Unformatted text preview: ns obtain equivalence at 100 mL is 0.1 = 0.01 . (b) The probability that one technician obtains equivalence between 98 and 104 mL is 0.7. So the probability that both technicians obtain equivalence between 98 and 104 mL is
2 0.7 2 = 0.49 .
(c) The probability that the average volume at equivalence from the technician is 100 mL is 9(0.12 ) = 0.09 . 219 2112. 10 6 = 10 −10 16 10 1 (b) P = 0.25 × ( ) = 0.020833 12
(a) P =
Let A denote the event that a sample is produced in cavity one of the mold. 1 a) By independence, P( A1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( )5 = 0.00003 8 b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P(B1 ∪ B 2 ∪...∪B 8 ) = P(B1) + P(B 2 )+...+P(B 8 ) 2113. 2114. 1 1 From part a., P(Bi ) = ( )5 . Therefore, the answer is 8( )5 = 0.00024 8 8 147 ' c) By independence, P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( ) ( ) . The number of sequences in 8 8 1 7 which four out of five samples are from cavity one is 5. Therefore, the answer is 5( ) 4 ( ) = 0.00107 . 8 8 Let A de...
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 Spring '10
 Feng
 Probability, Probability theory, success rate, large stone

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