Interpreting probability

musicgold

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.

SW VandeCarr

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)?

musicgold

SW VandeCarr,

Thanks.

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

SW VandeCarr

You're welcome.