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Class 02 Probability, Probability Distributions, Binomial Distribution

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.

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Class 02 Probability, Probability Distributions, Binomial Distribution

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  1. Class 02Probability, Probability Distributions, Binomial Distribution

  2. 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!

  3. 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.

  4. 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)

  5. Probability Examples

  6. Probability Fact: The Pr A will not happen is 1 minus the Pr it will happen (and vice versa). (4.5) 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)

  7. Consider Two Trials Prob of B given A 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)

  8. Conditional Probability People who switched to ALLSTATE saved on average $348 per year. P(Amount of Saving│Youswithed) does not equal P(Amount of Savings) “Amount of Saving” and “Switching” are NOT independent. http://www.couponsnapshot.com/merchant-Allstate-coupons-deals-5106.html

  9. Consider Two Trials 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)

  10. 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?

  11. 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?

  12. 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.

  13. Consider Two Traits and a randomly selected 2010 ND undergrad Any four numbesor %s allows you to fill in everything. 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

  14. Consider Two Traitsand a randomly selected ND undergrad Events A,F are NOT independent Also P(A)*P(F│A) 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

  15. Convert Probs to Table of Counts to make things easy to understand I have the D with Prob 1% Pr(Pos│DC)=20% Pr(Pos│D)=90% I tested positive. Do I have the disease? Pr(D│Pos) = 90/2010

  16. Convert Probs to Table of Counts to make things easy to understand I have the D with Prob 1% Pr(Pos│DC)=20% Pr(Pos│D)=90% Pr(D│Pos) = 90/2070 = 4.3%

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

  18. Consider 3 independent coin flips. This is a probability Distribution 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 It is a schedule that assigns the unit of probability to the set of possible numeric outcome. Addition law 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.

  19. The Addition Law I never use this. 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 use this instead... I figure out ALL the possible mutually exclusive outcomes and ADD the probabilities of those that apply.

  20. 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) Two correct ways • 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

  21. Five Probability Mass Functions P(x) is never negative. Sum of P(x) over all possible x values is = to 1.

  22. 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

  23. 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

  24. Five binomial pmf’sand their mode,mean,var,stddev P(x) is never negative. Sum of P(x) over all possible x values is = to 1.

  25. Probability Notation Pr(Ac) = Prob A does not happen = 1 – Pr(A) Just create a table of counts and go from there…..or maybe draw a probability tree to enumerate all possible outcomes 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)

  26. 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

  27. 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

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

  29. Tabular Approach to MONTY HALL Pr(Prize│MRG) = 100/100 = 1/3

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