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Class 02 Probability, Probability Distributions, Binomial Distribution

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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!

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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.

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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)

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Flip a Fair CoinDraw a Card from a well shuffled Deck Observe the weather tomorrow P(Head)=0.5P(Ace)=4/52P(R)= ? Probability Examples

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Probability Fact: The Pr A will not happen is 1 minus the Pr it will happen (and vice versa). Flip a Fair CoinDraw a Card from a well shuffled Deck Observe the weather tomorrow P(Head)=0.5P(Ace)=4/52P(R)= ? P(Tail)=1-0.5P(not an Ace) = 1-4/52P(R c )= 1-? Not A is denoted A c. So if it is difficult to find P(A), try finding P(A c ) 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)

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Consider Two Trials Flip a Fair CoinDraw a Card from a well shuffled Deck Observe the weather tomorrow P(H)=0.5P(Ace)=4/52P(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

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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.

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Consider Two Trials Flip a Fair CoinDraw a Card from a well shuffled Deck Observe the weather tomorrow Pr(H)=0.5Pr(Ace)=4/52Pr(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)

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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?

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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?

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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.

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Consider Two Traits and a randomly selected 2010 ND undergrad AcAc Atotal Female34793823861 Male39355554490 total74149378351 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.

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Consider Two Traits and a randomly selected ND undergrad AcAc Atotal Female34793823861 Male39355554490 total74149378351 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)

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Convert Probs to Table of Counts to make things easy to understand DCDC Dtotal Pos1980902070 Neg7920107930 total990010010,000 Pr(D│Pos) = 90/2010 I have the D with Prob 1% Pr(Pos│D)=90% Pr(Pos│D C )=20% I tested positive. Do I have the disease?

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Convert Probs to Table of Counts to make things easy to understand DCDC Dtotal Pos1980902070 Neg7920107930 total990010010,000 Pr(D│Pos) = 90/2070 = 4.3% I have the D with Prob 1% Pr(Pos│D)=90% Pr(Pos│D C )=20%

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We just used BAYES THEOREM!! See (4.17) or (4.18) or (4.19) to see what the formula looks like.

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Consider 3 independent coin flips. Pr(H,H,H) = 1/8 Pr(H,H,T) = 1/8 Pr(H,T,H) = 1/8 Pr(T,H,H) = 1/8 Pr(H,T,T) = 1/8 Pr(T,H,T) = 1/8 Pr(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.

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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 ways to 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.

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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

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Five Probability Mass Functions Number of Flips No. Heads12345 00.50.250.1250.06250.03125 10.5 0.3750.250.15625 2 0.250.375 0.3125 3 0.1250.250.3125 4 0.06250.15625 5 0.03125 P(x) is never negative. Sum of P(x) over all possible x values is = to 1.

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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 probabilities Pr(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

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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

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Five binomial pmf’s and their mode,mean,var,stddev Number of Flips No. Heads12345 00.50.250.1250.06250.03125 10.5 0.3750.250.15625 2 0.250.375 0.3125 3 0.1250.250.3125 4 0.06250.15625 5 0.03125 Mode0,111,222,3 Mean0.511.522.5 Var0.250.50.7511.25 Std dev0.50.7070.86711.118 P(x) is never negative. Sum of P(x) over all possible x values is = to 1.

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Probability Notation Pr(A c ) = 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

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

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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

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TA Office Hours Tuesday night 7 to 8:30 classroom 266 Assignment Due Next Class My “office” hours Every class day 3 to 430 In the classroom L051

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Tabular Approach to MONTY HALL not My Door Prize MRG200100300 Not000 200100300 Pr(Prize│MRG) = 100/100 = 1/3

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Chapter 5 Probability Distributions

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