hw3 solution

# hw3 solution - Chapter 3 1 Y = 3 SSS Y = 4 FSSS Y = 5 FFSSS...

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1. Y = 3 : SSS; Y = 4: FSSS; Y = 5: FFSSS, SFSSS; Y = 6: SSFSSS, SFFSSS, FSFSSS, FFFSSS; Y = 7: SSFFS, SFSFSSS< SFFFSSS, FSSFSSS, FSFFSSS, FFSFSSS, FFFFSSS 2. a. P(2) = P(Y = 2) = P(1 st 2 batteries are acceptable) = P(AA) = (.9)(.9) = .81 b. p(3) = P(Y = 3) = P(UAA or AUA) = (.1)(.9) 2 + (.1)(.9) 2 = 2[(.1)(.9) 2 ] = .162 c. The fifth battery must be an A, and one of the first four must also be an A. Thus, p(5) = P(AUUUA or UAUUA or UUAUA or UUUAA) = 4[(.1) 3 (.9) 2 ] = .00324 d. P(Y = y) = p(y) = P(the y th is an A and so is exactly one of the first y – 1) =(y – 1)(.1) y-2 (.9) 2 , y = 2,3,4,5,… 3. a. p(1) = P(M = 1 ) = P[(1,1)] = 36 1 p(2) = P(M = 2 ) = P[(1,2) or (2,1) or (2,2)] = 36 3 p(3) = P(M = 3 ) = P[(1,3) or (2,3) or (3,1) or (3,2) or (3,3)] = 36 5 Similarly, p(4) = 36 7 , p(5) = 36 9 , and p(6) = 36 11 b. F(m) = 0 for m < 1, 36 1 for 1 m < 2, F(m) = 1 0 36 25 36 16 36 9 36 4 36 1 6 6 5 5 4 4 3 3 2 2 1 1 m

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hw3 solution - Chapter 3 1 Y = 3 SSS Y = 4 FSSS Y = 5 FFSSS...

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