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 a) 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)=? 1) 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 2) 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 b) In the same testing results, if P(C)=0.3, then P(C/HPSA)=? P(C/LPSA)=? 1) By the same approach, P(C/HPSA)=0.46 2) P(C/LPSA)=0.266

4. Problem 37 a) P{(1-2-3)U(4-5)}=P(1-2-3)+P(4-5)-P(1-2-3- 4-5)=p 1 p 2 p 3 +p 4 p 5 -p 1 p 2 p 3 p 4 p 5, (it is also equal to 1 -P[(1-2-3) c ]P[(4-5) c ]=1-(1-p 1 p 2 p 3 )(1-p 4 p 5 ) b) P{[(1-2)U(3-4)] ∩5 }= p 5 X [ 1-(1-p 1 p 2 )(1-p 3 p 4 ) ] c) If the circuit 3 is open, then the probability will be the same as a): [1-(1-p 1 p 4 )(1-p 2 p 5 )]X(1-p 3 ); 1) In the other condition, if the circuit 3 is closing, then the probability will be [1-(1-p 1 )(1-p 2 )] X[1-(1-p 4 )(1-p 5 )] X p 3 2) The summation will be the probability of current flowing from A to B