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Homework #2

Homework #2. Chapter 3. Problem 18. Permutation between 5 boys and 10 girls The 4th position is a boy (14!/15!) X 5 =1/3 The 12th position is a boy The same reasoning= 1/3 The 3rd position is a particular boy (14!)/(15!)=1/15. Problem 34.

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Homework #2

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  1. Homework #2 Chapter 3

  2. Problem 18 • Permutation between 5 boys and 10 girls • The 4th position is a boy • (14!/15!) X 5 =1/3 • The 12th position is a boy • The same reasoning= 1/3 • The 3rd position is a particular boy • (14!)/(15!)=1/15

  3. Problem 34 • If P(HPSA/C)=0.268 (i.e., P(LPSA/C)=0.732), P(HPSA/NC)=0.135 (i.e., P(LPSA/NC)=0.865), and, P(C)=0.7, then P(C/HPSA)=? P(C/LPSA)=? • By using the Bayes’ rule, P(C/HPSA)= P(C∩HPSA) / P(HPSA) =[P(C)P(HPSA/C)]/ {[P(C)P(HPSA/C)]+[P(NC)P(HPSA/NC)]} =0.82 • P(C/LPSA)= P(C∩LPSA) / P(LPSA) =[P(C)P(LPSA/C)]/ {[P(C)P(LPSA/C)]+[P(NC)P(LPSA/NC)]}=0.69 • In the same testing results, if P(C)=0.3, then P(C/HPSA)=? P(C/LPSA)=? • By the same approach, P(C/HPSA)=0.46 • P(C/LPSA)=0.266

  4. Problem 37 • P{(1-2-3)U(4-5)}=P(1-2-3)+P(4-5)-P(1-2-3-4-5)=p1p2p3+p4p5-p1p2p3p4p5 , (it is also equal to 1 -P[(1-2-3)c]P[(4-5)c]=1-(1-p1p2p3)(1-p4p5) • P{[(1-2)U(3-4)]∩5}= p5X [1-(1-p1p2)(1-p3p4)] • If the circuit 3 is open, then the probability will be the same as a): [1-(1-p1p4)(1-p2p5)]X(1-p3); • In the other condition, if the circuit 3 is closing, then the probability will be [1-(1-p1)(1-p2)] X[1-(1-p4)(1-p5)] X p3 • The summation will be the probability of current flowing from A to B

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