Question

Let \(x\) be a Poisson random variable with mean \(\mu=2 .\) Calculate these probabilities: a. \(P(x=0)\) b. \(P(x=1)\) c. \(P(x>1)\) d. \(P(x=5)\)

Short answer

Question: Find the following probabilities for a Poisson random variable \(x\) with mean \(\mu = 2\): \(P(x = 0)\), \(P(x = 1)\), \(P(x > 1)\), and \(P(x = 5)\). Answer: The requested probabilities are: a. \(P(x=0) \approx 0.1353\) b. \(P(x=1) \approx 0.2707\) c. \(P(x>1) \approx 0.5940\) d. \(P(x=5) \approx 0.0361\)

Step-by-step solution

Calculate Probability for \(P(x=0)\)

To calculate \(P(x=0)\), we will plug in the values of \(k=0\) and \(\mu = 2\) into the Poisson PMF formula: \(P(x=0) = \frac{e^{-2} \cdot (2^0)}{0!} = \frac{e^{-2} \cdot (1)}{1} = e^{-2} \approx 0.1353\)

Calculate Probability for \(P(x=1)\)

To calculate \(P(x=1)\), we will plug in the values of \(k=1\) and \(\mu = 2\) into the Poisson PMF formula: \(P(x=1) = \frac{e^{-2} \cdot (2^1)}{1!} = \frac{e^{-2} \cdot (2)}{1} = 2e^{-2} \approx 0.2707\)

Calculate Probability for \(P(x>1)\)

To calculate \(P(x>1)\), we must first calculate \(P(x\leq1)\). Since we have already computed \(P(x=0)\) and \(P(x=1)\), we can find \(P(x\leq1)\) by summing these probabilities: \(P(x\leq1) = P(x=0) + P(x=1) \approx 0.1353 + 0.2707 \approx 0.4060\) Now, we will use the complementary rule to find \(P(x>1)\): \(P(x>1) = 1 - P(x\leq1) \approx 1 - 0.4060 \approx 0.5940\)

Calculate Probability for \(P(x=5)\)

To calculate \(P(x=5)\), we will plug in the values of \(k=5\) and \(\mu = 2\) into the Poisson PMF formula: \(P(x=5) = \frac{e^{-2} \cdot (2^5)}{5!} = \frac{e^{-2} \cdot (32)}{120} = \frac{32}{120}e^{-2} \approx 0.0361\)

Summarize Results

We have calculated the requested probabilities: a. \(P(x=0) \approx 0.1353\) b. \(P(x=1) \approx 0.2707\) c. \(P(x>1) \approx 0.5940\) d. \(P(x=5) \approx 0.0361\)

Key concepts

Probability Mass Function

In probability theory, the Probability Mass Function (PMF) is crucial when dealing with discrete random variables, such as in the Poisson distribution.
It gives the probability that a discrete random variable is exactly equal to some value.
For the Poisson distribution, the PMF is defined as follows:

• \(P(x=k) = \frac{e^{-\mu} \cdot \mu^k}{k!}\)

where:

• \(k\) is a non-negative integer, representing the number of events.
• \(\mu\) is the average number of events in a fixed interval (mean of the distribution).
• \(e\) is the base of the natural logarithm (approximately 2.71828).

This function is fundamental because it provides the probabilities needed to determine the likelihood of observing various event counts in a Poisson process.
For example, in an exercise with \(\mu=2\), the PMF helps calculate \(P(x=0), P(x=1),\) and \(P(x=5)\) using the given formula.

Mean of a Distribution

The mean of a distribution is a measure of central tendency.
For the Poisson distribution, it's denoted by \(\mu\) and significantly impacts the behavior of the distribution.

• In a Poisson process, \(\mu\) represents the average number of occurrences of an event in a fixed interval of time or space.

This mean value doesn't just tell us the average, but it's also the parameter that shapes the distribution.
A higher \(\mu\) means more frequent occurrences are likely, while lower \(\mu\) indicates fewer occurrences.
For instance, when the mean \(\mu=2\), it guides our expectations for how events are likely to be distributed.
So, when calculating probabilistic outcomes using the PMF, knowing the mean helps us understand the spread and concentration of probable values.

Complementary Rule

The complementary rule is a useful concept in probability that helps calculate the probability of an event by using the probability of its complement.
It's simple yet powerful:

• \(P(A^c) = 1 - P(A)\)

Where \(A\) represents an event, and \(A^c\) is the complement of \(A\), meaning all outcomes not in \(A\).
This rule is essential when dealing with probabilities that are easier to find by considering their opposites.
For example, in the Poisson distribution exercise, calculating \(P(x > 1)\) directly can be complex.
However, by finding \(P(x \leq 1)\) first and using the complementary rule, the task becomes simpler:

• \(P(x>1) = 1 - P(x \leq 1)\)

This approach is not only convenient but often necessary when addressing problems involving complementary probabilities in probability theory.

Discrete Random Variable

A discrete random variable is one that has a countable number of possible values.
This is in contrast to continuous random variables, which can take on any value within a range.

• In the context of the Poisson distribution, the discrete random variable \(x\) represents the number of events occurring within a fixed duration or space.

Discrete random variables are pivotal in various real-world settings where outcomes are counted in whole numbers, such as the number of email arrivals or customer visits.
In mathematical terms, events like "\(x\), the number of occurrences being 0, 1, 2,..." define the specific possibilities that the variable can take.
Understanding the nature of discrete random variables is essential for applying distributions like Poisson, where each possible outcome has its own probability dictated by the PMF.