
Introduction To Probability Solutions Manual 评分:
Introduction To Probability Solutions Manual(Charles M Grinstead, J Laurie Snell)
P(female lives to age c mumber of female survivors at age r 100.000 29. Solution by richard beigel a)In order to emerge from the interchange going west, the car must go straight at the first point cf decisioI, then make 47+1 right turILS, and finally go straight a second line. The probabilily P(r)of this occurring is P(m)=E(1n2n42+1=71m)2211 1+ 0≤r<1,btP(1)=0.SoP(1/2)=2/l5 (b) Using standard methods from calculus, one can show that P(r)attains a maximum at the value At this valuc of r, P(r)N.15 a)Assuming that each student gives any given Lire as anl ans wer with probability 1/4, then prob ability that they both give the same answer is 1/4 (b)In this case, they will both answer right front with probability (.58)2, etc. Thus, the probability that they both give the same answer is 39. 8% section 2.1 The problems in this section are all computer programs SECTION 2.2 (u)=1/80n(2,10 (b) fPC 6a 3 log 5 ≈.621 (b)P(a,b)(621)log(b/a) P(x>5) lo agJ P(x<7) log(7/2) ≈.778 log 5 log(25/7) ≈.791 log 5 ≈,632 7.()13,()1/2,(c)1/2,(d)1/3 13.2log21. 15.Y section 3.1 32 5.9,6 n2728 11 m3’271000 263×103 b)(3)×263×103 12·11……(12m+1) il n<12 and 127 21. They are the same 11 (b) She will get the best candidate if the second best candidate is in the first half and the best candidate is in the secon half. The probability that this happens is greater than 1/4 section 3.2 a)20 (b).0064 (e)0256 f) g)10 3() 36 5.98,965,729 7. b(n, p, j) pg (mj+1)!(j1)p b(n,p,j1) (nj+1)p But (j+1)p if and only if j< p(n +1), g=p(n+1 gives b(n, p, j) its largest value. If p(+1) is an integer there will be two possible values of j, namely j= p(n +1) and j=(7+1)1 9.n=15,y=7 11. Eight pieces of each kind of pie 13. The number of subsets of 2n objects of size j is(2n 2ni+1 >1→<+ 2m Thus i= n makes maximum 15.3143..441.181.027 17. There are()ways of putting a different objects into the 1st box, and then ("a)ways of putting e different objects into the 2nd and then one way to put the remaining objects into the 3rd box Thus the total number of ways is =723×108 52 (b) 1八2八4八3八3 044. 13\/13)/13\/13 315 21. 3(25)3=93( We subtract 3 because the three pure colors are each counled twice. 23. To make the boxes. you need n+1 bars. 2 on the ends and n1 for the divisions. The n1 bars and the r objects occupy n1+r places. You can choose any nl of these n1+r places for the bars and use the remaining r places for the objects. Thus the number of ways this can be done is 1+ 20(0)ym=n2 5 ≈.042 27. Ask John to make 42 trials and if he gets 27 or Inore correct accept his claim. Then the probability typ ∑b42,5,)=04, k>27 and the probability of a type Ii error is ∑b2,75,4)=0 29. b(n, p, m)=(m/pm(1p)nmm. Taking the derivative with respect to p and setting this equal to 71 a we obtain m(1p)=p(nm) and so p=m/n 96 33. By Stirling s formula. V4n(2n 2mne27)4 ml 4(4n)!(V2nn(nmmyen4v2r(4n)(4nyne4m VTn 2m 35. Consider an urn with n red balls and T, blue balls inside. The lelt s de of the identity ∑()=∑ counts Che nuInber of ways to choose n balls out of the 2n balls in Che url. The right hand counts the same thing but brcaks the counting into the sum of thc cascs where therc arc cxactly j red balls and n, blue balls 38. Consider the Pascal triangle(mod 3) for example 2 121 31001 4 11011 121121 61002001 1022011 121212121 91000000001 1011000000011 11121000000121 121001000001001 1311011000011011 121121000121121 51002001001002001 1611022011011022011 17121212121121212121 181000000002000000001 Note first that the entries in the third row are 0 for 0<3<3. Lucas notes that this will be true for iuny p. To see this assuine that 0<i<p. Note that PPIp +1 is an integer. Sincc p is primc and 0<3< p, p is not divisible by any of thc tcrms of j and so p1! must be divisible by j!. Thus for 0 <j< p we have 0 mod p. Let us call the triangle of the first three rows a basic triangle. The fact that the third row is 1001 produces two more basic triangles in the next three rows and an inverted triangle of 0's between these two basic triangles. This leads to the gth row 1002001 This produces a basic triangle, a basic triangle multiplied by 2(mod 3), and then another basic triangle in the next three rows. Again these triangles are separated by inverted 0 triangles. We can continue this way to construct the entire Pascal triangle as a bunch of multiples of basic triangles separated by inverted 0 triangles. We need only know what the mutiples are. The multiples in row mLp occur at positions 0, p, 2p,.p. Looking at the triangle we see that the multiple at position (mp, jp)is the sum of the multiples at positions 31)p and jp in the(mlp'th row. Thus these multiples satisfy the same recursion relation that determined the Pascal triangle. Therefore the multiple at position(mp, jp) in the triangle is m). Suppose we want to determine the value in the Pascal triangle mod p at the position(n, j) Let n= sp+ so and j=rp+ro, where so and To are p. Then the point(n, j)is at position 9o, ro) in a basic triangle multiplied by s Us But now we can repeat this proccss with the pair(, r)and continuc until s <p. This gives us thc resu 0+s1p+s2p2+…+skp 70+r1m2+r2p2+…+ Ifr,>s, for some j then the result is 0 since, in this case, the pair(s;, T,, )lies in one of the inverted ang Tf we consider the row ph1 then for all b, Sh=P1 and rr<p1 so the product, will be positive resulting in no zeros in the rows pk1. In particular for p=2 the rows pk1 will consist of all S. 39 b(2t,,n) 2(27 m!2n.2(n1) 2(71)…2 (271)(23)…1 n(2n section 3.3 3.(a)9699% (b)55.16 SECTION 4.1 (b)2/3 5.(a)(1)and(2) 7.(a) P(AnB)=P(ACC)=P(BnC) (b)(1) P(A)P(B)=P(A)P(C)=P(BP(C) 1 P(A∩BnC)=元≠P(A)F(B)PC) (O P(AC)=P(A)P(C)=, so C and A are independent P(CnB)=P(B)(C)=, soC and B are independent P(CA(AnB))=4*P(O)P(Anb)=8 so C and An B are not independent 8.P(A∩BC)=P({})=, P(4=P(B)=P()=2 Thus whilc p(A∩BnC)=P(A)P(B)P(C)=。, P(A∩B1=P(A∩C)=P(BC5 16 P(AP(B)P(AP(C)P(BP(C Therefore no two of these events are independent 9 (b)1/2 13.12 11八(2 307 13 483 11/1 17.(a)P(A∩B)=P(4)P(A∩B)=P)P(A)P(B P、A)(1P(B) PrAP(B (b) Use(a), replacing A by B and B by A 19.273 23. Put one white ball in one urr and all the rest in the other urn. this gives a probability of nearly 3/ 4, in particular greater thar. 1/2, for obtaining a white ball which is what you would have with an equal number of balls in each urn. Thus the best choice must have more white balls in one urn than the other. In the urn with more white balls, the best we can do is to have probability 1 of getting a white ball if this urn is chosen. In the urn with less white balls than black. the best we can do is to have one less white ball than black and then to have as many white balls as possible. Our solution is thus best for the urn with more white balls than black and also for the urn with morc black balls than white. Thcrcforc our solution is thc best we can do 25. We imust have  nk q This will be true if and only if mp=k. Thus p must equal k/n 27 a)P(Pickwick has no umbrella, given that it rains (b)P(It does not rain, given that he brings his umbrella)=S 12 P(Accepted by Dartmouth Accepted by Harvard The events ' Accepted by Dartmouth'and "Accepted by Harvard are not independent 31. The probability of a 60 year old male living to 80 is 41, and for a female it is 62 33. You have to make a lot of calculations all of which are like this: P(A11A2A3)=P(42)P(4a)P(A1)P(A2)P(A3) =P(A2)P(A3)(1P(A1) =P(41)P(A2)P(A3) 35. The random variables X1 and X2 have the same distributions, and in each case the range values are the integers between I and 10. The probability for each value is 1/10. They are independent If the first number is not replaced, the two distributions are the same as before but the two random variables are not independent. 37. P(max(X,r)=a)=P(X=a,Ysa)+P(XSa,Y=a)P(X=a,Y=a P(min(X,Y)=a=P(X=a,Y>a)+P(X>W,Y=a+P(X=a,y=a Thus P(max(X,)=a+P(min(X,Y)=a)=P(X=a)+P(Y=a and so u=t+sr 39.(a)1/9 (b)1/4 c) No 210124 43.,710. (a)The probability that, the first player wins under either service convention is equal to the proba ity that if a coin has probability p of coming up heads, and the cain is tossed 2N +1 times, then it comes up heads more often than tails. This probability is clearly greater than. 5 if and only if ( b) If the first team is serving on a given play, it will win the next point if and only if one of the folowing sequences of plays accurs (where ' means that the team that is serving wins the play and 'L' means that the team that is serving loses the play) W. LLW. LLLLW The probability that this happens is equal to mtaptqpt which equals 1q21+q Now, consider the game where a 'new play'is defined to be a sequence of plays that ends with a point being scored. Then the service convention is that at the beginning of a new play, the team that won the last new play serves. This is the same convention as the second convention in the preceding problem A FrOIn part a), we know that the first tealnl to serve under the second service convention will win the game more than half the time if and only if p>.5. In the present case, we use the new value20181211 上传 大小：15.03MB

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