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- 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?

[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?