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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(BA)
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)_ OZ _/>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: MultipleChoice Test A multiplechoice 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: TrueFalse Test Instead of a multiplechoice test, let’s now assume our ten question test consists only
of truefalse 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 ﬁvestep simulation method and the use of the random number table. Example 19: MultipleChoice 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% aiﬂﬂgcié 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 multiplestage 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 fivestep 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 ...
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This note was uploaded on 08/08/2009 for the course STAT 11 taught by Professor Hirtz during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
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