Cheating in Exam
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3. a) What is the probability that none of these vehicles requires warranty service?
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P(x=0) = 12 C0 (0.10)0 (1-0.10)12-0
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= (1) (1) (0.28243)
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=0.28243
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b) What is the probability that exactly nine of these vehicles require warranty service?
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P(x=9) = 12 C9 (0.10)9 (1-0.10)12-9
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= (220) (0.000000001) (0.729)
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=0.00000022 X 0.729
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=0.000000016038
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c) What is the probability that greater than nine of these vehicles require warranty service?
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P (x > 9) = p (x= 9), p (x=10), p (x =11), p (x = 12).
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= 12 C9 (0.10)9 (1-0.10)12-9+12 C10 (0.10) 10 (1-0.10)12-10+ 12 C11 (0.10) 11 (1-0.10)12-11+ 12 C12 (0.10)12 (1-0.10)12-12
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=0.000000016038+0.000000005346+0.000000000108+0.000000000001
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=0.000000021493
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d) What is the probability that three and fewer of these vehicles requires warranty service?
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P (x≤ = 3) =p (x =3), P(x= 2), p (x=1), p (x = 0)
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12 C3 (0.10)3 (0.9)9+12 C2 (0.10)2(0.9)10 + 12 C1 (0.10)1
3. a) What is the probability that none of these vehicles requires warranty service?
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P(x=0) = 12 C0 (0.10)0 (1-0.10)12-0
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= (1) (1) (0.28243)
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=0.28243
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b) What is the probability that exactly nine of these vehicles require warranty service?
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P(x=9) = 12 C9 (0.10)9 (1-0.10)12-9
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= (220) (0.000000001) (0.729)
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=0.00000022 X 0.729
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=0.000000016038
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c) What is the probability that greater than nine of these vehicles require warranty service?
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P (x > 9) = p (x= 9), p (x=10), p (x =11), p (x = 12).
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= 12 C9 (0.10)9 (1-0.10)12-9+12 C10 (0.10) 10 (1-0.10)12-10+ 12 C11 (0.10) 11 (1-0.10)12-11+ 12 C12 (0.10)12 (1-0.10)12-12
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=0.000000016038+0.000000005346+0.000000000108+0.000000000001
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=0.000000021493
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d) What is the probability that three and fewer of these vehicles requires warranty service?
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P (x≤ = 3) =p (x =3), P(x= 2), p (x=1), p (x = 0)
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12 C3 (0.10)3 (0.9)9+12 C2 (0.10)2(0.9)10 + 12 C1 (0.10)1