Binomial DistributionLecture 7(Lecture 7) 1 / 38Expected ValueE(X ) =Xall kk P(X = k )Interpretation1A probability-weighted average.2Long-run average of X.3The fair value of a gamble.4The balance point for a probability histogram/bargraph.(Lecture 7) 2 / 38Properties of E(X )1E(X + α) = E(X ) + α.2E(b X ) = b E(X ).3E(X + Y ) = E(X ) + E(Y )4If X and Y are independent, thenE(XY ) = E(X ) E(Y )(Lecture 7) 3 / 38Measuring the SpreadVariance = how spread out the pdf is.Variance = average squared distance from the mean:Var (X ) = E(X − E(X))2Var (X ) = E(X2) − (E(X))2Standard Deviationσ =qVar (X )(Lecture 7) 4 / 38PropertiesVar (αX ) = α2Var (X )For example,Var (3 · X ) = 32· Var (X ) = 9 · Var(X )Var (α + X ) = Var(X )For example,Var (3 + X ) = Var(X )If X and Y are independentVar (X + Y ) = Var (X ) + Var(Y )(Lecture 7) 5 / 38Independent EventsIf A and B are independentP(A and B) = P(A) P(B)If A1, A2, . . . , Anare independentP(A1, A2, . . . , and An) = P(A1) P(A2) . . . P(An)(Lecture 7) 6 / 38Flipping Fair CoinsEach outcome has probability 1/2Outcomes are independentIf we flip a coin n times12121212. . .12=12nP(16 Heads) =1216=165536= 0.00001526P(10 Heads then 6 Tails) =1216=165536= 0.00001526(Lecture 7) 7 / 38And Counting HeadsFlip a coin 4 timesX = number of HeadsP(X = 4) = (12)(12)(12)(12) =116P(X = 0) = P(4 Tails) =116P(X = 1) =???(Lecture 7) 8 / 38Sample SpaceHHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTTk 0 1 2 3 4P(X = k )(Lecture 7) 9 / 38Sample SpaceHHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTTk 0 1 2 3 4P(X = k ) 1/16 1/16(Lecture 7) 10 / 38Sample SpaceHHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTTk 0 1 2 3 4P(X = k ) 1/16 1/4 1/16(Lecture 7) 11 / 38Sample SpaceHHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTTk 0 1 2 3 4P(X = k ) 1/16 1/4 1/4 1/16(Lecture 7) 12 / 38Sample SpaceHHHH HTHH THHH TTHHHHHT HTHT THHT TTHTHHTH HTTH THTH TTTHHHTT HTTT THTT TTTTk 0 1 2 3 4P(X = k ) 1/16 1/4 3/8 1/4 1/16(Lecture 7) 13 / 38GenerallyFlip a coin n times2noutcomes in the sample spaceP(one sequence) = (1/2)nP(X = k ) = (1/2)n(# of seq. with k Heads)Question:How do we count these combinations???(Lecture 7) 14 / 38Choose an order(Lecture 7) 15 / 38Choose an order(Lecture 7) 16 / 38Choose an order(Lecture 7) 17 / 38Choose an order(Lecture 7) 18 / 38Choose an order(Lecture 7) 19 / 38Choose an order(Lecture 7) 20 / 38Choose an order(Lecture 7) 21 / 38Counting SequencesIn the above example:(5) (4) (3) (2) 1 = 5!More generallyIn factorialn! = n (n − 1) (n − 2) . . . (3)(2)1For example, 30! = 30 (29) (28) . . . (2)1 = 2.65 × 1032By definition0! = 1(Lecture 7) 22 / 385 choose 2(Lecture 7) 23 / 385 choose 25(Lecture 7) 24 / 385 choose 25 × 4 = 20(Lecture 7) 25 / 385 choose 25 × 4 = 20 ÷ 2 = 10(Lecture 7) 26 / 38n choose knCk=n!k!(n − k )!n = 5, k = 25C2=5!2! 3!=5(4)(3)(2)12(1) × (3)(2)1=1202 × 6= 10n = 6, k = 46C4=6!4! 2!=6(5)(4)(3)(2)1(4)(3)(2)(1) × (2)1=72048= 15n = 52, k = 352C3=52(51)50(3)(2)1=132, 6006= 22, 100(Lecture 7) 27 / 38Binomial Coefficients(Lecture 7) 28 / 38Probability of Flipping 6 of 10P(6 Heads in 10 Tries) =10C61210(Lecture 7) 29 / 38Probability of Flipping 6 of 10(Lecture 7) 30 / 38Probability of Flipping 6 of 1010!6! 4!11024=10(9)(8)(7)4(3)(2)111024=2101024= 0.2051(Lecture 7) 31 / 38Other ProbabilitiesWhat ifP(H) = 0.2 and P(T ) = 0.8Then, P(HT ) = 0.2 (0.8) = 0.16P(HTTHT ) = 0.2(0.8)(0.8)(0.2)(0.8) = (0.2)2(0.8)3= 0.02048P(TTHTH) = 0.8(0.8)(0.2)(0.8)(0.2) = (0.2)2(0.8)3= 0.02048P(TTTHH) = 0.8(0.8)(0.8)(0.2)(0.2) = (0.2)2(0.8)3= 0.02048P( 2 Heads ) =5C2(0.2)2(0.8)3= 10(0.02048) = 0.2048(Lecture 7) 32 / 38Sequence of Independent EventsA1, A2, . . . , Anare independent withP(Ai) = pP(Aci) = 1 − pThen...P(A1, Ac2, and A3) = p (1 − p) p = p2(1 − p)(Lecture 7) 33 / 38ExampleP( Dodgers win ) = 0.58P( Dodgers win twice ) = (0.58)(0.58) = 0.3364The probability to lose 2 games and then win 4:P( Lose 2 then Win 4 ) = (0.42)(0.42)(0.58)(0.58)(0.58)(0.58)= (0.42)2(0.58)4= 0.01996(Lecture 7) 34 / 38Binomial Experiment1n trials2Trials are independent3Two possible outcomesI“Success”, 1I“Failure”, 04P( “Success” ) = pP( “Failure” ) = 1 − p(Lecture 7) 35 / 38Example: Proposition 8Sample of 100 California voters1n = 1002Assume they are Independent (?)3“Support” = “success”4p = 0.44X = # of supporters(Lecture 7) 36 / 38Binomial ProbabilityP(X = k ) =nCkpk(1 − p)n−k- n = total number of trials- p = probability of successP( 40 supporters ) = P(X = 40)=100C40(0.44)40(0.56)60= (1.3746 × 1028)(5.47151 × 10−15)(7.78541 × 10−16)= 0.0586(Lecture 7) 37 / 38Binomial DistributionBinomial Experiment1n trials2Trials are independent3Two possible outcomes4P( “Success” ) = pBinomial ProbabilityP(X = k ) =nCkpk(1 − p)n−k=n!k! (n − k )!pk(1 − p)n−k(Lecture 7) 38 /