Introduction to Probability - Chapter 9 pps

... 1.2 .38 49 2.2 .4861 3.2 . 499 3 .3 .11 79 1.3 .4032 2.3 .4 893 3.3 . 499 5 .4 .1554 1.4 .4 192 2.4 . 491 8 3.4 . 499 7 .5 . 191 5 1.5 .4332 2.5 . 493 8 3.5 . 499 8 .6 .2257 1.6 .4452 2.6 . 495 3 3.6 . 499 8 .7 .2580 ... 1.7 .4554 2.7 . 496 5 3.7 . 499 9 .8 .2881 1.8 .4641 2.8 . 497 4 3.8 . 499 9 .9 .31 59 1 .9 .4713 2 .9 . 498 1 3 .9 .5000 Figure 9. 4: Table of values of NA(0,z),...

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... PROBABILITIES 3 .2033 09 .762057 .151121 .623868 .93 2052 .415178 .7167 19 .96 7412 .0 696 64 .67 098 2 .352320 .0 497 23 .750216 .784810 .0 897 34 .96 6730 .94 6708 .380365 .027381 .90 0 794 Table 1.1: Sample ... vol. 52, no. 3 (May 197 9), pp. 18 3-1 84. 26 M. vos Savant, Parade Magazine, 3 March 199 6, p. 14. 1.2. DISCRETE PROBABILITY DISTRIBUTIONS 25 First toss Second toss Thir...

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... itself determined by a probability vector. Theorem 11 .9 Let P be the transition matrix for a regular chain and v an arbi- trary probability vector. Then lim n→∞ vP n = w , where w is the unique fixed probability ... going to a certain four-year medical school in northern New England has, each year, a probability q of flunking out, a probability r of having to repeat the year, an...

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... 2532 3.1 596 Smith, 1855 .6 3204 1218.5 3.1553 De Morgan, c.1860 1.0 600 382.5 3.137 Fox, 1864 .75 1030 4 89 3.1 595 Lazzerini, 190 1 .83 3408 1808 3.141 592 9 Reina, 192 5 .54 19 2520 8 69 3.1 795 Table ... planning one would have to be extremely lucky to be able to stop so cleverly. The second author likes to trace his interest in probability theory to the Chicago World’s...

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... =  1234 2143  , 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 the permutation, we need only give the bottom row, with the understanding ... Griffin, 196 2), p. 146. 4 M. vos Savant, Ask Marilyn, Parade Magazine, Boston Globe, 21 August 199 4. 5 J. Riordan, An Introduction to Combinatorial Analysis...

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... 4.26. 148 CHAPTER 4. CONDITIONAL PROBABILITY .001 can not .01 .95 .05 + - .001 0 .05 .94 9 + - .051 .94 9 + - .98 1 1 0 can not .001 .05 0 .94 9 can not .0 19 Original Tree Reverse Tree .99 .99 9 Figure ... Appendix C) to compute a con- ditional probability. The number 93 ,753 in the table, corresponding to 4 0- year-old males, means that of all the males b...

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... 2 892 7 3657 8 3025 9 3362 10 298 5 11 3138 12 3043 13 2 690 14 2423 15 2556 16 2456 17 24 79 18 2276 19 2304 20 197 1 21 2543 22 2678 23 27 29 24 2414 25 2616 26 2426 27 2381 28 20 59 29 20 39 30 2 298 31 ... appears to grow without limit. We can now ask: How long will a customer have to wait in the queue for service? To examine this question, we let W i be the length of tim...

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... VARIANCE W L W L W L W L W L W L (2,3,12) L 10 9 8 6 5 4 (7,11) W 1/3 2/3 2/5 3/5 5/11 6/11 5/11 6/11 2/5 3/5 1/3 2/3 2 /9 1/12 1 /9 5/36 5/36 1 /9 1/12 1 /9 1/36 2/36 2/45 3/45 25/ 396 30/ 396 25/ 396 30/ 396 2/45 3/45 1/36 2/36 Figure ... the probability that the player wins 1 dollar; it is 244/ 495 . The probability of losing is then 251/ 495 . Thus if X is his winning for a...

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... variance 1, then (cf. Exam- ple 7.5) f X i (x)= 1 √ 2π e −x 2 /2 , 4 J. B. Uspensky, Introduction to Mathematical Probability (New York: McGraw-Hill, 193 7), p. 277. 290 CHAPTER 7. SUMS OF RANDOM ... 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 .400 in a four-ga...

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... if n = 1000, this is better than .99 and if n =10,000, this is better than .99 9. We can illustrate what the Law of Large Numbers says for this example graph- ically. The density for A n = S n /n ... <A 100 <.4) ≥ . 79 , P (.2 <A 1000 <.4) ≥ .97 9 . These values should be compared with the actual values, which are (to six decimal places) P (.2 <A 100 <.4) ≈ .96 25 49 P (...

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