Unformatted text preview: (d) P ( A'∩ B ) = 63 + 35 + 15 = 0.113 1000 0.032 P( A ∩ B) (e) P ( A  B ) = = = 0.2207 (25 + 63 + 15 + 7 + 35) / 1000 P( B) 5 (f) P = = 0.005 1000 2167. (a) Let A denote that a part conforms to specifications and let B denote for a part a simple component. For supplier 1, P(A)=P(AB)P(B)+ P(AB’)P(B’) = [(10002)/1000](1000/2000)+ [(100010)/1000](1000/2000) = 0.994 For supplier 2, P(A)= P(AB)P(B)+ P(AB’)P(B’) = [(16004)/1600](1600/2000)+ [(4006)/400](400/2000) = 0.995 (b) For supplier 1, For supplier 2, (c) For supplier 1, For supplier 2, P(AB’)=0.99 P(AB’)=0.985 P(AB)=0.998 P(AB)=0.9975 (d) The unusual result is that for both a simple component and for a complex assembly, supplier 1 has a greater probability that a part conforms to specifications. However, for overall parts, supplier 1 has a lower probability. The overall conforming probability is dependent on both the conforming probability conditioned on the part type and the probability of each part type. Supplier 1 produces more of the compelx parts so that overall conformance from supplier 1 is lower.
MindExpanding Exercises 2168. Let E denote a read error and let S, O, B, P denote skewed, offcenter, both, and proper alignments, respectively. P(E) = P(ES)P(S) + P(EO)P(O) + P(EB)P(B) + P(EP)P(P) = 0.01(0.10) + 0.02(0.05) + 0.06(0.01) + 0.001(0.84) = 0.00344 Let n denote the number of washers selected. a) The probability that non are thicker than, that is, all are less than the target is 0.4 n , by independence. n 0.4 n 1 0.4 2 0.16 3 0.064 Therefore, n = 3 b) The requested probability is the complement of the probability requested in part a. Therefore, n = 3 2169. 229 2170. Let x denote the number of kits produced. Revenue at each demand 50 100 200 100x 100x 100x 0 ≤ x ≤ 50 Mean profit = 100x(0.95)5x(0.05)20x 5x 100(50)5(x50) 100x 100x 50 ≤ x ≤ 100 Mean profit = [100(50)5(x50)](0.4) + 100x(0.55)5x(0.05)20x 5x 100(50)5(x50) 100(100)5(x100) 100x 100 ≤ x ≤ 200 Mean profit = [100(50)5(x50)](0.4) + [100(100)5(x100)](0.3) + 100x(0.25)  5x(0.05)  20x 0 5x Mean Profit Maximum Profit 74.75 x $ 3737.50 at x=50 32.75 x + 2100 $ 5375 at x=100 1.25 x + 5250 $ 5500 at x=200 Therefore, profit is maximized at 200 kits. However, the difference in profit over 100 kits is small. 0 ≤ x ≤ 50 50 ≤ x ≤ 100 100 ≤ x ≤ 200 2171. Let E denote the probability that none of the bolts are identified as incorrectly torqued. The requested probability is P(E'). Let X denote the number of bolts in the sample that are incorrect. Then, P(E) = P(EX=0)P(X=0) + P(EX=1) P(X=1) + P(EX=2) P(X=2) + P(EX=3) P(X=3) + P(EX=4)P(X=4) and P(X=0) = (15/20)(14/19)(13/18)(12/17) = 0.2817. The remaining probability for x can be determined from the counting methods in Appendix B1. Then, ( )( ) = ⎜⎝ 4 ! 1! ⎟⎠ ⎜⎝ 3 ! 12 ! ⎟⎠ = 5 ! 15 ! 4 ! 16 ! = 0.4696 P( X = 1) = ( ) ⎛⎜⎝ 420 ! ! ⎞⎟⎠ 4 ! 3 ! 12 ! 20 ! ! 16
5 15 13 20 4 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ ( )( ) = ⎜⎝ 3 ! 2 ! ⎟⎠ ⎜⎝ 2 ! 13 ! ⎟⎠ = 0.2167 P( X = 2) = ⎛ 20 ! ⎞ () ⎜ ⎟ ⎝ 4 ! 16 ! ⎠
5 15 22 20 4 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ ( )( ) = ⎜⎝ 3 ! 2 ! ⎟⎠ ⎜⎝ 1! 14 ! ⎟⎠ = 0.0309 P( X = 3) = ( ) ⎛⎜⎝ 420 ! ! ⎞⎟⎠ ! 16
5 15 31 20 4 ⎛ 5 ! ⎞ ⎛ 15 ! ⎞ P(X=4) = (5/20)(4/19)(3/18)(2/17) = 0.0010 and P(EX=0) = 1, P(EX=1) = 0.05, P(EX=2) = 0.05 2 = 0.0025 , P(EX=3) = 0.05 3 = 125 × 10 −4 , P(EX=4) = 0.05 4 = 6.25 × 10 −6 . Then, . P(E) = 1(0.2817) + 0.05(0.4696) + 0.0025(0.2167) + 125 × 10 −4 (0.0309) . +6.25 × 10 −6 (0.0010) = 0.306 and P(E') = 0.694
2172.
P( A '∩B') = 1 − P([ A '∩B' ]' ) = 1 − P( A ∪ B) = 1 − [P( A ) + P(B) − P( A ∩ B)] = 1 − P( A ) − P(B) + P( A )P(B) = [1 − P( A )][1 − P(B)] = P( A ' )P(B' ) 230 2173. The total sample size is ka + a + kb + b = (k + 1)a + (k +1)b. ka + a k (a + b) , P ( B) = P(A ) = ( k + 1) a + ( k + 1) b ( k + 1) a + ( k + 1) b and ka ka P ( A ∩ B) = = ( k + 1) a + ( k + 1) b ( k + 1)( a + b ) Then , k ( a + b )( k + 1) a ka k ( a + b )( ka + a ) P ( A ) P ( B) = = = = P ( A ∩ B) 2 2 2 ( k + 1)( a + b ) [( k + 1) a + ( k + 1) b ] ( k + 1) ( a + b ) 231 ...
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 Spring '10
 Feng
 Probability, Harshad number, Mean Profit

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