Question

Find the probabilities for \(x\) using the Poisson formula. $$ \mu=2 ; P(x=0), P(x=1), P(x>1), \text { and } P(x=5) $$

Short answer

Answer: The probabilities for the given Poisson distribution are: P(x=0) ≈ 0.135, P(x=1) ≈ 0.270, P(x>1) ≈ 0.595, and P(x=5) ≈ 0.036.

Step-by-step solution

Calculate P(x=0)

To calculate the probability P(x=0) using the Poisson formula, we plug in the values λ=2 and x=0: $$ P(x=0) = \frac{e^{-2} \cdot 2^0}{0!} = \frac{e^{-2} \cdot 1}{1} = e^{-2} \approx 0.135 $$

Calculate P(x=1)

Now we will calculate the probability P(x=1) using the Poisson formula, with λ=2 and x=1: $$ P(x=1) = \frac{e^{-2} \cdot 2^1}{1!} = \frac{e^{-2} \cdot 2}{1} = 2e^{-2} \approx 0.270 $$

Calculate P(x>1)

We need to find the probability for x > 1, which means P(x=2) + P(x=3) + ... and so on to infinity. However, we can simplify it by using the complementary probability. Since the sum of all probabilities equals 1, we can write P(x > 1) as 1 - P(x=0) - P(x=1): $$ P(x>1) = 1 - P(x=0) - P(x=1) \approx 1 - 0.135 - 0.270 \approx 0.595 $$

Calculate P(x=5)

Finally, we will calculate the probability for P(x=5) using the Poisson formula, with λ=2 and x=5: $$ P(x=5) = \frac{e^{-2} \cdot 2^5}{5!} = \frac{e^{-2} \cdot 32}{120} \approx 0.036 $$ So, we have obtained the following probabilities for the given Poisson distribution: P(x=0) ≈ 0.135 P(x=1) ≈ 0.270 P(x>1) ≈ 0.595 P(x=5) ≈ 0.036

Key concepts

probability theory

In our daily lives, we often talk about the 'chance' or 'likelihood' of events occurring. Probability theory is the branch of mathematics concerned with analyzing such random phenomena and quantifying the chances of various outcomes. At the heart of probability theory lies the concept of probability, which is a mathematical way of expressing the likelihood that a specific event will occur. It is always a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

For any given set of events, the sum of the probabilities must equal 1, ensuring that one of the possible outcomes will definitely occur. When solving probability problems, especially in homework exercises, we often utilize specific probability distributions, such as the Poisson distribution, to model and understand the behavior of rare events in a fixed interval of time or space.

Poisson formula

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known average rate and independently of the time since the last event. The formula for the Poisson probability mass function is given by:

\[\begin{equation} P(x;\text{mean}) = \frac{e^{-\text{mean}} \times \text{mean}^x}{x!},\tag{1}\refstepcounter{equation}\text{Equation (\theequation)}\[\begin{equation}where \(e\) is the base of the natural logarithm, \(\text{mean}\) is the average number of occurrences (often denoted by \(\lambda\)), and \(x!\) is the factorial of \(x\), representing the number of events for which we want to calculate the probability. The power of the Poisson formula lies in its ability to handle situations that involve counting occurrences of events and is handy in fields such as telecommunications, astronomy, finance, and even traffic flow analysis.

complementary probability

Complementary probability is a powerful concept used in probability theory to simplify the calculation of probabilities. It is based on the principle that the probability of an event occurring and the event not occurring must sum up to 1. In other words, the probability of an event's complement is equal to one minus the probability of the event itself.

When calculating probabilities, particularly when dealing with a large number or an infinite sequence of events, using the complement can be more straightforward. For instance, if it's hard to calculate the probability of 'more than a certain number' of events occurring, we can instead calculate the probability of 'that number or fewer' events occurring and subtract from 1. This approach is seen in the calculation of \(P(x>1)\) in our exercise, where calculating the probability of 2 or more events was accomplished by subtracting the probabilities of 0 and 1 events from 1, using the formula:

\[\begin{equation}P(A') = 1 - P(A),\tag{2}\refstepcounter{equation}\text{Equation (\theequation)}\[\begin{equation}where \(A\) is the event in question, and \(A'\) is its complement. This fundamental concept not only simplifies calculations but also reveals a more intuitive understanding of how probabilities relate to the concept of certainty and total possibility space.