The Binomial Distribution

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1 The Binomial Distibution A. It would be vey tedious if, evey time we had a slightly diffeent poblem, we had to detemine the pobability distibutions fom scatch. Luckily, thee ae enough similaities between cetain types, o families, of expeiments, to make it possible to develop fomulas epesenting thei geneal chaacteistics. Fo example, many expeiments shae the common element that thei outcomes can be classified into one of two events, e.g. a coin can come up heads o tails; a child can be male o female; a peson can die o not die; a peson can be employed o unemployed. These outcomes ae often labeled as success o failue. Note that thee is no connotation of goodness hee - fo example, when looking at biths, the statistician might label the bith of a boy as a success and the bith of a gil as a failue, but the paents wouldn t necessaily see things that way. The usual notation is p pobability of success, pobability of failue 1 - p. Note that p + 1. In statistical tems, A Benoulli tial is each epetition of an expeiment involving only 2 outcomes. We ae often inteested in the esult of independent, epeated benoulli tials, i.e. the numbe of successes in epeated tials. 1. independent - the esult of one tial does not affect the esult of anothe tial. 2. epeated - conditions ae the same fo each tial, i.e. p and emain constant acoss tials. Hayes efes to this as a stationay pocess. If p and can change fom tial to tial, the pocess is nonstationay. The tem identically distibuted is also often used. B. A binomial distibution gives us the pobabilities associated with independent, epeated Benoulli tials. In a binomial distibution the pobabilities of inteest ae those of eceiving a cetain numbe of successes,, in n independent tials each having only two possible outcomes and the same pobability, p, of success. So, fo example, using a binomial distibution, we can detemine the pobability of getting 4 heads in 10 coin tosses. How does the binomial distibution do this? Basically, a two pat pocess is involved. Fist, we have to detemine the pobability of one possible way the event can occu, and then detemine the numbe of diffeent ways the event can occu. That is, P(Event) (Numbe of ways event can occu) * P(One occuence). Suppose, fo example, we want to find the pobability of getting 4 heads in 10 tosses. In this case, we ll call getting a heads a success. Also, in this case, n 10, the numbe of successes is 4, and the numbe of failues (tails) is n One way this can occu is if the fist 4 tosses ae heads and the last 6 ae tails, i.e. The Binomial Distibution - Page 1

2 S S S S F F F F F F The likelihood of this occuing is P(S) * P(S) * P(S) * P(S) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) Moe geneally, if p pobability of success and 1 p pobability of failue, the pobability of a specific seuence of outcomes whee thee ae successes and n- failues is n p So, in this paticula case, p.5, 4, n- 6, so the pobability of 4 staight heads followed by 6 staight tails is (o 1 out of 1024). Of couse, this is just one of many ways that you can get 4 heads; futhe, because the epeated tials ae all independent and identically distibuted, each way of getting 4 heads is eually likely, e.g. the seuence S S S S F F F F F F is just as likely as the seuence S F S F F S F F S F. So, we also need to know how many diffeent combinations poduce 4 heads. Well, we could just wite them all out but life will be much simple if we take advantage of two counting ules: 1. The numbe of diffeent ways that N distinct things may be aanged in ode is (1)(2)(3)...(N-1)(N), (whee 0! 1). An aangement in ode is called a pemutation, so that the total numbe of pemutations of N objects is. The symbol is called N factoial. EXAMPLE. Rank candidates A, B, and C in ode. The possible pemutations ae: ABC ACB BAC BCA CBA CAB. Hence, thee ae 6 possible odeings. Note that 3! (1)(2)(3) 6. NOTE: Appendix E, Table 6, p. 19 contains a Table of the factoials fo the integes 1 though 50. Fo example, 12! * (O see Hayes Table 8, p. 947). You calculato may have a factoial function labeled something like x! 2. The total numbe of ways of selecting distinct combinations of N objects, iespective of ode, is N N!(N - )! N We efe to this as N choose. Sometimes the numbe of combinations is known as a binomial coefficient, and sometimes the notation N C is used. So, in the pesent example, The Binomial Distibution - Page 2

3 N 10 4!(N - )! 10! 10*9*8* ( 10 4 )! 4*3* 2*1 24 Note that, fo 10!, I stopped once I got to 7; and I didn t wite out 6! in the denominato. This is because both numeato and denominato have 6! in them, so they cancel out. So, thee ae 210 ways you can toss a coin 10 times and get 4 heads. EXAMPLE. Candidates A, B, C and D ae unning fo office. Vote fo two. The possible choices ae: AB AC AD BC BD CD, i.e. thee ae 6 possible combinations. Confiming this with the above fomula, we get!(n - )! 2 2!(4-2)! (4)(3)(2)(1) 12 6 (2)(1)(2)(1) 2 EXAMPLE. Thee ae 100 applicants fo 3 job openings. The numbe of possible combinations is ! 100*99*98 970, ,700!(N - )! 3 3!97! 3*2 6 Again, note that, if you didn t take advantage of 97! appeaing on both top and bottom, you d have a much lengthie calculation. See Appendix E, Table 7, page 20 fo N C values fo vaious values of N and. (O see Hayes, Appendix E, Table IX, p. 948). You calculato may have a function labeled nc o something simila. C. So putting eveything togethe now: we know that any specific seuence that poduces 4 heads in 10 tosses has a pobability of Futhe, we now know that thee ae 210 such seuences. Ego, the pobability of 4 heads in 10 tosses is 210 * We can now wite out the complete fomula fo the binomial distibution: In sampling fom a stationay Benoulli pocess, with the pobability of success eual to p, the pobability of obseving exactly successes in N independent tials is N p N -!(N - )! p N - Once again, N choose tells you the numbe of seuences that will poduce successes in N ties, while p N- tells you what the pobability of each individual seuence is. The Binomial Distibution - Page 3

4 To put it anothe way, the andom vaiable X in a binomial distibution can be defined as follows: Let Xi 1 if the ith benoulli tial is successful, 0 othewise. Then, X ΣX i, whee the X i s ae independent and identically distibuted (iid). That is, X the # of successes. Hence, Any andom vaiable X with pobability function given by p(x ; N, N p) p N -!(N - )! p N -, X 0,1,2,..., N is said to have a binomial distibution with paametes N and p. EXAMPLE. In each of 4 aces, the Democats have a 60% chance of winning. Assuming that the aces ae independent of each othe, what is the pobability that: a. The Democats will win 0 aces, 1 ace, 2 aces, 3 aces, o all 4 aces? b. The Democats will win at least 1 ace c. The Democats will win a majoity of the aces SOLUTION. Let X eual the numbe of aces the Democats win. a. Using the fomula fo the binomial distibution, 0 p 0 1 p 1 p p 3 p !(4-0)! 1!(4-1)! 2!(4-2)! *.60* , * * , 3!(4-3)! (4-4)! , , b. P(at least 1) P(X 1) 1 - P(none) 1 - P(0) O, P(1) + P(2) + P(3) + P(4) c. P(Democats will win a majoity) P(X 3) P(3) + P(4) The Binomial Distibution - Page 4

5 EXAMPLE. In a family of 11 childen, what is the pobability that thee will be moe boys than gils? Solve this poblem WITHOUT using the complements ule. SOLUTION. You could go though the same tedious pocess descibed above, which is what most students did when I fist asked this uestion on an exam. You would compute P(6), P(7), P(8), P(9), P(10), and P(11). O, you can look at Appendix E, Table II (o Hays pp ). Hee, both Hayes and I list binomial pobabilities fo values of N and fom 1 though 20, and fo values of p that ange fom.05 though.50. Thus, on page E-5, we see that fo N 11 and p.50, P(6) + P(7) + P(8) + P(9) + P(10) + P(11) NOTE: Undestanding the tables in Appendix E can make things a lot simple fo you! EXAMPLE. [WE MAY SKIP THIS EXAMPLE IF WE RUN SHORT OF TIME, BUT YOU SHOULD STILL GO OVER IT AND MAKE SURE YOU UNDERSTAND IT] Use Appendix E, Table II, to once again solve this poblem: In each of 4 aces, the Democats have a 60% chance of winning. Assuming that the aces ae independent of each othe, what is the pobability that: a. The Democats will win 0 aces, 1 ace, 2 aces, 3 aces, o all 4 aces? b. The Democats will win at least 1 ace c. The Democats will win a majoity of the aces SOLUTION. It may seem like you can t do this, since the table doesn t list p.60. Howeve, all you have to do is edefine success and failue. Let success P(opponents win a ace).40. The uestion can then be ecast as finding the pobability that a. The opponents will win 4 aces, 3 aces, 2 aces, 1 ace, o none of the aces? b. The opponents will win 0, 1, 2, o 3 aces; o, the opponents will not win all the aces c. The opponents will not win a majoity of the aces We theefoe look at page E-4 (o Hayes, p. 927), N 4 and p.40, and find that a. P(4).0256, P(3).1536, P(2).3456, P(1).3456, and P(0) b. P(0) + P(1) + P(2) + P(3) 1 - P(4).9744 c. P(1) + P(0) In geneal, fo p >.50: To use Table II, substitute 1 - p fo p, and substitute N - fo. Thus, fo p.60 and N 4, the pobability of 1 success can be found by looking up p.40 and 3. D. Mean of the binomial distibution. Recall that, fo any discete andom vaiable, E(X) Σxp(x). Theefoe, E(X i ) Σxp(x) 0 * (1 - p) + 1 * p p, that is, the mean of any The Binomial Distibution - Page 5

6 benoulli tial is p, the pobability of success. By applying ou theoems fo expectations, we find that E(X) E(X 1 + X X N ) E(X 1 ) + E(X 2 ) E(X N ) Np. NOTE: Hayes povides an altenative poof in section 4.8; I think my poof is much simple. EXAMPLE. If we toss a fai coin 3 times (N 3, p.50), we expect to get 1.5 heads. E. Vaiance of the binomial distibution. Given ou above definitions, note that, fo any Benoulli tial, X i 2 X i. Thus, V(X i ) E(X i 2 ) - E(X i ) 2 p - p 2 p(1 - p) p. Because tials ae independent, we find that V(X) V(X 1 ) + V(X 2 ) +...+V(X N ) Np. EXAMPLE. If you toss a fai coin 3 times (N 3, p.50,.50) the vaiance of the expected numbe of heads is 3 *.50 * NOTE. These last two points mean that the mean and vaiance of the binomial distibution ae dependent on only two paametes, N and p. WARNING: The symbol X gets used many diffeent ways in statistics. In this case, X the sum of all the Benoulli tials, but in othe instances it might efe to an individual Benoulli tial. F. Shape of the binomial distibution. When p.5, the binomial distibution is symmetical - the mean and median ae eual. Even when p <>.5, the shape of the distibution becomes moe and moe symmetical the lage the value of N. This is vey impotant, because the binomial distibution can uickly become unwieldy - as we will late see, thee ae appoximations to the binomial that can be much easie to use when N is lage. The Binomial Distibution - Page 6

7 g. A MORE COMPLETE LISTING OF THE COUNTING RULES FOR PERMUTATIONS AND COMBINATIONS (OPTIONAL). Hee is a moe extensive set of counting ules that can be useful fo vaious poblems in pobability. They aen t essential fo ou immediate puposes, so we pobably won t go ove them in class unless we have exta time. But, these aen t vey had, and they may come in handy fo you some day, so I am including them hee. 1. NUMBER OF POSSIBLE SEQUENCES FOR N TRIALS. Suppose that a seies of N tials wee caied out, and that on each tial any of K events might occu. Then the following ule holds: If any one of K mutually exclusive and exhaustive events can occu on each of N tials, then thee ae K n diffeent seuences that may esult fom a set of such tials. EXAMPLE. If you toss a die once, any of 6 numbes can show up (K 6). Ego, if you toss it 3 times, any of seuences ae possible (e.g., 111, 342, 652, etc.). [You calculato pobably has a y x function o something simila; on mine, I pess 6, then y x, then 3, then.] 2. SEQUENCES. Sometimes the numbe of possible events in the fist tial of a seies is diffeent fom the numbe possible in the second, the second diffeent fom the thid, etc. That is, K 1 K 2 K 3, etc. Unde such conditions, If K 1,...,K N ae the numbes of distinct events that can occu on tials 1,..., N in a seies, then the numbe of diffeent seuences of N events that can occu is (K 1 )(K 2 )...(K N ). EXAMPLE. Two occupations and thee eligions yield 6 combinations of occupation and eligion. Tossing a coin (2 outcomes) and tossing a die (6 outcomes) yield 12 possible outcomes. Note that, when K i K fo all i, then ule 1 becomes a special case of ule 2. Note also that, when K 1 1, K 2 2,..., K N N, then ule 3 becomes a special case of ule PERMUTATIONS. A ule of exteme impotance in pobability computations concens the numbe of ways that objects may be aanged in ode. The numbe of diffeent ways that N distinct things may be aanged in ode is (1)(2)(3)...(N-1)(N), (whee 0! 1). An aangement in ode is called a pemutation, so that the total numbe of pemutations of N objects is. The symbol is called N factoial. The notation N P N is also sometimes used fo, fo easons which will be clea in a moment. EXAMPLE. Rank candidates A, B, and C in ode. The possible pemutations ae: ABC ACB BAC BCA CBA CAB The Binomial Distibution - Page 7

8 Thee ae 6 possible odeings. Note that 3! (1)(2)(3) 6. NOTE: Appendix E, Table 6, p. 19 contains a Table of the factoials fo the integes 1 though 50. Fo example, 12! * (O see Hayes Table 8, p. 947). You calculato may have a factoial function labeled something like x! 3B. PERMUTATIONS OF SIMILAR OBJECTS. Suppose we have N objects, N 1 alike, N 2 alike,..., N k alike (ΣN i N). Then, the numbe of ways of aanging these objects is N 1! N 2!...N k! EXAMPLE. We have 4 balls, 2 ed and 2 blue. The possible ways of aanging the balls ae BBRR, BRBR, BRRB, RRBB, RBRB, RBBR, o 6 ways altogethe. To confim that thee ae 6 ways, 6 N 1! N 2!...N k! 2!2! EXAMPLE. If we have 6 balls, 2 ed, 2 blue, and 2 geen, the numbe of possible ways of aanging them is 6! N 1! N 2!...N k! 2!2!2! 8 NOTE: I add this ule to Hayes s list because many of the othe ules ae special cases of it. When N i 1 fo all i, Rules 3 (Pemutations) and 3B ae euivalent. When N i 1 fo i 1 to, and N +1 N -, then ules 4 (Odeed combinations) and 3B ae the same. When N 1 and N 2 N -, Rules 5 (Combinations) and 3B ae the same. 4. ORDERED COMBINATIONS; o, PERMUTATIONS OF N OBJECTS TAKEN AT A TIME. Sometimes it is necessay to count the numbe of ways that objects might be selected fom among some N objects in all ( N). Futhe, each diffeent aangement of the objects is consideed sepaately. The numbe of ways of selecting & aanging objects fom among N distinct objects is (N - )! Vebally, we efe to this as odeed combinations of N things taken at a time. The notation N P is sometimes used, and may appea on you calculato using simila notation. Note also that N P N. The Binomial Distibution - Page 8

9 EXAMPLE. Candidates A, B, C, and D ae unning fo office. Indicate you fist and second choice. Thee ae 12 possible choices: AB BA AC CA AD DA BC CB BD DB CD DC. To confim this, using counting Rule #4, we get (N - )! (4-2)! (4)(3)(2)(1) 12 (2)(1) EXAMPLE. Thee ae 100 applicants fo 3 job openings. Indicate you fist, second, and thid choices. If you ty to list all the possible pemutations, you will be busy fo a vey long time. If you ty to calculate 100! and 97! by hand, you d bette plan on being in gad school a long time. And, even a calculato with a factoial function may not help you; mine just poduces an eo message when I give it numbes this big. If, on the othe hand, you take advantage of the fact that most tems in the numeato and denominato cancel out, e.g. (N - )! 100! (100)(99)(98)(97)(96)...(1) 100* 99* ,200 (100-3)! (97)(96)...(1) you will find that it is petty easy to figue out the numbe of odeed combinations. NOTE: You could also use Rule 3B. People ae divided into 4 categoies: the fist best (N 1 1), the 2nd best (N 2 1), the 3d best (N 3 1) and all the est (N 4 97). Hence, ule 3B yields 100!/(1!1!1!97!) 100!/97! 5. COMBINATIONS. Often, we ae not inteested in the ode of events, but only in the numbe of ways things could be selected fom N things, iespective of ode. Fom ule 4, we know that the total numbe of ways of selecting things fom N and odeing them is /(N)! Fom ule 3, we know that each set of objects has! odeings. Theefoe, The total numbe of ways of selecting distinct combinations of N objects, iespective of ode, is N N!(N - )! N We efe to this as N choose. Sometimes the numbe of combinations is known as a binomial coefficient, and sometimes the notation N C is used. Note that the numbe of pemutations is! times lage than the numbe of combinations. EXAMPLE. Candidates A, B, C and D ae unning fo office. Vote fo two. The possible choices ae: The Binomial Distibution - Page 9

10 AB AC AD BC BD CD, i.e. thee ae 6 possible combinations. Confiming this with ule #5, we get (4)(3)(2)(1) 12 6!(N - )! 2 2!(4-2)! (2)(1)(2)(1) 2 EXAMPLE. Thee ae 100 applicants fo 3 job openings. The numbe of possible combinations is ! 970, ,700!(N - )! 3 3!97! 6 NOTE: You could also use ule 3B. People ae divided into 2 goups: the 3 best (N 1 3) and all the est (N 2 97). Hence, ule 3B yields 100!/(3!97!) See Appendix E, Table 7, page 20 fo N C values fo vaious values of N and. (O see Hayes, Appendix E, Table IX, p. 948). You calculato may have a function labeled nc o something simila. 6. Binomial distibution. In sampling fom a stationay Benoulli pocess, with the pobability of success eual to p, the pobability of obseving exactly successes in N independent tials is N p N -!(N - )! p N - 7. Binomially distibuted vaiables. Let Xi 1 if the ith benoulli tial is successful, 0 othewise. If X ΣX i, whee the X i s ae independent and identically distibuted (iid), then X has a binomial distibution, and E(X) Np, V(X) Np. The Binomial Distibution - Page 10

11 SUMMARY OF HIGHLIGHTS 1. NUMBER OF POSSIBLE SEQUENCES FOR N TRIALS. If any one of K mutually exclusive and exhaustive events can occu on each of N tials, then thee ae K n diffeent seuences that may esult fom a set of such tials. 2. SEQUENCES. If K 1,...,K N ae the numbes of distinct events that can occu on tials 1,..., N in a seies, then the numbe of diffeent seuences of N events that can occu is (K 1 )(K 2 )...(K N ). 3. PERMUTATIONS The numbe of diffeent ways that N distinct things may be aanged in ode is (1)(2)(3)...(N-1)(N), (whee 0! 1). 3B. PERMUTATIONS OF SIMILAR OBJECTS. Suppose we have N objects, N 1 alike, N 2 alike,..., N k alike (ΣN i N). Then, the numbe of ways of aanging these objects is N 1! N 2!...N k! 4. ORDERED COMBINATIONS; o, PERMUTATIONS OF N OBJECTS TAKEN AT A TIME. The numbe of ways of selecting & aanging objects fom among N distinct objects is (N - )! 5. COMBINATIONS. The total numbe of ways of selecting distinct combinations of N objects, iespective of ode, is N N!(N - )! N - 6. Binomial distibution. In sampling fom a stationay Benoulli pocess, with the pobability of success eual to p, the pobability of obseving exactly successes in N independent tials is N p N -!(N - )! 7. Binomially distibuted vaiables. Let Xi 1 if the ith benoulli tial is successful, 0 othewise. If X ΣX i, whee the X i s ae independent and identically distibuted (iid), then X has a binomial distibution, and E(X) Np, V(X) Np. p N - The Binomial Distibution - Page 11

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