Introduction to Probability - Chapter 4 doc

... Smith Unconditional probability b bb b g g g b 1 /4 1/8 1/8 1/8 1/8 1 /4 1 /4 1 /4 1 /4 1 /4 1 1/2 1/2 1/2 1/2 1 bg gb gg b bb b b 1 /4 1/8 1/8 1 /4 1 /4 1 /4 1 1/2 1/2 bg gb Conditional probability 1/2 1 /4 1 /4 Figure 4. 13: ... drawn is red? 4. 3. PARADOXES 175 13 Write a program to allow you to compare the strategies play-the-winner and play-the-best-machine for the...

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... 64 CHAPTER 2. CONTINUOUS PROBABILITY DENSITIES -1 1 2 3 0.2 0 .4 0.6 0.8 1 -1 1 2 3 0.2 0 .4 0.6 0.8 1 F Z (z) f (z) Z Figure 2.15: Distribution and density functions for Example 2. 14. 1 E Z Figure ... the average, you will have to wait. Write a program to see if your conjecture is right. 2.2. CONTINUOUS DENSITY FUNCTIONS 65 -1 -0 .5 0.5 1 1.5 2 0.2 0 .4 0.6 0.8...

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... 24 241 4 25 2616 26 242 6 27 2381 28 2059 29 2039 30 2298 31 2081 32 1508 33 1887 34 146 3 35 15 94 36 13 54 37 1 049 38 1165 39 1 248 40 149 3 41 1322 42 142 3 43 1207 44 1259 45 12 24 Table 5.7: Numbers ... Chosen 1 2 646 2 29 34 3 3352 4 3000 5 3357 6 2892 7 3657 8 3025 9 3362 10 2985 11 3138 12 3 043 13 2690 14 242 3 15 2556 16 245 6 17 247 9 18 2276 19 23 04 20 1971 21...

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... is equally likely to fall in any one of these years, 258 CHAPTER 6. EXPECTED VALUE AND VARIANCE xm(x)(x −7/2) 2 1 1/6 25 /4 2 1/6 9 /4 3 1/6 1 /4 4 1/6 1 /4 5 1/6 9 /4 6 1/6 25 /4 Table 6.6: Variance ... constant probability of about 5/8 of surviving for another decade. 8 For example, of the 64 people who survive to age 6, 5/8 of 64 or 40 survive to 16, 5/8 of these 40...

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... how large to take n to get a desired accuracy. ✷ 8.1. DISCRETE RANDOM VARIABLES 309 0 0.2 0 .4 0.6 0.8 1 0 0.02 0. 04 0.06 0.08 0.1 0 0.2 0 .4 0.6 0.8 1 0 0.02 0. 04 0.06 0.08 0 0.2 0 .4 0.6 0.8 ... 1 0 0.02 0. 04 0.06 0.08 0 0.2 0 .4 0.6 0.8 1 0 0.02 0. 04 0.06 0.08 0.1 0.12 0. 14 0 0.2 0 .4 0.6 0.8 1 0 0.02 0. 04 0.06 0.08 0.1 0.12 0 0.2 0 .4 0.6 0.8 1 0 0.05 0.1 0.15 0.2...

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... RANDOM WALKS IN EUCLIDEAN SPACE 47 3 5 10 15 20 25 30 35 40 -1 0 -8 -6 -4 -2 2 4 6 8 10 Figure 12.1: A random walk of length 40 . Theorem 12.1 The probability of a return to the origin at time 2m is ... that S(x)=2R(x) − 1 2  1 1 −4x  + 1 2  √ 1 −4x  and R(x)=2xS(x)+x  1 1 −4x  . (d) Show that R(x)= x (1 −4x) 3/2 , and S(x)= 1 2  1 (1 −4x) 3/2  − 1 2  1 1 −4x...

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... 35 40 -1 0 -8 -6 -4 -2 2 4 6 8 10 Figure 1.1: Peter’s winnings in 40 plays of heads or tails. One can understand this calculation as follows: The probability that no 6 turns up on the first toss ... more appropriate to assign a distribution function which assigns probability .513 to the outcome boy and probability .48 7 to the outcome girl than to assign prob...

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... {a 1 ,a 2 ,a 3 ,a 4 } can be written in the form σ =  12 34 2 143  , indicating that a 1 went to a 2 , a 2 to a 1 , a 3 to a 4 , and a 4 to a 3 . Ifwealwayschoosethetoprowtobe1234then, to prescribe ... distributions. 78 CHAPTER 3. COMBINATORICS Number of people Probability that all birthdays are different 20 .5885616 21 .5563117 22 .5 243 047 23 .49 27028 24 .46 16...

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... VARIABLES 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0. 04 0.05 0.06 0.07 0.08 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0. 04 0.05 0.06 0.07 0.08 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0. 04 0.05 0.06 0.07 0.08 n ... series lengths: four-game, five-game, six-game, seven-game. (b) Using one of the distribution found in part (a), find the probability that his batting average exceeds .4...

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... 1.1 .3 643 2.1 .48 21 3.1 .49 90 .2 .0793 1.2 .3 849 2.2 .48 61 3.2 .49 93 .3 .1179 1.3 .40 32 2.3 .48 93 3.3 .49 95 .4 .15 54 1 .4 .41 92 2 .4 .49 18 3 .4 .49 97 .5 .1915 1.5 .43 32 2.5 .49 38 3.5 .49 98 .6 ... .49 98 .6 .2257 1.6 .44 52 2.6 .49 53 3.6 .49 98 .7 .2580 1.7 .45 54 2.7 .49 65 3.7 .49 99 .8 .2881 1.8 .46 41 2.8 .49 74 3.8 .49 99 .9 .3159 1.9 .47 13 2.9 .49 81 3.9 .5000...

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