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Suppose event A implies B (i.e. A B ). Show that if the pair { A , B } is independent, then either P ( A ) = 0 or P ( B ) = 1 .

A B implies P ( A B ) = P ( A ) ; independence implies P ( A B ) = P ( A ) P ( B ) . P ( A ) = P ( A ) P ( B ) only if P ( B ) = 1 or P ( A ) = 0 .

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A company has three task forces trying to meet a deadline for a new device. The groups work independently, with respective probabilities0.8, 0.9, 0.75 of completing on time. What is the probability at least one group completes on time? (Think. Then solve “by hand.”)

At least one completes iff not all fail. P = 1 - 0 . 2 0 . 1 0 . 25 = 0 . 9950

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Two salesmen work differently. Roland spends more time with his customers than does Betty, hence tends to see fewer customers. On a given day Roland seesfive customers and Betty sees six. The customers make decisions independently. If the probabilities for success on Roland's customers are 0 . 7 , 0 . 8 , 0 . 8 , 0 . 6 , 0 . 7 and for Betty's customers are 0 . 6 , 0 . 5 , 0 . 4 , 0 . 6 , 0 . 6 , 0 . 4 , what is the probability Roland makes more sales than Betty? What is the probability that Roland will makethree or more sales? What is the probability that Betty will make three or more sales?

PR = 0.1*[7 8 8 6 7];PB = 0.1*[6 5 4 6 6 4];PR3 = ckn(PR,3) PR3 = 0.8662PB3 = ckn(PB,3) PB3 = 0.6906PRgB = ikn(PB,0:4)*ckn(PR,1:5)' PRgB = 0.5065
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Two teams of students take a probability exam. The entire group performs individually and independently. Team 1 has five members and Team 2has six members. They have the following indivudal probabilities of making an `”A” on the exam.

Team 1: 0.83 0.87 0.92 0.77 0.86 Team 2: 0.68 0.91 0.74 0.68 0.73 0.83

  1. What is the probability team 1 will make at least as many A's as team 2?
  2. What is the probability team 1 will make more A's than team 2?
P1 = 0.01*[83 87 92 77 86];P2 = 0.01*[68 91 74 68 73 83];P1geq = ikn(P2,0:5)*ckn(P1,0:5)' P1geq = 0.5527P1g = ikn(P2,0:4)*ckn(P1,1:5)' P1g = 0.2561
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A system has five components which fail independently. Their respective reliabilities are 0.93, 0.91, 0.78, 0.88, 0.92. Units 1 and 2 operate as a “series”combination. Units 3, 4, 5 operate as a two of three subsytem. The two subsystems operate as a parallel combination to make the complete system. What is reliability of the completesystem?

R = 0.01*[93 91 78 88 92];Ra = prod(R(1:2)) Ra = 0.8463Rb = ckn(R(3:5),2) Rb = 0.9506Rs = parallel([Ra Rb])Rs = 0.9924
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A system has eight components with respective probabilities

0 . 96 0 . 90 0 . 93 0 . 82 0 . 85 0 . 97 0 . 88 0 . 80

Units 1 and 2 form a parallel subsytem in series with unit 3 and a three of five combination of units 4 through 8. What is the reliability of the complete system?

R = 0.01*[96 90 93 82 85 97 88 80];Ra = parallel(R(1:2)) Ra = 0.9960Rb = ckn(R(4:8),3) Rb = 0.9821Rs = prod([Ra R(3) Rb])Rs = 0.9097
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How would the reliability of the system in [link] change if units 1, 2, and 3 formed a parallel combination in series with the three of five combination?

Rc = parallel(R(1:3)) Rc = 0.9997Rss = prod([Rb Rc])Rss = 0.9818
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How would the reliability of the system in [link] change if the reliability of unit 3 were changed from 0.93 to 0.96? What change if the reliabilityof unit 2 were changed from 0.90 to 0.95 (with unit 3 unchanged)?

R1 = R; R1(3) =0.96;Ra = parallel(R1(1:2)) Ra = 0.9960Rb = ckn(R1(4:8),3) Rb = 0.9821Rs3 = prod([Ra R1(3) Rb])Rs3 = 0.9390 R2 = R;R2(2) = 0.95; Ra = parallel(R2(1:2))Ra = 0.9980 Rb = ckn(R2(4:8),3)Rb = 0.9821 Rs4 = prod([Ra R2(3) Rb]) Rs4 = 0.9115
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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