Geometric Distribution Problem Clarification

Click For Summary
SUMMARY

The discussion centers on calculating the probability of a driver being seriously injured in a car crash, given a probability of 0.1% per hour over 1200 hours in a year. The formula used is P(at least one success) = 1 - P(failure in one trial)^n, leading to a calculated probability of approximately 0.699. Key points include the independence of hourly crash probabilities and the need for clear assumptions regarding the average driver and their risk factors. The expected value for the number of hours before an injury occurs is also discussed, with confusion around the correct formula.

PREREQUISITES
  • Understanding of geometric distribution and its applications
  • Familiarity with probability concepts, including independent events
  • Knowledge of expected value calculations in probability
  • Basic statistics terminology and notation
NEXT STEPS
  • Study the derivation and application of the geometric distribution in real-world scenarios
  • Learn about the implications of independent versus dependent events in probability
  • Explore advanced probability concepts, such as conditional probability and Bayes' theorem
  • Review statistical assumptions and their impact on probability calculations
USEFUL FOR

Statisticians, data analysts, risk assessment professionals, and anyone interested in applying probability theory to real-world situations, particularly in safety and risk management contexts.

knowLittle
Messages
307
Reaction score
3
TL;DR
I some hints or confirmation if my solution is correct.
I am bit confused about different formulas I have found involving geometric distribution.
And functions from R-language.
(Geometric). The probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. A driver is required to traverse this area for 1200 hours in the course of a year. What is the probability that the driver will be seriously injured during the course of the year? In the course of 15 months? What is the expected number of hours that a driver will drive before being seriously injured? Given that a driver has driven 1200 hours, what is the probability that he or she will be injured in the next 100 hours?What is the probability that the driver will be seriously injured during the course of the year?
<br /> \text{This means what is the probability that the person gets seriously injured at least once in a year.}<br />

<br /> P(\text{at least one success}) = 1 - P(\text{failure in one trial})^n<br />

<br /> \text{In the formula above, n represents the total number of trials.}<br />

<br /> \text{We will be cruel and label an accident as a success.}<br />

<br /> 1 - (1 - \frac{.1}{100})^{1200} = 0.6989866<br /> <br />

prob of accident is p=.1/100 = 0.001
prob of non-accident is non_accident = 1-p

Would this answer be correct?
 
Physics news on Phys.org
The idea is correct, but there are some things that you should get in the habit of watching for.
knowLittle said:
(Geometric). The probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. A driver
Is this an average driver with the average probability of a crash?
knowLittle said:
is required to traverse this area for 1200 hours in the course of a year. What is the probability that the driver will be seriously injured during the course of the year?
Are the odds of a crash on each hour independent of each other?
I believe that these are aspects that you should make some clear statements about to make your equation correct.
 
  • Like
Likes   Reactions: knowLittle
FactChecker said:
The idea is correct, but there are some things that you should get in the habit of watching for.

Is this an average driver with the average probability of a crash?
I think I lack the intuition. I assume it's an average driver with an average probability, yes.

Does it mean that my p value is wrong?
I am confused about this different values for mean :
\text{Expected val }= \frac{1}{p}
or
\frac{1-p}{p}
?
FactChecker said:
Are the odds of a crash on each hour independent of each other?
I believe that these are aspects that you should make some clear statements about to make your equation correct.
The odds of crashing each hour are independent according to the problem description.
 
knowLittle said:
I think I lack the intuition. I assume it's an average driver with an average probability, yes.
That is the kind of thing that you should clearly state. And be careful if the problem doesn't state it.
knowLittle said:
Does it mean that my p value is wrong?
I am confused about this different values for mean :
\text{Expected val }= \frac{1}{p}
or
\frac{1-p}{p}
?
It's ok if you clearly state that the person you are talking about has that probability.
knowLittle said:
The odds of crashing each hour are independent according to the problem description.
Nothing that I see says that, especially for one person doing all the driving. It would be a natural assumption, but it is better if you state it clearly.
Suppose that one person who drives through safely a thousand times would tend to get careless? That is a very real-world situation. I am not trying to be picky, I am just trying to encourage you to specifically state if you are assuming some probabilities are independent (and/or to question that assumption). These are mistakes that I see all the time.
 
  • Like
Likes   Reactions: knowLittle
Thank you, Fact Checker.
Besides the assumptions statement. Is my solution correct?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K