Problem 91 - Calculus for Biology and Medicine

Question

\(X\) and \(Y\) are independent and Poisson with mean 3 . (a) Find \(P(X+Y=2)\). (b) Given that \(X+Y=2\), find the probability that \(X=k\) for \(k=0,1\), and \(2 .\)

Step-by-Step Solution

Answer

(a) \(P(X+Y=2) = 18e^{-6}\). (b) \(P(X=0|X+Y=2)=\frac{1}{4}, \ P(X=1|X+Y=2)=\frac{1}{2}, \ P(X=2|X+Y=2)=\frac{1}{4}.\)

Step 1: Understanding the Problem

We are given that both random variables \(X\) and \(Y\) are independent Poisson random variables with a mean of 3. We need to find two things: first, the probability that \(X+Y=2\), and second, the conditional probabilities of \(X=k\) given that \(X+Y=2\) where \(k=0, 1,\) and \(2\).

Step 1: Probability Distribution of Poisson Random Variable

The probability of a Poisson random variable with mean \(\lambda\) taking the value \(n\) is given by the formula \[P(X=n) = \frac{e^{-\lambda} \lambda^n}{n!}.\] Since both \(X\) and \(Y\) are Poisson with mean 3, we'll use \(\lambda = 3\) for both.

Step 2: Determining Distribution of \(X+Y\)

The sum \(X+Y\) of two independent Poisson random variables with parameters \(\lambda_1\) and \(\lambda_2\) is itself a Poisson random variable with parameter \(\lambda_1 + \lambda_2\). Thus, \(X+Y\) is Poisson with mean \(3 + 3 = 6\).

Step 3: Finding \(P(X+Y=2)\)

Using the formula for the Poisson distribution, we find \(P(X+Y=2) = \frac{e^{-6} 6^2}{2!} = \frac{e^{-6} \cdot 36}{2} = 18e^{-6}.\)

Step 4: Conditional Probability \(P(X=k | X+Y=2)\)

The conditional probability \(P(X=k | X+Y=2)\) can be determined by using the formula \(P(X=k | X+Y=2) = \frac{P(X=k, Y=2-k)}{P(X+Y=2)}\).

Step 5: Calculating \(P(X=k, Y=2-k)\) for \(k=0, 1,\) and \(2\)

For each \(k\): - \(k=0\): \(P(X=0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3}\), \(P(Y=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9e^{-3}}{2}\), so \(P(X=0, Y=2) = e^{-6} \cdot \frac{9}{2}\).- \(k=1\): \(P(X=1) = \frac{e^{-3} \cdot 3^1}{1!} = 3e^{-3}\), \(P(Y=1) = 3e^{-3}\), so \(P(X=1, Y=1) = 9e^{-6}\).- \(k=2\): \(P(X=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9e^{-3}}{2}\), \(P(Y=0) = e^{-3}\), so \(P(X=2, Y=0) = \frac{9e^{-6}}{2}\).

Step 6: Calculating Each Conditional Probability

Now substitute back to calculate each conditional probability:- \(P(X=0 | X+Y=2) = \frac{9e^{-6}/2}{18e^{-6}} = \frac{1}{4}.\)- \(P(X=1 | X+Y=2) = \frac{9e^{-6}}{18e^{-6}} = \frac{1}{2}.\)- \(P(X=2 | X+Y=2) = \frac{9e^{-6}/2}{18e^{-6}} = \frac{1}{4}.\)

Key Concepts

Conditional Probability

Conditional probability is a fundamental concept in probability theory. It refers to the likelihood of an event occurring given that another event has already happened. In our exercise, we are tasked with finding the probability that \(X = k\) given that \(X+Y=2\). This scenario is particularly useful in focusing on specific outcomes and their relationships, rather than looking at all possibilities.

To calculate conditional probability, we can use the formula:

• \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

Here, \(P(A|B)\) is the probability of event \(A\) given event \(B\) has occurred, \(P(A \cap B)\) is the joint probability of both events happening, and \(P(B)\) is the probability of \(B\) occurring. In our case, we determine \(P(X=k | X+Y=2)\) by finding the joint probability of each \(X = k\) and \(Y = 2-k\), divided by \(P(X+Y=2)\).

This method allows us to narrow down the broader scenario to a more specific and detailed context.

Independent Random Variables

In probability, two random variables are independent if the occurrence of one does not affect the probability distribution of the other. This concept is essential when dealing with complex probability problems, as it allows one to break down multi-variable problems into simpler, separate functions.

For independent random variables \(X\) and \(Y\), the joint probability is the product of their individual probabilities:

• \(P(X=n, Y=m) = P(X=n) \times P(Y=m)\)

In the given exercise, \(X\) and \(Y\) are independent Poisson random variables. This independence simplifies our calculations because we can treat each variable separately when calculating probabilities. By knowing the probabilities of \(X\) and \(Y\) individually, we can find joint or combined probabilities, such as \(P(X=k, Y=2-k)\), without needing to consider how the variables might interact with one another.

This simplifies solving problems that involve the sum of random variables, as seen in our exercise.

Probability Distribution

Probability distributions describe how the outcomes of a random variable are spread out over different possible values. In our exercise, we're dealing with the Poisson distribution, which is commonly used for modeling counts of events that occur independently over a given period.

The Poisson distribution is characterized by its mean, or parameter \(\lambda\), which reflects the average rate of occurrence. The probability that a Poisson random variable \(X\) takes on a specific value \(n\) is given by:

• \(P(X=n) = \frac{e^{-\lambda} \lambda^n}{n!}\)

This formula gives us the distribution of probabilities across all possible values of \(X\). For any pair of independent Poisson variables, the sum \(X+Y\) follows another Poisson distribution with a parameter equal to the sum of their individual means.

In this exercise, both \(X\) and \(Y\) have means of 3. Therefore, when determining \(P(X+Y=2)\), we treat \(X+Y\) as a new Poisson variable with a mean \(6\), which allows us to find the probabilities associated with specific outcomes using the Poisson probability formula. By understanding these distributions, we can predict the likelihood of different outcomes in random processes.