How Do We Accurately Interpret Probability in Different Scenarios?

[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]

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.

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)?

 

[musicgold]

SW VandeCarr,

Thanks.

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

 

[SW VandeCarr]

SW VandeCarr,

Thanks.

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

You're welcome.

 

[blue_raver22]

I would like to clarify the concept of interpreting probability in this scenario. Probability is a measure of the likelihood of an event occurring, and it is often expressed as a number between 0 and 1. In this case, the probability of all three drivers wearing their seat belts is calculated by multiplying the individual probabilities of each driver wearing their seat belt (0.8 * 0.8 * 0.8). This is because each driver's behavior is independent of the others in this scenario, and thus can be treated as separate Bernoulli trials.

However, if we were to say that 80% of the sample (of the three people) will be wearing seat belts, this would be a different scenario. In this case, we are not looking at the individual probabilities of each driver, but rather the overall probability of the entire sample. This would require a different approach to calculating the probability and cannot be considered as Bernoulli trials.

It is important to understand the context and the specific question being asked in order to accurately interpret probability. In this case, we are looking at the probability of all three drivers wearing seat belts, and thus it is appropriate to use the individual probabilities of each driver. However, if the question were to change to the overall probability of the sample, then a different approach would be needed.