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Interpreting probability

  1. Jan 23, 2012 #1

    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?

  2. jcsd
  3. Jan 23, 2012 #2
    The short answer is yes. As you take larger and larger samples the expectation is that the sample distributions would approach P(B)=0.8; P(~B)=(1-P(B))=0.2

    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)?
    Last edited: Jan 23, 2012
  4. Jan 25, 2012 #3
    SW VandeCarr,


    Yes, I do know how to calculate the probabilities of 0,1 or 2 drivers wearing seat belts.
  5. Jan 25, 2012 #4
    You're welcome.
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