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CIE A-Level Maths Study Notes

5.1.1 Poisson Probability Calculations

The Poisson distribution is a vital concept in probability theory, particularly useful for modeling the frequency of events occurring in a fixed interval of time or space.

The Basics of Poisson Distribution

The Poisson distribution helps us understand how likely it is for a certain number of events to happen in a fixed interval, given the average number of times these events usually occur. It's useful when we're looking at events that happen independently and at a constant average rate.

Formula

The probability of observing exactly kk events is given by:

P(X=k)=eλλkk!P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}

  • ee is the base of the natural logarithm (about 2.71828).
  • kk is the number of events we're interested in.
  • λ\lambda is the average number of events.
  • k!k! is kk factorial, the product of all positive integers up to kk.

Key Points

  • Discrete: It deals with counts of events, so kk can be 0, 1, 2, and so on.
  • Versatile: Used in various fields for analyzing rare events over time or space.

Application Steps

1. Identify λ\lambda : Find the average event rate.

2. Determine kk: Decide the specific number of events you're interested in.

3. Calculate Probability: Use the formula to calculate the likelihood of (k) 3. events.

Examples

Example 1: Bookstore Customers

Suppose a bookstore averages 3 customers per hour. What is the probability exactly 5 customers arrive in an hour?

Solution:

  • Average Rate (λ)(\lambda): 3 customers/hour.
  • Interested in (k)(k): 5 customers.
  • Probability Calculation: Using the formula with λ=3\lambda = 3 and k=5k = 5, we find the probability is about 10.08%.
Probability Graph

Example 2: Network Faults

In a network operation centre, an average of 2 faults occur per day. What is the probability that there will be no faults in a day?

Solution:

  • Average Rate (λ)(\lambda): 2 faults/day.
  • Interested in (k)(k): 0 faults.
  • Probability Calculation: Using the formula with λ=2\lambda = 2 and k=0k = 0, we find the probability is about 13.53%.
Probability Graph
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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