ch_02 - CHAPTER 2 2—1 We use De Morgan's law: (a) K+§+...

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Unformatted text preview: CHAPTER 2 2—1 We use De Morgan's law: (a) K+§+ 3+3 = AB + A3 = A(B+'B) = A (b) (A+B)(KF) = (A+B)(A‘+§) = A§+ BK because A; = m m? = w} ' WW”... 2-2 IfA={2f_x_<_5} B={3fx:6} S={-°°<x<°°} then A+B={2§x§6} AB={3§x§5} (“mm = {29:56} [{x<3} + {x>5}] ={2§x<3}+{5<x_<_6} 2-3 If AB = {m} then ACE hence P(A) 5w?) 2—4 (a) P(A) = P(AB) + MAE) pas) = P(AB) + P(KB) If, therefore, P(A)==P(B) - P(AB) then P(Afi) = 0 P(KB) = 0 hence P(EBHE) = P(KB) + P(Ai) - o (b) If P(A) = P(B) = 1 then 1 = P(A)§ P(A-FB) hence I = P(A+B) = P(A) + P(B) - P(AB) = 2 - P(AB) This yields P(AB) = l 2-5 From (2—13) it follows that P(A+B+C) = P(A) + P(B+C) - P[A(B+C)] P(B+C) = P(B) + P(C) - P(BC) P[A(B+C)] = P(AB) + P(AC) - P(ABC) because ABAC = ABC. Combining, we obtain the desired result. Using induction, we can show similarly that P(A1+A2+ ---+An) = P(Al) + P(A2)+ ...+P(An) - P(AlAz) — _ P(An_1An) + P(A ) + ...+-P(An-2An) 1A2A3 *_*1’(A1A2 - ‘ - An) w--. ..-.... w— m... V... n... .-W 2-6 Any subset of S contains a countable number of elements, hence, it can be written as a countable union of elementary events. It is therefore an event. 2-7 Forming all unions, intersections, and complements of the sets {1} and {2,3}, we obtain the following sets: {6}, {1}, {4}, {2,3}, {1,4}, {1,2,3}, {2,3,4}, {1,2,3,4} 2-8 If ACB,P(A) = 1/4, and P(B) - 1/3, then P(AB) a pm _, 1/4 : 3 P(AIB)=_P(B) Pm m z _ P(AB) a P(A) _ P(B|A) _ P(A) P(A) ' 1 2-9 P(A|BC)P(B|C) , %£%&%1 zflfigg P(ABC) a P(C) = P(MIC) P(ABC) P(BC) P(BC) P(C) P(A|BC)P(B|C)P(C) = P(C) = P(ABC) 2-10 We use induction. The formula is true for n==2 because P(A1A2) = P(A2IA1)P(A1). Suppose that it is true for n. Since p(A ...A1) = P(A ...A2A1)p(A1 ...An) n+lAn n+l|An we conclude that it must be true forr1+1. 2-11 First solution. The total number of m element subsets equals (3) (see Probl. 2-26). The total number of m element subsets containing :0 equals n—l (m_l). Hence n n-l m p a(m)/(rn--l) = E Second solution. Clearly, P{coIAm} = m/n is the probability that to is in a specific Am. Hence (total probability) m m p = X PicolAm}p(Am) = a Z P(Am) = n where the summation is over all sets Am. v“ 2—12 (a) P{6ft§8}=1% (b) P{6:t§8|t>5} , 2136:1258} 3% 2-13 From (2-27) it follows that to+t1 P{toft§to+tlltzto} 'J aunt/I a(t)dt t t o 0 t1 PM: 5 t1}= I a(t)dt O Equating the two sides and setting tl-to+ At we obtain a(to)/J a(t)dt = «1(0) t o for every to. Hence, ” ” «(0):o - in J a(t)dt = a(0)to I a(t)dt - e t t o o Differentiating the setting c=a(0), we chclude that -ct Ct P{t§t1}-=1-e 1 a(to) = c e W 2-14 If A and B are independent, then P(AB) =P(A)P(B). If they are mutually exclusive, then P(AB) - 0. Hence, A and B are mutually exclusive and independent iff P(A)P(B) - 0. WWW 1 = AlAZ + AlA2 hence P(A1) = P(A1A2) + P(A1A2) 2-15 Clearly, A If the events Al and K2 are independent, then P(A1A2) = P(A1) — P(A1A2) = P(A1) - P(A1)P(A2) - P(A1)[1 —P(A2)] = P<A1)P(K2) hence, the events A1 and K2 are independent. Furthermore, S is independent with any A because SA - A. This Yields P(SA) = NA) = P(S)P(A) Hence, the theorem is true for n=2. To prove it in general we use induction: Suppose that An+1 is independent of A An. Clearly, 1,000, An+1 and An+1 are independent of Bl....,Bn. Therefore P(Bl N . BnAn+l) I P(Bl ' O . Bn)P(An+l) 1:031 ... BnAn+1) = P(Bl 'H Bn)P(An+1) 2.16 The desired probabilities are given by m -~ 1 k: — 1 n k 2.17 Let 141,142 and A3 represent the events A1 2 “ball numbered less than 07" equal to m is drawn” A2 : “ball numbered m is drawn” A3 :: “ball numbered greater than m is drawn” P(A1 occurs n1 2 k — 17 A2 occurs n2 2 1 and A3 occurs n3 2 0) (m + 71-2 + _ ml 1712 ng t“ m" " ‘ P p I): n1! n2! n3! 1 2 5 I fh)! (Swill—1 (k <33)“ | l 2.18 All cars are equally likely so that the first car is selected with probability p : 1/3. This gives the desired probability to be (if) (if (if = 0’26 2.19 P{“d7‘awmg a white ball ’7} 2 P(“atleat one white ball m k trials ” n ) z 1 — P(“ull black balls in k trials”) (“1 : 1....j... m+n ( k > 2.20 Let D = 27’ represent the penny diameter. 80 long as the center of the penny is at a distance of 7“ away from any side of the square, the penny will be entirely inside the square. This gives the desired probability to be @3341“?le 2.21 Refer to Example 3.14. (a) Using (3.39), we get (b) P(“tw0 one—digit and four two—digit numbers”) 2 P(“all one — digit numbers”) 2 . n The number of equatlons of the form P(AiAk) P(A1)P(Ak) equals (2). The number of equations involving r sets equals (g). Hence the total number N of such equations equals n n n . = +... v <2)+(3) +(n) And since n n n _ :1: n (0)+(1)+---+(n) - (1+1) 2 we conclude that n=2“- (3)-<‘1‘)=2“-1-n 1 and B2 by R the set of red balls. We denote by B respectively the balls in boxes 1 and 2 and We have (assumption) P(Bl)=P(B2)=0.5 P(RIBl)=O.999 P(R|32)=o.001 Hence (Bayes' theorem) P(RI31)P(B1) 0.999 “31‘” " P(R|31')‘p"(n'1‘)+“'p'(“]n nzmnz) g o.999+o.oo1 = 0'9” WWW 6 2—24 We denote by Bl and 32 respectively the ball in boxes 1 and 2 and by D all pairs of defective parts. We have (assumption) P(B1) = P(BZ) = 0.5 To find P(D[Bl) we proceed as in Example 2—10: First solution. In box B1 there are 1000 x999 pairs. The number of pairs with both elements defective equals 100 x 99. Hence, 100 X 99 1000 X 999 Second solution. The probability that the first bulb selected from Bl is defective equals 100/1000. The probability that the second is defective assuming the first was effective equals 99/999. Hence, P(D|Bl) = 100 99 P(Dl31) 1000 x 999 We similarly find _ 100 99 P(D|Bz) ' 2000"1999 (a) P(D) = P(D|Bl)P(Bl) + P(DI32)P(BZ) = 0.0062 P(D|Bl)P(Bl) 9(0) _____n_______________,_______________________________________________________ (b) P(BllD) = = 0.80 2—25 Reasoning as in Example 2-13, we conclude that the probability that the bus and the train meet equals 2 2 10 x (10+x)60---—--2 -—-—2 Equating with 0.5, we find x= 60 -lO/l-l. 2—26 We wish to show that the number Nn(k) of the element Subsets of S equals a n(n—l) "' (n-k+l) Nn(k) "ITTfTTT"T?" This is true for k-l because the number of l-element subsets equals n. Using induction in k, we shall show that n-k k+1 We attach to each k-element subset of S one of the remaining n-k elements Nn(k+1) = Nn(k) 1<k<n (1) of S. We, then, form Nn(k) (n-k) k+1-element subsets. However, these subsets are not all different. They form groups each of which has k+l identical elements. We must, therefore, divide by k+1. '7 2—27 In this experiment we have 8 outcomes. Each outcome is a selection of a particular coin and a specific sequence of heads or tails; for example fhh is the outcome "weselected the fair coin and we observed hh". The event F = {the selected coin is fair} consists of the four outcomes fhh, fht, fth and fhh. Its complement F is the selection of the two— headead coin. The event HH = {heads at both tosses} consists of two outcomes. Clearly, P(F) = P(_) = —;— P(HH|F) = 7:— P(HH|F) = 1 Our problem is to find P(F|HH). From (2-41) and (2-43) it follows that P(HH) = P(HH|F)P(F) + P(HH|-I-:)P(F) = —3- P(HH|F)P(F) _ 1/4 x 1/2 _ l P(F'HH)=—Hfifi3— ‘—‘s/T "—5" ...
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue University-West Lafayette.

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ch_02 - CHAPTER 2 2—1 We use De Morgan's law: (a) K+§+...

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