Binomial Distribution (continued)Lecture 8(Lecture 8) 1 / 19Counting Sequencesn Factorialn! = n (n − 1) (n − 2) . . . (3)(2)10! = 1Examples:4! = 4 (3) (2) 1 = 245! = 5 (4) (3) (2) 1 = (5) 4! = 12010! = 10 (9) (8) (7) (6) 5! = 3, 628, 800(Lecture 8) 2 / 19Combinationsn choose knCk=n!k! (n − k)!n = 6, k = 46C4=6!4! (6 − 4)!=6(5)(4)(3)(2)1(4)(3)(2)(1) × (2)1=72048= 15n = 20, k = 420C4=20!4! (20 − 4)!=20 (19) (18) (17) 16!4(3)(2)1 × 16!=116, 28024= 4, 845(Lecture 8) 3 / 19ExampleWe have a pool of- 15 true-false (T/F) questions and- 20 multiple-choice questions.Question: We want to make an exam with 10 questions exactly 4 ofwhich are T/F. How many exams with exactly 4 T/F questions can beconstructed?Note: Ignore the order of the questions.(Lecture 8) 4 / 19ExampleThe total number of ways to choose 4 T/F questions is15C4=15!4! 11!= 1, 365The total number of ways to choose 6 multiple choice questions is20C6=20!6! 14!= 38, 760The total number of exams with 4 T/F and 6 multiple choicequestions are15C4×20C6= 1, 365 × 38, 760 = 52, 907, 400(Lecture 8) 5 / 19Another ExampleWe have a pool of 15 true-false questions and 20 multiple-choicequestions. Among exams with 10 questions, what is the probability topick an exam with 4 T/F questions (and 6 multiple choice)?Recall the classical definition of probability:P(4 T/F among 10 Questions) =# of exams with 4 T/F Questions# of exams with 10 Questions=15C4×20C635C10=52, 907, 400183, 579, 396= 0.2882(Lecture 8) 6 / 19Probability of Flipping 6 of 10P(6 Heads in 10 Tries) =10C61210(Lecture 8) 7 / 19Probability of Flipping 6 of 10(Lecture 8) 8 / 19Probability of Flipping 6 of 1010!6! 4!11024=10(9)(8)(7)4(3)(2)111024=2101024= 0.2051(Lecture 8) 9 / 19Other ProbabilitiesFlip a Biased Coin 5 timesP(H) = 0.2 and P(T) = 0.8P(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 8) 10 / 19Binomial Experiment1n trials2Trials are independent3Two possible outcomesI“Success”, 1I“Failure”, 04P( “Success” ) = pP( “Failure” ) = 1 − pp is constant for all trials.(Lecture 8) 11 / 19ExampleYou flip a biased coin 100 times1n = 1002Tosses are Independent3“Heads” = “success”“Tails” = “failure”4p = P( “Heads” ) = 0.44 (this is given)1 − p = P( “Tails” ) = 0.56X = # of Heads(Lecture 8) 12 / 19Binomial ProbabilityP(X = k) =nCkpk(1 − p)n−kWhat is the probability that 40 Heads come up in the 100 tosses?P( 40 Heads ) = P(X = 40)=100C40(0.44)40(0.56)60= (1.3746 × 1028) (5.47151 × 10−15) (7.78541 × 10−16)= 0.0586(Lecture 8) 13 / 19Binomial 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 8) 14 / 19Example: The sock drawerA drawer contains 6 red and 4 black socks. What is the probability thatyou select 2 red socks when you pick 5 out of the drawer (Withreplacement)?1There are 5 trials.2Each draw is independent of the other, since we choose a sockeach time with replacement.3Since we are interested in selecting red socks, we call this asuccess and selecting a black sock is a failure.4For each trial, the probability of success is p = 6/10 = 0.6 isconstant.This is a Binomial experiment.(Lecture 8) 15 / 19Example: The sock drawerA drawer contains 6 red and 4 black socks. What is the probability thatyou select 2 red socks when you pick 5 out of the drawer (Withreplacement)?X : # of Red socks (i.e. successes) in 5 trials.P(X = 2) =5C2·6102·1 −6105−2= 10 · (0.6)2· (0.4)3= 0.2304.(Lecture 8) 16 / 19Example: The sock drawerA drawer contains 6 red and 4 black socks. What is the probability thatyou select at least 4 red socks when you pick 5 out of the drawer (withreplacement)?P(X ≥ 4) = P(X = 4) + P(X = 5)=5C4· (0.6)4· (0.4)1+5C5· (0.6)5· (0.4)0= 5 · (0.6)4· (0.4)1+ (0.6)5· (0.4)0= 0.2592 + 0.0778 = 0.3369(Lecture 8) 17 / 19“At least one...”I play the pick-3 (out of 10 with replacement) every day for 300 days.What is the probability I win at least once?(the probability of success is calculated based on selecting 3 out of 10with replacement)P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + . . .TrickP(X ≥ 1) = 1 − P(X = 0)= 1 −300C0(0.001)0(0.999)300−0= 0.2593(Lecture 8) 18 / 19Take the shorter roadStéphane is going to work 7 days a week and on any given day he islate with probability 0.643 . What is the probability that he is late nomore than 5 days next week?P(X ≤ 5) = P(X = 0) + P(X = 1) + P(X = 2)+ P(X = 3) + P(X = 4) + P(X = 5)P(X ≤ 5) = 1 − P(X > 5)= 1 − [P(X = 6) + P(X = 7)]= 1 −7C6(0.643)6(0.357)1−7C7(0.643)7(0.357)0= 0.7779(Lecture 8) 19 /