Problem 93 - Calculus for Biology and Medicine

Question

Let \(X\) and \(Y\) be independent Poisson distributed variables with means 4 and 2 respectively. Calculate \(P(X=2 \mid X+Y=3)\).

Step-by-Step Solution

Answer

\(P(X=2 \mid X+Y=3) = \frac{4}{9}\).

Step 1: Understand the problem

We need to calculate the conditional probability \(P(X = 2 \mid X + Y = 3)\). Both \(X\) and \(Y\) are independent Poisson distributed variables, where \(X\) has a mean of 4 and \(Y\) has a mean of 2.

Step 2: Apply the law of total probability

The probability \(P(X = 2 \mid X + Y = 3)\) can be calculated using the formula \(P(X = 2 \mid X + Y = 3) = \frac{P(X = 2, Y = 1)}{P(X + Y = 3)}\). Therefore, we need to find \(P(X = 2, Y = 1)\) and \(P(X + Y = 3)\).

Step 3: Calculate \(P(X=2, Y=1)\)

Since \(X\) and \(Y\) are independent, \(P(X = 2, Y = 1) = P(X = 2) \cdot P(Y = 1)\). For a Poisson distribution with mean \(\lambda\), \(P(Z = k) = \frac{e^{-\lambda} \lambda^k}{k!}\).\[P(X=2) = \frac{e^{-4} 4^2}{2!} = \frac{e^{-4} \times 16}{2} = 8e^{-4}\]\[P(Y=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}\]So, \(P(X=2, Y=1) = 8e^{-4} \cdot 2e^{-2} = 16e^{-6}\).

Step 4: Calculate \(P(X+Y=3)\)

For independent Poisson variables, \(X + Y\) is also a Poisson variable with mean \(4 + 2 = 6\). So, \(P(X+Y=3) = \frac{e^{-6} 6^3}{3!} = \frac{e^{-6} \times 216}{6} = 36e^{-6}\).

Step 5: Calculate the Conditional Probability

Using the values from the previous steps, \(P(X = 2 \mid X + Y = 3) = \frac{16e^{-6}}{36e^{-6}} = \frac{16}{36} = \frac{4}{9}\).

Key Concepts

Poisson Distribution

The Poisson distribution is a probability distribution that describes the likelihood of a given number of events happening in a fixed interval of time or space. This distribution is particularly used when these events occur with a known constant mean rate and are independent of the time since the last event. For a random variable \(Z\) following a Poisson distribution with mean \(\lambda\), the probability that \(Z\) takes on a value \(k\) is given by:

• \(P(Z = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)

In the given exercise, the random variable \(X\) follows a Poisson distribution with a mean of 4, and random variable \(Y\) has a mean of 2. This means:

• For \(X\), \(P(X = 2)\) is calculated as \(\frac{e^{-4} \times 4^2}{2!}\).
• For \(Y\), \(P(Y = 1)\) is computed as \(\frac{e^{-2} \times 2^1}{1!}\).

These calculations are essential because they provide the building blocks to solve problems involving Poisson distributed variables, like the one in this exercise. The formula includes exponential terms \(e^{-\lambda}\) and factorials, which cater to the characteristics of events happening independently and randomly.

Independent Random Variables

In probability theory, two random variables are said to be independent if the occurrence of one does not affect the probability of occurrence of the other. Independence simplifies the computation of joint probabilities of such random events.
For independent Poisson distributed variables like \(X\) and \(Y\) in our problem, the joint probability \(P(X = 2, Y = 1)\) can be expressed as the product of their individual probabilities:

• \(P(X = 2, Y = 1) = P(X = 2) \cdot P(Y = 1)\)

This is a crucial property utilized in many probability problems because it allows us to handle complex scenarios by breaking them down into simpler independent parts. Since \(X\) and \(Y\) have no influence over each other, their joint probability becomes the product of their individual probabilities. This characteristic of independence plays a significant role in calculating the necessary terms to eventually find the solution for the conditional probability in our exercise.

Law of Total Probability

The Law of Total Probability is a fundamental concept that relates marginal probabilities to conditional probabilities. It is used to find the probability of an event based on known probabilities of other disjoint events.
In the context of our exercise, we utilized this law to help compute the conditional probability \(P(X = 2 \mid X + Y = 3)\). This can be expressed by:

• \(P(X = 2 \mid X + Y = 3) = \frac{P(X = 2, Y = 1)}{P(X + Y = 3)}\)

Here, \(P(X = 2, Y = 1)\) is the joint probability that both \(X\) and \(Y\) take particular values, while \(P(X + Y = 3)\) is the total probability that the sum of \(X\) and \(Y\) equals 3.
The Law of Total Probability is extremely useful in scenarios where you need to account for all possible ways an event can occur. By putting it to use, we could arrive at the solution required in this exercise in a streamlined and logical manner, allowing us to understand how individual probabilities affect the conditional outcome.