Introduction to Probability - Chapter 3 ppt

... {a1,a2,a 3 ,a4} can be written in theformσ =1 234 21 43 ,indicating that a1went to a2, a2 to a1, a 3 to a4, and a4 to a 3 .Ifwealwayschoosethetoprowtobe1 234 then, to prescribe ... ω2, ω 3 , and ω5. Each ofthese paths has the same probability p2q.Thusb (3, p,2)=3p2q. Considering allpossible numbers of successes we haveb (3, p,0) = q 3 ,b (3, p,1)=3pq2,b (3, p,2)=3p2q,b (3, p ,3) ... +1possibilities. Hence, the total number of sequences with no duplications is 36 5 36 4 · 36 3 · · (36 5 −r +1).Thus, assuming that each sequence is equally likely,pr= 36 5 36 4 · · (36 5 − r +1) 36 5r.We denote...

Introduction to Probability - Chapter 7 pptx

... vol. 204 (1 937 ), pp. 39 7 39 9. 7.1. SUMS OF DISCRETE RANDOM VARIABLES 287=1 36 ·16=1216,P (S 3 =4) = P(S2 =3) P (X 3 =1)+P (S2=2)P (X 3 =2)=2 36 ·16+1 36 ·16= 3 216,and ... (S2=8)=5 /36 , P (S2=9)=4 /36 , P (S2=10) =3/ 36,P (S2=11)=2 /36 , and P(S2=12)=1 /36 .The distribution for S 3 would then be the convolution of the distribution for S2with the distribution for X 3 .ThusP ... distribution for S2with the distribution for X 3 .ThusP (S 3 =3) = P(S2=2)P (X 3 =1) 30 0 CHAPTER 7. SUMS OF RANDOM VARIABLES -1 5 -1 0 -5 510 150.0250.050.0750.10.1250.150.175n = 5n...

Introduction to Probability - Chapter 10 pptx

... -5 000000000000 0-1 0011111124979 730 0 233 420000000 0100000000000 -5 0210000000000 0 33 471117141111101625 130 000000000000 0-1 001221 131 00000 -5 000000000000 0-1 00 231 000000000 0 31 0000000000 50100000000000 -5 0 34 4710119111214 131 0 ... BRANCHING PROCESSES 38 3Generation Probability of dying out1.22 .31 2 3 .38 52 03 4 . 437 1165 .4758796 .5058787 .5297 13 8 .549 035 9 .56494910 .57822511 .58941612 .598 931 Table 10.1: Probability of ... BRANCHING PROCESSES 38 7GeometricpjData Model0 .2092 .18161 .2584 .36 662 . 236 0 .2028 3 .15 93 .11224 .0828 .06215 . 035 7 . 034 46 .0 133 .01907 .0042 .01058 .0011 .00589 .0002 .0 032 10 .0000 .0018Table...

Introduction to Probability - Answers Exercises ppt

... (a)P = 34 512 30 2 /30 1 /30 41 /30 2 /30 0502 /30 01 /3 100010200001.(b)N = 34 5 35 /32 4 /3 4 132 52 /32 7 /3 ,t= 35 4655,B=12 35 /94/941 /32 /3 52/97/9.(c) ... (a)41 13 105210=7. 23 × 10−8.(b)41 3 2 13 4 13 3 13 35210= .044.(c)4! 13 4 13 3 13 2 13 152 13 = .31 5.21. 3( 25) − 3 = 93 (We subtract 3 because ... Gg Gg, ggGG, Gg 8 /31 /64 /32 /3 GG, gg 4 /34 /38 /34 /3 Gg, Gg 4 /31 /38 /34 /3 Gg, gg 2 /31 /64 /38 /3 .From this we obtaint = Nc =GG, Gg 29/6GG, gg 20 /3 Gg, Gg 17 /3 Gg, gg 29/6,andB...

Introduction to Probability - Chapter 1 pps

... DISCRETE PROBABILITIES 3 .2 033 09 .762057 .151121 .6 238 68. 932 052 .415178 .716719 .967412.069664 .670982 .35 232 0 .0497 23 .750216 .784810 .089 734 .966 730 .946708 .38 036 5 .02 738 1 .900794Table 1.1: ... .5 13. Thus, it is more appropriate to assign a distributionfunction which assigns probability .5 13 to the outcome boy and probability .487 to the outcome girl than to assign probability 1/2 to ... time, Chestnut 10 percent of the time, and Dolby 30 percent 1.2. DISCRETE PROBABILITY DISTRIBUTIONS 33 012 3 012 3 00081 632 6420 32 486464 32 4456Number of games A has wonNumber of...

Introduction to Probability - Chapter 2 docx

... 5000 2 532 3. 1596Smith, 1855 .6 32 04 1218.5 3. 15 53 De Morgan, c.1860 1.0 600 38 2.5 3. 137 Fox, 1864 .75 1 030 489 3. 1595Lazzerini, 1901 . 83 3408 1808 3. 1415929Reina, 1925 .5419 2520 869 3. 1795Table ... = 35 5/1 13: L =πP(E)2=12 35 51 13 1808 34 08=56= . 833 3 .Even with careful planning one would have to be extremely lucky to be able to stopso cleverly.The second author likes to ... L>√3if|r| < 1/2, which occurs with probability 1/2 −(−1/2)1 −(−1)=12. 3. L>√3if2π /3 <α<4π /3, which occurs with probability 4π /3 − 2π /3 2π − 0=1 3 .We see that our simulations...

Introduction to Probability - Chapter 4 doc

... precisebelow. ✷ 146 CHAPTER 4. CONDITIONAL PROBABILITY Number having The resultsDisease this disease ++ +– –+ ––d1 32 15 2110 30 1 704 100d22125 39 6 132 1187 410d 3 4660 510 35 68 73 509Total 10000Table ... Appendix C) to compute a con-ditional probability. The number 93, 7 53 in the table, corresponding to 4 0- year-old males, means that of all the males born in the United States in 1950, 93. 7 53% were ... thesecond card drawn is red? 4 .3. PARADOXES 175 13 Write a program to allow you to compare the strategies play-the-winner andplay-the-best-machine for the two-armed bandit problem of Example...

Introduction to Probability - Chapter 5 docx

... 28927 36 57 8 30 25 9 33 6210 2985 11 31 38 12 30 43 13 2690 14 24 23 15 255616 2456 17 2479 18 227619 230 4 20 1971 21 25 43 22 2678 23 2729 24 241425 2616 26 2426 27 238 128 2059 29 2 039 30 2298 31 ... 2616 26 2426 27 238 128 2059 29 2 039 30 2298 31 2081 32 1508 33 1887 34 14 63 35 1594 36 135 4 37 1049 38 1165 39 124840 14 93 41 132 2 42 14 23 43 1207 44 1259 45 1224Table 5.7: Numbers chosen by ... 216 CHAPTER 5. DISTRIBUTIONS AND DENSITIESFemale MaleA 37 56 93 B 63 60 1 23 C 47 43 90Below C 5 8 13 152 167 31 9Table 5.8: Calculus class data.Female MaleA 44 .3 48.7 93 B 58.6 64.4 1 23 C...

Introduction to Probability - Chapter 6 doc

... VARIANCEWLWLWLWLWLWL(2 ,3, 12) L1098654(7,11) W1 /3 2 /3 2/5 3/ 55/116/115/116/112/5 3/ 51 /3 2 /3 2/91/121/95 /36 5 /36 1/91/121/91 /36 2 /36 2/45 3/ 4525 /39 6 30 /39 625 /39 6 30 /39 62/45 3/ 451 /36 2 /36 Figure ... 36 possibilities. Thus,x = 31 36 · y.But when Huygens rolls he wins on 6 out of the 36 possible outcomes, and in theother 30 , he is led back to where his chances are x.Thusy =6 36 + 30 36 · ... 17 .17 1681 .1681 -2 17 .17 1678 .1678 3 16 .16 1626 .1626 -4 18 .18 1696 .16965 16 .16 1686 .1686 -6 16 .16 1 633 .1 633 Table 6.1: Frequencies for dice game.=96−126= − 3 6= −.5 .This...

Introduction to Probability - Chapter 8 docx

... 8.2. CONTINUOUS RANDOM VARIABLES 31 9nP (|Sn/n|≥.1) Chebyshev100 .31 731 1.00000200.15 730 .50000 30 0.0 832 6 .33 333 400.04550 .25000500.02 535 .20000600.01 431 .16667700.00815 .14286800.00468 ... x) to compute the exact probability that you estimated in Exercise 1. Compare the two results. 3 Write a program to toss a coin 10,000 times. Let Snbe the number of headsin the ﬁrst n tosses. ... its mean.9Private communication. 31 0 CHAPTER 8. LAW OF LARGE NUMBERSExample 8 .3 Let X1, X2, ,Xnbe a Bernoulli trials process with probability .3 for success and .7 for failure. Let...

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