monty Hall Problem• Three doors. Prize behind one, goats behind the

other two.• I pick a door.• Monty (who knows where the prize is) reveals a

goat. (Assume he ALWAYS reveals a goat).• What is the probability the prize is behind my

door?

INDEPENDENCE solves the Monty Hall Problem

• P(Monty reveals a goat) = 1• P(Monty reveals a goat │ my door has prize) = 1• Events “Monty reveals a goat” “my door has prize”

are INDEPENDENT.• P(my door has prize) = 1/3• P(my door has prize │Monty reveals a goat) = 1/3• So….if I switch to the other unopened door…I win the

prize with probability 2/3.

Consider Two Traits and a randomly selected 2010 ND undergrad

Ac A total

Female 3479 382 3861

Male 3935 555 4490

total 7414 937 8351

Pr(A) = 937/8351

Pr(F) = 3861/8351

Pr(A│F) = 382/3861

Pr(F│A) = 382/937

Pr(A∩F) = 382/8351

Pr(AUF) = (3479+382+555)/8351

Any four numbes or %s allows

you to fill in everything.

Consider Two Traitsand a randomly selected ND undergrad

Ac A total

Female 3479 382 3861

Male 3935 555 4490

total 7414 937 8351

Pr(A) = 937/8351

Pr(F) = 3861/8351

Pr(A│F) = 382/3861

Pr(F│A) = 382/937

Pr(A∩F) = 382/8351

Pr(AUF) = (3479+382+555)/8351

Events A,F are NOT

independent

Also P(A)*P(F│A)

Convert Probs to Table of Counts to make things easy to understand

DC D total

Pos 1980 90 2070

Neg 7920 10 7930

total 9900 100 10,000

Pr(D│Pos) = 90/2010

I have the D with Prob 1%

Pr(Pos│D)=90%

Pr(Pos│DC)=20%

I tested positive. Do I have the disease?

Convert Probs to Table of Counts to make things easy to understand

DC D total

Pos 1980 90 2070

Neg 7920 10 7930

total 9900 100 10,000

Pr(D│Pos) = 90/2070 = 4.3%

I have the D with Prob 1%

Pr(Pos│D)=90%

Pr(Pos│DC)=20%

We just used BAYES THEOREM!!

See (4.17) or (4.18) or (4.19) to see what the formula looks like.

Consider 3 independent coin flips.

Pr(H,H,H) = 1/8

Pr(H,H,T) = 1/8Pr(H,T,H) = 1/8Pr(T,H,H) = 1/8

Pr(H,T,T) = 1/8Pr(T,H,T) = 1/8Pr(T,T,H) = 1/8

Pr(T,T,T) = 1/8

Pr(3H) = 1/8

Pr(2H) = 3/8

Pr(1H) = 3/8

Pr(0H) = 1/8

Addition law

This is a probability Distribution

It is a schedule that assigns the unit of

probability to the set of possible numeric

outcome.

Random Variable X is the number of heads in

3 flips.X is discrete (takes on

only a few values), and this is a probability

MASS function.

The Addition Law

P(AUB) = P(A) + P(B) – P(A∩B) (4.6) = P(A) + P(B) if A,B are MUTUALLY EXCLUSIVE

A and B are mutually exclusive if P(A∩B)=0

So P(1H in 3 tosses) = P(H,T,T) + P(T,H,T) + P(T,T,H)because there are three mutually exclusive waysto throw 1 H in three flips.

I never use this.

I use this instead... I figure out ALL the possible mutually exclusive outcomes and ADD the

probabilities of those that apply.

Don’t Make this mistake• P(H1UH2) = P(H1) + P(H2) = ½ + ½ = 1– Because H1 H2 are not mutually excusive (both can

happen….neither can happen)

• P(H1UH2) = P(H1)+P(H2)-P(H1∩H2) = ½ + ½ - ¼.• P(H1UH2) = P(H1,T2) + P(H1,H2) + P(T1,H2)• = ¼ + ¼ + ¼

Two correct ways

Five Probability Mass Functions

Number of FlipsNo.

Heads 1 2 3 4 50 0.5 0.25 0.125 0.0625 0.031251 0.5 0.5 0.375 0.25 0.156252 0.25 0.375 0.375 0.31253 0.125 0.25 0.31254 0.0625 0.156255 0.03125

P(x) is never negative.

Sum of P(x) over all possible x

values is = to 1.

The Binomial (family) of pmf’s.

• Assumptions– Random variable X is the number of successes in n

independent trials with p(success) = p on each trial.

• Parameters– The binomial has two parameters: n and p

• Calculation of the probabilitiesPr(x successes) = BINOMDIST(x,n,p,false)Pr(x or fewer successes) = BINOMDIST(x,n,p,true)

Important word

p can be any number between 0 ad 1

EMBS: 5.4

Characteristics of any pmf• MODE (most likely). The x value with the highest probability.

– For the binomial, table the pmf to find the mode.• MEAN (or expected value). The probability-weighted average X

– Sum over all possible x values of x*P(x)– For the binomial, the mean will be n*p

• VARIANCE. The probability-weighted average squared distance from the mean.– Sum of (x-mean)^2 * p(x)– For the binomial, VAR(X) = n*p*(1-p)

• STANDARD DEVIATION. The square root of the variance.– Since VARIANCE is average squared distance, STANDARD DEVIATION will be

an “average distance”.It is okay if, at this point, you do not appreciate VARIANCE and STANDARD DEVIATION

EMBS: 5.2, 5.3

Five binomial pmf’sand their mode,mean,var,stddev

Number of FlipsNo.

Heads 1 2 3 4 50 0.5 0.25 0.125 0.0625 0.031251 0.5 0.5 0.375 0.25 0.156252 0.25 0.375 0.375 0.31253 0.125 0.25 0.31254 0.0625 0.156255 0.03125

Mode 0,1 1 1,2 2 2,3Mean 0.5 1 1.5 2 2.5

Var 0.25 0.5 0.75 1 1.25Std dev 0.5 0.707 0.867 1 1.118

P(x) is never negative.

Sum of P(x) over all possible x

values is = to 1.

Probability NotationPr(Ac) = Prob A does not happen = 1 – Pr(A)

Pr(A│B) = Prob A given B = Pr(A∩B)/Pr(B)

Pr(A∩B) = Prob A and B = Pr(A) *Pr(B│A) = Pr(B)*Pr(A│B)

Pr(AUB) = Prob A or B = Pr(A) + Pr(B) – Pr(A∩B)

Just create a table of counts and go from there…..or maybe draw a probability

tree to enumerate all possible outcomes

A Probability DistributionA schedule that assigns the unit of probability to the possible values taken on by a random variable (number)

A Probability Mass FunctionWhen the random variable is discrete, it’s probability distribution is a probability MASS function because probability MASSES on each possible discrete outcome value.

Characteristics of any probability distribution

Mode (most likely), Mean (expected value), variance, standard deviation.

EMBS: 5.1, 5.2, 5.3

The Binomial Pmf

• Applies to the number of success in n independent trials.

• Parameters are n and p.• Mean (expected value) is n*p• Variance is n*p*(1-p)• Standard deviation is sqrt(n*p*(1-p))• =binomdist(X,n,p,false) to find a probability the

binomial random variable =‘s X.• = binomdist(X,n,p,true) to find the probabilit the

binomial random variable is <= X.EMBS: 5.4

TA Office HoursTuesday night

7 to 8:30 classroom 266

Assignment Due Next Class

My “office” hoursEvery class day 3 to 430In the classroom L051

Tabular Approach to MONTY HALL

not My Door Prize

MRG 200 100 300

Not 0 0 0

200 100 300

Pr(Prize│MRG) = 100/100 = 1/3