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Interpreting probability 
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#1
Jan2312, 08:48 PM

P: 180

Hi,
My question is related to the following problem. “80% of all California drivers wear seat belts. If three drivers are pulled over, what is the probability that all would be wearing their seat belts?” Now I know that the answer of this problem is = 0.8 * 0.8 * 0.8 = (Probability of the first person wearing belt x prob. of the second person wearing belt x…) However, there is another question that comes to my mind. What if we say that 80% of the sample (of the three people) will be wearing seat belts? Or do we have to always treat them as Bernoulli trials? Thanks. 


#2
Jan2312, 09:27 PM

P: 2,501

However the probability that all individuals in the sample were wearing seat belts would approach [itex]P=(0.8)^n[/itex] where n is the sample size. For a sample of size 3, there will be considerable variability with 2 or 3 being more likely than 0 or 1 wearing seat belts. Do you know how to calculate the exact probabilities of 0,1 or 2 drivers wearing seat belts (assuming P(B) holds for the population)? 


#3
Jan2512, 12:37 PM

P: 180

SW VandeCarr,
Thanks. Yes, I do know how to calculate the probabilities of 0,1 or 2 drivers wearing seat belts. 


#4
Jan2512, 02:15 PM

P: 2,501

Interpreting probability



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