Geometric and Poisson Distributions: Finding Probabilities

Discrete Probability Distributions

Geometric Distribution

The geometric distribution is a discrete probability distribution that models the number of trials needed to achieve the first success in a sequence of independent trials, each with the same probability of success. It is commonly used in scenarios where repeated attempts are made until a desired outcome occurs.

• Definition: The geometric distribution describes the probability that the first success occurs on the x-th trial.

• Conditions:

  • A trial is repeated until a success occurs.

  • Each trial is independent of the others.

  • The probability of success, p, is constant for each trial.

  • The random variable x represents the trial number of the first success.

• Probability Formula:

The probability that the first success occurs on trial number x is given by:

$P(x) = p \times q^{x-1}$ where $q = 1 - p$ is the probability of failure.

• Example: Suppose the failure rate of businesses after five years is 20% (so p = 0.2). Six businesses are selected at random. What is the probability that the sixth business selected is the first one to have failed?

• Solution:

  • Check conditions: Trials are repeated until a business fails, each trial is independent, and the probability of failure is constant.

  • Let p = 0.2 (probability of failure), q = 0.8 (probability of success), and x = 6.

  • Apply the formula:

  $P(6) = 0.2 \times 0.8^{6-1} = 0.2 \times 0.8^{5} \approx 0.0655$

  • Interpretation: There is approximately a 6.55% chance that the sixth business selected is the first to have failed.

Poisson Distribution

The Poisson distribution is a discrete probability distribution used to model the number of times an event occurs in a fixed interval of time, area, or volume. It is particularly useful for rare events and when the mean number of occurrences is known.

• Definition: The Poisson distribution describes the probability of observing exactly x occurrences of an event in a given interval.

• Conditions:

  • The experiment consists of counting the number of times x an event occurs in a given interval (time, area, volume).

  • The probability of the event occurring is the same for each interval.

  • The number of occurrences in one interval is independent of the number in other intervals.

• Probability Formula:

The probability of exactly x occurrences in an interval is:

$P(x) = \frac{\mu^{x} \times e^{-\mu}}{x!}$ where $\mu$ is the mean number of occurrences per interval, $e \approx 2.71828$ is Euler's number, and $x!$ is the factorial of x.

• Example: The mean number of accidents per month at a certain intersection is 3. What is the probability that more than 4 accidents will occur in any given month?

• Solution:

  • Check conditions: Counting accidents per month, probability is constant, and occurrences are independent.

  • Let $\mu = 3$.

  • We want $P(x > 4)$.

  • Since Poisson probabilities for $x$ go from 0 to infinity, calculate $P(x > 4) = 1 - P(x \leq 4)$.

  • Compute $P(x = 0), P(x = 1), P(x = 2), P(x = 3), P(x = 4)$ using the formula:

  $P(0) = \frac{3^{0} \times e^{-3}}{0!} = e^{-3} \approx 0.0498$ $P(1) = \frac{3^{1} \times e^{-3}}{1!} = 3 \times e^{-3} \approx 0.1494$ $P(2) = \frac{3^{2} \times e^{-3}}{2!} = 4.5 \times e^{-3} \approx 0.2240$ $P(3) = \frac{3^{3} \times e^{-3}}{3!} = 4.5 \times e^{-3} \approx 0.2240$ $P(4) = \frac{3^{4} \times e^{-3}}{4!} = 3.375 \times e^{-3} \approx 0.1680$

  • Sum probabilities for $x = 0$ to $x = 4$: $0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 = 0.8152$

  • Calculate $P(x > 4) = 1 - 0.8152 = 0.1848$

  • Interpretation: There is an 18.48% chance of more than 4 accidents in a month at this intersection.

Poisson Probability Table

The Poisson probability table provides values for $P(x)$ for different values of $x$ and $\mu$. It is useful for quickly finding probabilities without manual calculation.

x	P(x)
0	0.0498
1	0.1494
2	0.2240
3	0.2240
4	0.1680
5	Additional info: Not calculated, but would be $\frac{3^{5} \times e^{-3}}{5!}$
6	Additional info: Not calculated, but would be $\frac{3^{6} \times e^{-3}}{6!}$

Note: For probabilities involving "more than" or "at least" a certain number of events, sum the relevant probabilities or use the complement rule as shown above.

Summary

• The geometric distribution models the number of trials until the first success, with a constant probability of success.

• The Poisson distribution models the number of occurrences of an event in a fixed interval, given a known mean rate.

• Both distributions are discrete and require specific conditions to be met for their application.

Additional info: The Poisson distribution is often used in fields such as traffic analysis, biology, and telecommunications to model rare events. The geometric distribution is commonly used in reliability engineering and quality control.