Poisson Process Probability Question

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SUMMARY

The discussion centers on calculating probabilities using the Poisson process, specifically for an average of 0.4 accidents per day. The probability of the time until the next accident exceeding 3 days is derived from the formula for no accidents occurring over that period. Additionally, the probability of the time until the third accident exceeding 5 days can be calculated using similar principles of the Poisson distribution. The key takeaway is the application of the exponential distribution to model the time between events in a Poisson process.

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  • Understanding of Poisson processes and their properties
  • Familiarity with exponential distribution and its applications
  • Basic knowledge of probability theory and calculations
  • Ability to manipulate and apply probability formulas
NEXT STEPS
  • Study the properties of the Poisson distribution in detail
  • Learn how to derive probabilities using the exponential distribution
  • Explore real-world applications of Poisson processes in risk assessment
  • Practice solving problems involving multiple events in a Poisson framework
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Students in statistics, data analysts, and professionals in fields requiring risk assessment and event modeling, particularly those interested in accident frequency and time-to-event analysis.

sakebu
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Hello,

I have this one problem but have no idea how to get started.

Avg. number of accidents is .4 accidents / day (Poisson Process)

What is the probability that the time from now to the next accident will be more than 3 days?

What is the probability that the the time from now to the 3rd accident will be more than 5 days?

Any information / formulas would be appreciated.
 
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The trick to finding the probability of a given event is to express it in terms of probabilities you know. For example, [itex]\begin{eqnarray*}<br /> P(\text{next accident is the day after tomorrow or later}) &=& P(\text{no accident today or tomorrow}) \\<br /> &=& P(\text{no accident today}) \cdot P(\text{no accident tomorrow}|\text{no accident today}) \\<br /> &=& P(\text{no accident today}) \cdot P(\text{no accident tomorrow}) \\<br /> &=& P(\text{no accident on a given day})^2 \\<br /> &=& [1- P(\text{accident on a given day})]^2<br /> \end{eqnarray*}[/itex]
 

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