HW5 - 540:210:01 ENG PROBABILITY HW 5 2-83 a 20/100 b 19/99...

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540:210:01: ENG PROBABILITY HW 5 2-83. a) 20/100 b) 19/99 c) (20/100)(19/99) = 0.038 d) If the chips are replaced, the probability would be (20/100) = 0.2 2-89. a) P A B P AB P B ( ) ( ) ( ) ( . )( . ) . 04 05 020 b) P A B P A B P B ( ) ( ) ( ) ( . )( . ) .   06 05 030 2-90. P A P A B P A B P AB P B P AB P B ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( . )( . ) ( . )( . ) . . . 0 2 0 8 0 3 0 2 016 0 06 0 22 2-93. Let R denote the event that a product exhibits surface roughness. Let N,A, and W denote the events that the blades are new, average, and worn, respectively. Then, P(R)= P(R|N)P(N) + P(R|A)P(A) + P(R|W)P(W) = (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15) = 0.028 2-95. a) (0.88)(0.27) = 0.2376 b) (0.12)(0.13+0.52) = 0.0.078 2-97. Let A denote a event that the first part selected has excessive shrinkage. Let B denote the event that the second part selected has excessive shrinkage. a) P(B)= P( B A )P(A) + P( B A ')P(A') = (4/24)(5/25) + (5/24)(20/25) = 0.20 b) Let C denote the event that the third part selected has excessive shrinkage.
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20 . 0 25 20 24 19 23 5 25 20 24 5 23 4 25 5 24 20 23 4 25 5 24 4 23 3 ) ' ' ( ) ' ' ( ) ' ( ) ' ( ) ' ( ) ' ( ) ( ) ( ) (
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