Geometric Distribution Problem Clarification

In summary, the probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. For an average driver who traverses this area for 1200 hours in the course of a year, the probability of being seriously injured at least once is 0.6989866. The expected number of hours that a driver will drive before being seriously injured is 1000. Given that the driver has already driven 1200 hours, the probability of being injured in the next 100 hours is 0.00990099. These calculations assume that the odds of crashing each hour are independent.
  • #1
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TL;DR Summary
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?
[itex]
\text{This means what is the probability that the person gets seriously injured at least once in a year.}
[/itex]

[itex]
P(\text{at least one success}) = 1 - P(\text{failure in one trial})^n
[/itex]

[itex]
\text{In the formula above, n represents the total number of trials.}
[/itex]

[itex]
\text{We will be cruel and label an accident as a success.}
[/itex]

[itex]
1 - (1 - \frac{.1}{100})^{1200} = 0.6989866

[/itex]

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

Would this answer be correct?
 
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  • #2
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.
 
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  • #3
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 :
[tex] \text{Expected val }= \frac{1}{p}[/tex]
or
[tex] \frac{1-p}{p}[/tex]
?
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.
 
  • #4
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 :
[tex] \text{Expected val }= \frac{1}{p}[/tex]
or
[tex] \frac{1-p}{p}[/tex]
?
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.
 
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  • #5
Thank you, Fact Checker.
Besides the assumptions statement. Is my solution correct?
 

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