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# ch_03 - CHAPTER 3 3.1(a P(A occurs atleast twice in n...

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Unformatted text preview: CHAPTER 3 3.1 (a) P(A occurs atleast twice in n trials) = 1 — P(A never occurs in 11. trials) — P(A occurs once in 11. trials] = 1 - (1 - p)" - np(1 - 10)"—1 (b) P(A occurs atleast thrice in 71. trials) = 1 — P(A never occurs in n trials) — P(A occurs once in 17. trials) —P(A occurs twice in 17. trials) = 1 — <1 —p)" — npa —p)'--1 — Wm —p)'--2 3.2 . 1 1 1 P(doublesiz) — E x a — g P( “double six atleast three times in 11 trials”) 1 - (56’) (316)° (3%)“) - (51°) (3%) (3%)” - (52°) (3%)2 (3%)“ 3—3 If A - {seven}, then P(A) - 36-6- P(K) =% If the dice are tossed 10 times, then the probability that X will occur 10 times equals (5/6)“). Hence, the probability p that {seven} will show at least once equals 1 - (5/6)1° 3—4 3—5 If k is the number of heads, then P{even} - P{k - 01+ pm a 2} + — 4 n-4 ' qn + (3)9261 2 4.61:)? q + But :1 n n-l n 2n-2 ls(q+q)n=q +(1)Pq +(2)Pq + -1 n n-Z (p - q)n - q“ - (:)p q n 2 - C I U + (2)9 q Adding, we obtain 1 + (p —. q)n s 2 P{even} In this experiment, the total number of outcomes is the number ( ,1: ) of ways of picking n out of N objects. The number of ways of picking k out of the K good components equals ( f) and the number of ways of picking n—k Out of the N—K defective components equals ( if ). Hence, the number of ways of picking k good components and n—k deafective components equals ( f) ( if). From this and (2-25) it follows that p=(f)(SII‘)/(§> —_---____..——--____--—-————--_...--___-—---_-—-—-__—._——-—————_..---__~----__——---- 191 =1— (2)6 = 0.665 1— <2>”— (If) (a (at 1 _ (2)18 _ (118) (%> (3)17 _ (128) (92 (2)16 Z 0597 10 3.7 (a) Let n represent the number of Wins required in 50 games so that the net gain or loss does not exceed \$1. This gives the net gain to be 50—71 16<n<17.3 71217 17 33 P(net gain does not exceed \$1) = (‘1’?) (21!) (i) = 0.432 P(net gain or loss exceeds \$1) = 1 — 0.432 = 0.568 (b) Let n represent the number of wins required so that the net gain or loss does not exceed \$5. This gives (50—71) —5<n—T<5 13.3<n<20 P(net gain does not exceed \$5) = 27119: 14 (5;?) (Elan (\$50—71 = 0.349 P(net gain or loss exceeds \$5) = 1 — 0.349 = 0.651 3.8 Deﬁne the events A2“ 7“ successes in n Bernoulli trials” B=“success at the ith Bernoulli trial” 0’: “r — 1 successes in the remaining 72 — 1 Bernoulli trials excluding the ith trial” We need P(AB) 13(30) P(B)P(C) 7» P“%”=_HA)=_HA)= .mA) n' 3.9 There are 8%) ways of selecting 13 cards out of 52 cards. The number of ways to select 13 cards of any suit (out of 13 cards) equals (g) = 1. Four such (mutually exclusive) suits give the total number of favorable outcomes to be 4. Thus the desired probability is given by i = 6.3 x 10—12 (i3) 12 3.10 Using the hint, we obtain 19(Nk+1 — Nk) = (1(Nk — Nk—1)_ 1 Let Mk+l = Nk+1 — Nk so that the above iteration gives Mk+1 =ng—l H H: | I? ’B M Q This gives Thus 13 which gives for 2' = a a+b l-MMV a p—q' _ a+b—p-q’m’éq M2: 1 WM) M, p=q 1) (1+1) . 1—(Q/p)b a %—1 MF4-1-MMV” p_q’p#q w, p=q 3.11 Pn :an+a+an—ﬁ Arguing as in (3.43), we get the corresponding iteration equation Pn:Pn+a+an—ﬁ and proceed as in Example 3.15. 3.12 Suppose one bet on k = 1, 2, - - - , 6. ’Then p. = m 0.. M.) = o e) (at p2==ka:appear(nztwo¢hce)==(3) (%)2 (g) 293 = P(k appear on all the tree dice) = (if p0==rmk:appau~none)==(%)3 'Thus,VK3get Net gain 2 2p1 + 3192 + 4193 — p0 = 0.343. 14 ...
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