Chap_2_Supplement_Ross_Examples

Chap_2_Supplement_Ross_Examples - taining 6 white and 5...

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Unformatted text preview: taining 6 white and 5 Jails is white and the :lected as being rele- -outcomes. Further- ball selected is white 1 the first is black, the hich the first two are .y drawn” means that r, we see that the de- 1g the outcome of the 5 point of view, there h set of 3 balls corre- ed. As a result, if all lection is noted, then ne is taken to be the resentation of the ex- 'eviously. l 6 men and 9 women. ' that the committee at each of the ice the desired prob- \Wnr " Noles fin-n it tram Ross . . . . .. . .. . H _. .,_.. _ .. .i _: . :.. ,...,:-,..,__.._.._., ,. .I.-.-.._ “WW-nary" '.-!‘ Clad? 2 Lee f6 Section 2.5 Sample Spaces Having Equally Likely Outcomes 37 Exam 1 An urn contains n balls, of which one is s ecial. If k of these balls EXowiqlg p p are ' drawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is chosen? Solution Since all of the balls are treated in an identical manner, it follows that the set of k balls selected is equally likely to be any of the sets of k balls. n Therefore, k (:>(:::) 5 P{special ball is selected} = —-—~—— = n We could also have obtained the preceding result by letting A ,- denote the event that the special ball is the ith ball to be chosen, 1' = 1, . .. , k. Then, since each one of the n balls is equally likely to be the ith ball chosen, it follows that P(Ai) = 1/n. Hence, since these events are clearly mutually exclusive, we have that k k P{special ball is selected} = P(U A) = 2 P(Ai) = 5 i=1 i=1 We could have argued that P(Ai) = 1/n, by noting that there are n(n - 1) (n — k + 1) = nl/(n — k)! equally likely outcomes ofthe exper- iment, of which (n —1)(n - 2)---(n — i + 1)(1)(n — i)-~(n — k + 1) = (n — 1)!/ (n - k)! result in the special ball being the ith one chosen. From this it follows that -1l P(Ai)=g1‘-’E*L=% I Example 5e. Suppose that n + m balls, of which n are red and m are blue, are arranged in a linear order in such a way that all (n + m)! possible orderings are equally likely. If we record the result of this experiment by only listing the colors of the successive balls, show that all the possible results remain equally likely. Solution Consider any one of the (n + m)! possible orderings and note that any permutation of the red balls among themselves and of the blue balls among themselves does not change the sequence of colors. As a result, every ordering of colorings corresponds to n! ml different orderings of the n + m balls, so n! m! (n + m)! For example, suppose that there are 2 red balls, numbered r1, r2, and 2 blue balls, numbered b1, b2. Then, of the 4! possible orderings, there will be 2! 2! or- derings that result in any specified color combination. For instance, the follow- ing orderings result in the successive balls alternating in color with a red ball first: rhbiarzabz riabzarzabi ’bbprhbz rzabza’hbl every ordering of the colors has probability of occurring. Hence each of the possible orderings of the colors has probability 514 = % of occurring. I 38 Chapter 2 Axioms of Probability Example 5f. A poker hand consists of 5 cards. If the cards have distinct consecu- tive values and are not all of the same suit, we say that the hand is a straight. For Soluti instance, a hand consisting of the five of spades, six of spades, seven of spades, ' eight of spades, and nine of hearts is a straight. What is the probability that one is dealt a straight? the 4‘ Solution We start by assuming that all (552) possible poker hands are equal- ly likely. To determine the number of outcomes that are straights, let us first de- termine the number of possible outcomes for which the poker hand consists of an ace, two, three, four, and five (the suits being irrelevant). Since the ace can be any 1 of the 4 possible aces, and similarly for the two, three, four, and five, it follows that there are 45 outcomes leading to exactly one ace, two, three, four, and five. Hence, since in 4 of these outcomes all the cards will be of the same .7 ( suit (such a hand is called a straight flush), it follows that there are 45 — 4 hands la er that make up a straight of the form ace, two, three, four, and five. Similarly, 3;: p y there are 45 - 4 hands that make up a straight of the form ten, jack, queen, - < king, and ace. Hence there are 10(45 — 4) hands that are straights Thus the de- _ 12’ 13 sired probability is receiw 10(45 — 4) ceives % .0039 I 52 receivr 5 j; Exampl A 5-card poker hand is said to be a full house if it consists of 3 cards oft ' ame denomination and 2 cards of the same denomination. (That is, a full house is three of a kind plus a pair.) What is the probability that one is dealt a full house? 52 5 determine the number of possible full houses, we first note that there are Solution Again we assume that all < > possible hands are equally likely. To Some 1 next two ex; 4 4 . . < > different combinations of, say, 2 tens and 3 Jacks. Because there are 2 3 Example Si. 13 different choices for the kind of pair and, after a pair has been chosen, there ._. Ofbthe" are 12 other choices for the denomination of the remaining 3 cards, it follows n C so that the probability of a full house is solutio 4 4 days, t1 13 - 12- <2)(3) bility o —— z .0014 . come is 52 (365 - 5 _, ability probab Example 5h. In the game of bridge the entire deck of 52 cards is dealt out to 4 people players. What is the probability that to 365. (a) one of the players receives all 13 spades; probab (b) each player receives 1 ace? have distinct consecu- hand is a straight. For )ades, seven of spades, 1e probability that one lOkCI’ hands are equal- straights, let us first de- poker hand consists of nt). Since the ace can three, four, and five, it .e ace, two, three, four, 'ds will be of the same there are 45 — 4 hands 1r, and five. Similarly, form ten, jack, queen, straights. Thus the de- if it consists of 3 cards iination. (That is,a full ility that one is dealt a s are equally likely. To 5t note that there are ks. Because there are has been chosen, there ning 3 cards, it follows cards is dealt out to 4 ”“_’TEWJ-ITJJH'_'V\V’ r- _ Section 2.5 Sample Spaces Having Equally Likely Outcomes 39 52 13, 13’ 13, 13> possrble d1v1s10ns of the cards among Solution (a) There are< 39 13, 13, 13 leading to a fixed player having all 13 spades, it follows that the desired proba- bility is given by 39 4(13, 13, 13) 52 13, 13, 13, 13 (b) To determine the number of outcomes in which each of the distinct players receives exactly 1 ace, put aside the aces and note that there are the 4 distinct players. As there are < > possible divisions of the cards a 6.3 X 10"2 12, 12, 12, 12 receive 12. As there are 4! ways of dividing the 4 aces so that each player re- ceives 1, we see that the number of possible outcomes in which each player 48 12, 12, 12, 12 4! 48 12, 12, 12, 12 52 13, 13, 13, 13 Some results in probability are quite surprising when initially encountered. Our 4 ( 8 > possible divisions of the other 48 cards when each player is to receives exactly 1 ace is 4l< Hence the desired probability is z .105 l next two examples illustrate this phenomenon. Exampl If n people are present in a room, what is the probability that no two of celebrate their birthday on the same day of the year? How large need n be so that this probability is less than i? Solution As each person can celebrate his or her birthday on any one of 365 days, there is a total of (365)" possible outcomes. (We are ignoring the possi- bility of someone’s having been born on February 29.) Assuming that each out- come is equally likely, we see that the desired probability is (365)(364)(363) ~- (365 — n + 1) / (365)”. It is a rather surprising fact that when n 2 23, this prob- ability is less than That is, if there are 23 or more people in a room, then the probability that at least two of them have the same birthday exceeds Many people are initially surprised by this result, since 23 seems so small in relation to 365, the number of days of the year. However, every pair of individuals has 5 (3:5)2 = 32—5 of having the same birthday, and in a group of 23 probability 40 Chapter 2 Axioms of Probability 23 _ 4 aces i people there are < 2 > = 253 different pairs of individuals. Looked at this way, . the tw< . plete a the result no longer seems so surprising. When there are 50 people in the room, the probability that at least two '- Example 5k. share the same birthday is approximately .970. And with 100 persons in the - players room, the odds are better than 3,000,000:1 (that is, the probability is greater - mates. 3 x 106 than —-—6 ) that at least two people have the same birthday. I n? Offe 3 X 10 + 1 21 offei Exampl A deck of 52 playing cards is shuffled and the cards turned up one at -' 801mm a t ' - until the first ace appears. Is the next card—that is, the card following - the first ace—more likely to be the ace of spades or the two of clubs? Solution To determine the probability that the card following the first ace is ways 0 the ace of spades, we need to calculate how many of the (52)! possible order- " are (4C ings of the cards have the ace of spades immediately following the first ace. To _ out] H begin, note that each ordering of the 52 cards can be obtained by first ordering " orderel the 51 cards different from the ace of spades and then inserting the ace of spades Sive_d( into that ordering. Furthermore, for each of the (51)! orderings of the other themse cards, there is only one place where the ace of spades can be placed so that it probah follows the first ace. For instance, if the ordering of the other 51 cards is 4c, 6h,Jd, 5s, Ac, 7d, , Kh then the only insertion of the ace of spades into this ordering that results in its following the first ace is 4c, 6h, Id, 53, Ac, As, 7d, , Kh To dete Therefore, we see that there are (51)! orderings that result in the ace of spades ' following the first ace, so __ flrSt no 51 ! ' 2' d f P{the ace of spades follows the first ace} = = é P11213535: is so e In fact, by exactly the same argument, it follows that the probability that the two ' second of clubs (or any other specified card) follows the first ace is also In other remain words, each of the 52 cards of the deck is equally likely to be the one that fol- selves , lows the first ace! ’ Many people find this result rather surprising. Indeed, a common reaction is to suppose initially that it is more likely that the two of clubs (rather than the ace of spades) follows the first ace, since that first ace might itself be the ace of divisim spades. This reaction is often followed by the realization that the two of clubs might itself appear before the first ace, thus negating its chance of immediate- ly following the first ace. However, as there is one chance in four that the ace of spades will be the first ace (because all 4 aces are equally likely to be first) and only one chance in five that the two of clubs will appear before the first ace (because each of the set of 5 cards consisting of the two of clubs and the s Looked at this way, tility that at least two th 100 persons in the probability is greater 3 birthday. I ards turned up one at I is, the card following wo of clubs? lowing the first ace is (52)! possible order- uwing the first ace. To ined by first ordering ’ting the ace of spades rderings of the other 11 be placed so that it )ther 51 cards is ring that results in its it in the ace of spades 51)! _ 1 52)! ' 52 obability that the two :e is also In other 0 be the one that fol- d, a common reaction :lubs (rather than the ;ht itself be the ace of that the two of clubs :hance of immediate- e in four that the ace ally likely to be first) )pear before the first two of clubs and the ._ 'WHWITPHFWTF ‘wr_'.7+_-. _.’.g.. my Avg-whim” unn- ""-'-'-"-. w-“avr-rr-rmwmwrmammm, 1-. - t -. ._ - ,.- - .: , r -. _ . . Section 2.5 Sample Spaces Having Equally Likely Outcomes 41 4 aces is equally likely to be the first of this set to appear), it again appears that the two of clubs is more likely. However, this is not the case and a more com- plete analysis shows that they are equally likely. I Exampl® A football team consists of 20 offensive and 20 defensive players. The pla s are to be paired in groups of 2 for the purpose of determining room- mates. If the pairing is done at random, what is the probability that there are no offensive—defensive roommate pairs? What is the probability that there are 2i offensive—defensive roommate pairs,i = 1, 2, . .. , 10? 4o _ (40)! 2,2,...,2 _ (2020 ways of dividing the 40 players into 20 ordered pairs of two each. [That is, there are (40) !/220 ways of dividing the players into a first pair, a second pair, and so on.] Hence there are (40)!/22°(20)! ways of dividing the players into (un- ordered) pairs of 2 each. Furthermore, since a division will result in no offen- sive—defensive pairs if the offensive (and defensive) players are paired among themselves, it follows that there are [(20) !/21°(10)!]2 such divisions. Hence the probability of no offensive—defensive roommate pairs, call it P0, is given by (20)! 2 <2‘°(10)!> [(20)!]3 Solution There are ° _ (40)! [(10):]2(40)! 220(20)! To determine P2,- , the probability that there are 2i offensive—defensive pairs, we 2 ways of selecting the 21' offensive players and the first note that there are < 2 1 2i defensive players who are to be in the offensive—defensive pairs. These 4i players can then be paired up into (21')! possible offensive—defensive pairs (This is so because the first offensive can be paired with any of the 21' defensives, the second offensive with any of the remaining 21' — 1 defensives, and so on.) As the remaining 20 — 2i offensives (and defensives) must be paired among them- selves, it follows that there are 20 2 _ (20—21)! 2 (2i>(21)!l2‘°"(10 — ml divisions which lead to 2i offensive—defensive pairs. Hence 20 2 _ (20—21)! 2 (2i) (Miro-"(10 — ml Pzi=—————-(4O)! l=0,1,...,10 220(20)! 42 Chapter 2 Axioms of Probability The P2,, 1‘ = 0, 1, , 10, can now be computed or they can be approximated by making use of a result of Stirling which shows that n! can be approximated by n"+"2 + ("X/277. For instance, we obtain that P0 z 1.3403 X 10—6 P10 z .345861 P20 z X 10—6 I Our next three examples illustrate the usefulness of Proposition 4.4. In Ex- ample 5], the introduction of probability enables us to obtain a quick solution to a counting problem. Example 5]. A total of 36 members of a club play tennis, 28 play squash, and 18 play badminton. Furthermore, 22 of the members play both tennis and squash, 12 play both tennis and badminton, 9 play both squash and badminton, and 4 play all three sports. How many members of this club play at least one of these sports? Solution Let N denote the number of members of the club, and introduce probability by assuming that a member of the club is randomly selected. If for any subset C of members of the club, we let P(C) denote the probability that the selected member is contained in C, then _ number of members in C P(C) N Now, with T being the set of members that plays tennis, S being the set that plays squash, and B being the set that plays badminton, we have from Proposi— tion 4.4 that P(TUSU B) = P(T) + P(S) + P(B) — P(TS) — P(TB) — P(SB) + P(TSB) =36+28+18—22—12—9+4 N _ E N Hence we can conclude that 43 members play at least one of the sports. I The next example in this section not only possesses the virtue of giving rise to a somewhat surprising answer but is also of theoretical interest. Example 5m. The matching problem. Suppose that each of N men at a party throws his hat into the center of the room. The hats are first mixed up, and then each man randomly selects a hat. What is the probability that (a) none of the men selects his own hat; (b) exactly k of the men select their own hats? Sol ma the abi If v ith sibl SC]( 1'] , l firs hat 1y, \ Als 311C He whi N 1 mm abil hat: ...
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This note was uploaded on 01/14/2011 for the course EECS 501 taught by Professor Tecszs.. during the Fall '08 term at University of Michigan.

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Chap_2_Supplement_Ross_Examples - taining 6 white and 5...

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