Question 765554: 1.Four students who commute together claim that they missed an exam due to a flat tire on the same vehicle. On the make-up exam, the instructor asks the students to identify the particular tire that went flat. If they really did not have a flat tire and randomly select one of the four tires on the vehicle, what is the probability that they all select the same tire?
2.The Orange County Department of Public Health tests water for contamination because of the presence of E. coli bacteria. To reduce laboratory costs, water samples from six public swimming areas are combined for one test, and further testing is only done if the combined sample fails. Based on past results, there is 2% chase of finding E.coli in a public swimming area. Find the probability that a combined sample from six public swimming areas will reveal the presence of E. coli bacteria.
I cannot get the right answer for these two questions no matter what way I tried. Please help me solve it! Thank You
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1.Four students who commute together claim that they missed an exam due to a flat tire on the same vehicle. On the make-up exam, the instructor asks the students to identify the particular tire that went flat. If they really did not have a flat tire and randomly select one of the four tires on the vehicle, what is the probability that they all select the same tire?
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Ans: (1/4)^4 = 1/2^8 = 1/256
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2.The Orange County Department of Public Health tests water for contamination because of the presence of E. coli bacteria. To reduce laboratory costs, water samples from six public swimming areas are combined for one test, and further testing is only done if the combined sample fails. Based on past results, there is 2% chance of finding E.coli in a public swimming area. Find the probability that a combined sample from six public swimming areas will reveal the presence of E. coli bacteria.
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Note:: If any one of the samples has E. coli, the combined sample will have
indicate E. coli.
The only way the sample will pass is if all 6 are clean of E. coli.
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P(E. coli in combined sample) = 1 - P(no E.coli in all 6 samples)
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= 1 - (0.98)^6
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= 1 - 0.8858
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= 0.1142
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Cheers,
Stan H.
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