P0071_Soln - S 3 S 4 S 5 ( 29 Each of those possibilities...

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P0071_Soln.doc Probability that a sample is not contaminated is one minus the probability that it is contaminated (complement): P S ( 29 = 1 - PS ( 29 = 1 - 0 .10 = 0 .90 Since the samples are assumed to be independent, the probabilities of contamination/non-contamination can be multiplied, e.g. , P S 1 S 2 S 3 S 2 S 3 ( 29 = P S 1 ( 29 P S 2 ( 29 P S 3 ( 29 P S 4 ( 29 P S 5 ( 29 We wish to find the probability that exactly one of the five samples will be contaminated. Based on the information provided, we don’t seem to care whether the one contaminated sample is the first sample tested, the second sample tested, the third sample tested, the fourth sample tested, or the fifth sample tested. PS 1 S 2 S 3 S 4 S 5 ( 29 P S 1 S 2 S 3 S 4 S 5 ( 29 P S 1 S 2 S 3 S 4 S 5 ( 29 P S 1 S 2 S 3 S 4 S 5 ( 29 P S 1 S 2
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Unformatted text preview: S 3 S 4 S 5 ( 29 Each of those possibilities is disjoint from the others, so the probabilities add: PS 1 S 2 S 3 S 4 S 5 ( 29 + P S 1 S 2 S 3 S 4 S 5 ( 29 + P S 1 S 2 S 3 S 4 S 5 ( 29 + P S 1 S 2 S 3 S 4 S 5 ( 29 + P S 1 S 2 S 3 S 4 S 5 ( 29 PS 1 ( 29 P S 2 ( 29 P S 3 ( 29 P S 4 ( 29 P S 5 ( 29 + P S 1 ( 29 PS 2 ( 29 P S 3 ( 29 P S 4 ( 29 P S 5 ( 29 + P S 1 ( 29 P S 2 ( 29 PS 3 ( 29 P S 4 ( 29 P S 5 ( 29 + P S 1 ( 29 P S 2 ( 29 P S 3 ( 29 PS 4 ( 29 P S 5 ( 29 + P S 1 ( 29 P S 2 ( 29 P S 3 ( 29 P S 4 ( 29 PS 5 ( 29 51-.10 ( 29 4 .10 ( 29 = .33...
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This note was uploaded on 02/01/2011 for the course BME 335 taught by Professor Dunn during the Spring '10 term at University of Texas at Austin.

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