2-18 - Chapter 4 e Probabilities and Simulation General...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 4 e Probabilities and Simulation General multiplication rule for two events and for three events For two events A and B . . P(Aand B)=P(A)xP(B|A) For three events A, B, and C P(A and Band Call occur) = P(A ) x P(B IA) x P (C M and Boccur) Example 14: Deck of Cards, revisited Suppose a regular deck of cards is shuffled Well and that three cards are dealt one at a time to Daniel. Find the following probabilities. a. P (first two cards are clubs) d. P (ten, five, two) b. P (all 3 cards are clubs) e. P (jack, queen, king) \3 7, l l v c. P (all 3 cards are tens) f. P (spade, heart, diamond) -. , :5, .2. _ " 51'5,'eo'0‘000l8 ‘— £.L1.E:O.ol7. $2 at 50 Definition of Independent Events Events A and Bare independent if P(B M) = P(B ) Example 15: Skipping Class Melanie likes to skip class. Suppose the probability that Melanie attends class on' Tuesday is P(attending class Tuesday) = 0.25 and the probability that she attends class on Wednesday is P(attending class Wednesday) = 0.25. Also, the probability that she attends class both days is P(attending class both days) = 0.20. Are the events A = attends class on Tuesday and B = attends class on Wednesday independent? PCBlA): P(A «nAB)_ O-Z _/>P<\53: 0.7.5 - PCA) 0.25 ' o'?’ i No‘ Hes: ‘tv‘c no'l': fno'cpcnalcvvé edefl 1'35. The two fundamental rules for probabilities are 0 s P (outcome) 5 1, and 2 P (outcome) = 1 all outcomes in 5 Also, the probability that any particular event A of a random experiment will occur is the sum of all probabilities of the outcomes in the event, or r P (A) = 2 P (outcome) all outComes in A " V54 Chapter 4 '— Probabilities and Simulation 4.4 Simulating random sampling via a box model The probability models for many experiments can be represented using a box model. This concept is illustrated in the next example. Example 16: Multiple-Choice Test A multiple-choice test has ten questions. For each question there are five options to choose from, one of the options is correct. Imagine that you have no clue whatsoever what the correct answers are and blindly, or randomly, guess the answer to each of the ten questions. Eventually we will want to find the probability distribution of our random variable, the number of correct answers. For now, illustrate this scenario with a box model. CDBOX O O O O] CZ) #ofdrqws (finmPle ‘32:) = lO @ 04/ or LJ/O replacemefli‘q = w/ replacernenf 63 what ismx sfnii'szi'g’é SUM Is this an example of random sampling with replacement? Yes .r Example 17: True-False Test Instead of a multiple-choice test, let’s now assume our ten question test consists only of true-false questions. We are interested in counting the number of times you will get a correct answer if you randomly guess on each question. Draw the box model. ® [1 91 ® n=lo 6) w/ replaceMen‘fi' G) SUM Is this an example of a counting random variable? Yes ~ We\(1 coun‘él‘fl3 Ouf it .04 (ohfec‘l: aflSerSg Example 18: Flipping a Fair Coin . Jimmy wants to count the number of heads he gets when flipping a fair coin. Draw the box model that represents this scenario? Is this an example of a counting random variable? ©[lo] C237. é) (/5/ rep \«ceMenf G) Sum 55 Chapter 4 — Probabilities and Simulation 4.5 Random sampling with or without replacement When studying complex chance experiments that are built from simpler underlying chance experiments it is useful do simulations. You use Monte Carlo simulations to solve probability problems by repeatedly simulating the real—world experiment being studied. » One way to do simulations is with a random number table. The following example explains the five-step simulation method and the use of the random number table. Example 19: Multiple-Choice Test, revisited Consider again our multiple—choice test has ten questions. How likely are you to geto, 1, 2, 3, 4, 5, 6, 7, 8, 9, orglO correct answers? That is, find the probability distribution of our random variable, the number of correct answers. ' Step 1. Choose a (Probability) Model . Since there are five answers to choose from, the probability that we guess the correct answer to any particular question is (/5 v _[:LOOOQ] So guessing an answer is similar to randomly drawing one of the numbers 1, 2, 3, 4, or 5 at random with equal probability. Let’s identify drawing a 1 with guessing the correct answer. . ‘ ‘ Step2. Define One Simulation Trying to guess the answers to the ten questions corresponds to selecting ten random numbers as described above. Each 1 represents a correct answer. ft: to draws UJ/(‘CP‘QCCMcfl't Step 4. Repeat the Simulation - ~ Let’s use the first two columns of Table 8.3 of Appendix B to simulate 40 values of the statistic of interest, the correct answer. Since there are ten digits in the table and we only need five digits, let 1 and 6 correspond to our “1”, 2 and 7 correspond to our “2”, 3 and 8 correspond to our “3”, 4 and 9 correspond to our “4”, and 5 and 0 correspond to our List the first five simulations below. - 56 Chapter 4 - Probabilities and Simulation \— Repetition Digits Number of Correct Answers MR 1 3223;: i228% 3 2 405353 50958 7* 3 gag—me 43739; 1 4 H.807 3:2qu ”3:. 5 (seem 6127i”; ’5 m“— SKlP Ti—Hs'. Step 5.' Find the Experimental Probability of the Event (Statistic) of Interest. Using the above setup, the results of the 40 simulations are shown in the table below. How do the experimental and theoretical probabilities compare? - Number of correct answers Experimental Experimental fre uenc ' Theoretical . robabili ' ' unflblli Snooowcncnhwmi—Lo Note: The theoretical probabilities are provided for comparison — to know how to do these calculationsm (3ch F Instead of using the random number table, simulation can also be'done using the LifeStats program. 57 Chapter 4 -- Probabilities and Simulation Example 20: Drawing Balls, revisited Reconsider our bag with 4 green balls (G) and 6 black balls (B). Suppose the bag also contains five yellow balls (Y). Allison picks a ball at random from the bag, then puts it back in the bag, then draws a second ball, then puts it back in the bag, and then finally draws a third bali. Draw a probability tree to ill'ustrate this scenario, scenario A. 1‘I <7 3 5 f Y 5 ‘5 5 H a . , ii :5 é a Y a a Y a e 1 9/] t; H 6_ 15 I 9,; 6! ‘45 5 1.725 I 55%,; If GBYeBYGBYéfiYfiEYGBYGflY®$Y®EY After Allison is done, Ben take the bag of 15 balls and draws three balls. However, Ben. doesn’t replace each ball after it is drawn. Draw a probability tree to illustrate this scenario, call it scenario B. “1/15 [IA/l5 g/ls (7 B ' Y 3/1 5 5 .. . \( . B Y H z/e / 36/ (75 5/ 4/ (3H / E» ‘f \ 3 3 (a; 3, Li/, 3, /, w; “27> 15 .9 /, ‘1 i5 Li/I HM a 4N //,J\ Ham j “5/6 3 W5 4K5 W5 (7 a Y eg‘r (951 6‘13Y‘6EY‘C66Y @BYGBYOBY In which scenario are the events independent? gay/Mfr}; ’ Om: draw) 00085 140% aifl-flgcié 014%PI’5 In which scenario do we have random sampling without replacement? igiflédriio B \ Regardless of whether we are drawing with or without replacement, there are usually- too many outcomes to draw a probability tree. Instead, use the following rule to calculate the number of possible outcomes. Multiplication counting rule for multiple-stage experiments Consider an experiment with kstages. Suppose Stage 1 can be performed in N1 ways; then Stage 2 can be performed in N2 ways; then Stage 3 can be performed in N3 ways; and so on and so forth. Then there are ’ _ N1XN2XN3X...XNk ways of performing the kstages of the experiment. 58 Chapter 4 — Probabilities and Simulation Example 21: Practicing the Multiplication Rule 1. If we toss a die six times, what is the total number of possible outcomes? exoxexexoxé=d=qatéee 2. Johnny is a magician. He draws five cards, with replacement, for a magic trick. What is the total number of possible outcomes for these five cards? ‘31 ’< 52. x ‘32 K 5 Z X 52 -: 52:) :: . 380\Zotl,o’3z 3. For a regular deck of cards, what is the total number of ways of dealing 5 cards? 52x5| x50 X‘lclxgt-(g = 3il,875.200 4. Erin has seven pairs of pants, ten shirts, three sweaters,'six pairs of socks, and four pairs of shoes. Assuming she picks one of each time, what is the total number of outfits Erin can create? Kl‘nol, , 7xlO>< 3*Q‘L‘Il 5‘0'40, Example 22: Drawing Jacks What is the probability of getting at least one jack in a hand of 5 cards dealt from an ordinary deck of 52 playing cards? Use the five-step procedure and a box model to answer this question. Then try this problem with the simulation software. ' Step 1. Choose a Model Draw a box model for this problem. Is it a counting variable box model? [1 111000.01: D [33 WE] Step 2. Define a Simulation What does one simulation consist of? Are we drawing with or without replacement? {L25 draws w/o replacemcnf Step 3. Define the Event of Interest What is a successful simulation? $UM> O are 60M 2 :L Step 4. Repeat the Simulation How many times should werepeat the simulation? 59 ...
View Full Document

Page1 / 6

2-18 - Chapter 4 e Probabilities and Simulation General...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online