Presentation on theme: "Class 02 Probability, Probability Distributions, Binomial Distribution."— Presentation transcript:
Class 02 Probability, 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 CoinDraw a Card from a well shuffled Deck Observe the weather tomorrow P(Head)=0.5P(Ace)=4/52P(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 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)
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
Conditional Probability People who switched to ALLSTATE saved on average $348 per year. 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 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)
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 AcAc Atotal Female Male total 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) = ( )/8351 Any four numbes or %s allows you to fill in everything.
Consider Two Traits and a randomly selected ND undergrad AcAc Atotal Female Male total 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) = ( )/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 DCDC Dtotal Pos Neg total ,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?
Convert Probs to Table of Counts to make things easy to understand DCDC Dtotal Pos Neg total ,000 Pr(D│Pos) = 90/2070 = 4.3% I have the D with Prob 1% Pr(Pos│D)=90% Pr(Pos│D C )=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) = 1/8 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.
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.
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 Flips No. Heads 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 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
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’s and their mode,mean,var,stddev Number of Flips No. Heads Mode0,111,222,3 Mean Var Std dev P(x) is never negative. Sum of P(x) over all possible x values is = to 1.
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
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
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 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
Tabular Approach to MONTY HALL not My Door Prize MRG Not Pr(Prize│MRG) = 100/100 = 1/3