How Do We Accurately Interpret Probability in Different Scenarios?

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Discussion Overview

The discussion revolves around interpreting probability in the context of a problem involving California drivers and seat belt usage. Participants explore the implications of treating the scenario as Bernoulli trials and consider the variability in sample outcomes.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents a probability problem regarding the likelihood of three drivers wearing seat belts, initially calculating it as (0.8)^3.
  • Another participant suggests that as sample sizes increase, the sample distributions would approach the population probability of 0.8 for wearing seat belts, while also noting variability in smaller samples.
  • There is a question about whether the sample of three drivers should always be treated as independent Bernoulli trials.
  • A later reply confirms that the participant knows how to calculate the probabilities for different outcomes (0, 1, or 2 drivers wearing seat belts).

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the sample should always be treated as Bernoulli trials, and the discussion remains open regarding the interpretation of probability in this context.

Contextual Notes

The discussion highlights the variability in smaller sample sizes and the assumptions underlying probability calculations, but does not resolve these issues.

musicgold
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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.
 
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musicgold said:
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 [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:
SW VandeCarr,

Thanks.

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

Thanks.

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

You're welcome.
 

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