HW6 - 540:210:01 ENG PROBABILITY HW 6 2-109(a 3(0.2 4...

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540:210:01: ENG PROBABILITY HW 6 2-109. (a) ) 2 . 0 ( 3 4 =0.0048 (b) ) 8 . 0 * 2 . 0 * 4 ( 3 3 =0.0768 2-113. Let A denote the event that a sample is produced in cavity one of the mold. a) By independence, P A A A A A ( ) ( ) . 1 2 3 4 5 5 1 8 0 00003 b) Let B i be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P B B B P B P B P B ( ... ) ( ) ( ) ... ( ) 1 2 8 1 2 8 From part a., P B i ( ) ( ) 1 8 5 . Therefore, the answer is 8 1 8 0 00024 5 ( ) . c) By independence, P A A A A A ( ) ( ) ( ) ' 1 2 3 4 5 4 1 8 7 8 . The number of sequences in which four out of five samples are from cavity one is 5. Therefore, the answer is 5 1 8 7 8 0 00107 4 ( ) ( ) . . 2-114. Let A denote the upper devices function. Let B denote the lower devices function. P(A) = (0.9)(0.8)(0.7) = 0.504 P(B) = (0.95)(0.95)(0.95) = 0.8574 P(A B) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(A
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This note was uploaded on 08/03/2011 for the course PHYS 750:116 taught by Professor Brahmia during the Spring '11 term at Rutgers.

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HW6 - 540:210:01 ENG PROBABILITY HW 6 2-109(a 3(0.2 4...

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