# Interpreting probability

1. Jan 23, 2012

### 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.

2. Jan 23, 2012

### 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 $P=(0.8)^n$ 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
3. Jan 25, 2012

### musicgold

SW VandeCarr,

Thanks.

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

4. Jan 25, 2012

### SW VandeCarr

You're welcome.