Asst3-suggested

# Asst3-suggested - Statistics 427 Homework#3 Solutions to...

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Solutions to Suggested Problems 2.84 We know that the probability that a vehicle passes inspection is .70 and that that results for each vehicle are mutually independent events a. P( all three pass) = (0.70)(0.70)(0.70) = 0.343 (By independence) = 0.343 b. 1 – P(all pass) = 1 - 0.343 = 0.657 c. P(exactly one passes) = (0.70)(0.30)(0.30) + (0.30)(0.70)(0.30) + (0.30)(0.30)(0.70) = 0.189 d. P(# pass 1) = P(0 pass) + P(exactly one passes) = (0.3) 3 + 0.189 = 0.216 e. P(3 pass | 1 or more pass) 353 . 973 . 343 . ) . 1 ( ) . 3 ( ) . 1 ( ) . 1 . 3 ( = = = = pass P pass P pass P pass pass P 2.91 a. P(line 1) = 333 . 1500 500 = ; P(Crack) = () ( ) ( ) 444 . 1500 666 1500 600 40 . 400 44 . 500 50 . = = + + b. P(Blemish | line 1) = 0.15 c. P(Surface Defect) = ( ) ( ) ( ) 1500 172 1500 600 15 . 400 08 . 500 10 . = + + P(line 1 and Surface Defect) = ( ) 1500 50 1500 500 10 . = So P(line 1 | Surface Defect) =

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## This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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Asst3-suggested - Statistics 427 Homework#3 Solutions to...

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