Problem 93 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The probability is \( \frac{4}{9} \).

Step 1: Understand Poisson Distribution

In a Poisson distribution, the probability of exactly k events occurring in a fixed interval of time or space is given by \( P(Z = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( \lambda \) is the average number of events. Here, \( X \sim Pois(4) \) and \( Y \sim Pois(2) \).

Step 2: Calculate Joint Distribution

Since \( X \) and \( Y \) are independent Poisson variables, \( X+Y \) follows a Poisson distribution with mean \( \lambda = 4 + 2 = 6 \). Therefore, \( X+Y \sim Pois(6) \).

Step 3: Conditional Probability Formula

We need to calculate \( P(X=2 \mid X+Y=3) \). Using conditional probability, it can be written as \( P(X=2 \mid X+Y=3) = \frac{P(X=2 \cap X+Y=3)}{P(X+Y=3)} \).

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

Using the Poisson formula, we find the probability for \( X+Y = 3 \):\[ P(X+Y=3) = \frac{e^{-6} \cdot 6^3}{3!} = \frac{e^{-6} \cdot 216}{6} = 36e^{-6}. \]

Step 5: Calculate \( P(X=2 \cap X+Y=3) \)

If \( X = 2 \) and \( X+Y = 3 \), then \( Y \) must be 1. Compute \( P(X=2) \) and \( P(Y=1) \) separately: \[ P(X=2) = \frac{e^{-4} \cdot 4^2}{2!} = 8e^{-4} \] and \[ P(Y=1) = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2}. \] The joint probability \( P(X=2 \cap Y=1) \) is \[ P(X=2 \cap Y=1) = P(X=2) \cdot P(Y=1) = 8e^{-4} \cdot 2e^{-2} = 16e^{-6}. \]

Step 6: Calculate Conditional Probability

Substitute the values into the conditional probability formula: \[ P(X=2 \mid X+Y=3) = \frac{P(X=2 \cap X+Y=3)}{P(X+Y=3)} = \frac{16e^{-6}}{36e^{-6}} = \frac{16}{36} = \frac{4}{9}. \]

Step 7: Conclusion

Thus, the probability that \( X=2 \) given that the sum \( X+Y=3 \) is \( \frac{4}{9} \).

Key Concepts

conditional probability

Conditional probability helps us measure the likelihood of an event, given that another event has occurred. In this exact problem, we calculate the probability that the number of occurrences of one event (say, Event X = 2) occurs alongside another related event (Event X+Y = 3). This is denoted as \( P(A \mid B) \), which stands for the probability of event A occurring given event B has occurred.
For our exercise, we've used the formula:

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

Here, it's crucial to understand both parts:

• The numerator, \( P(X = 2 \cap X + Y = 3) \), is the probability that both X equals 2 and the joint event sum \( X + Y \) equals 3.
• The denominator, \( P(X + Y = 3) \), is the probability of the total sum of X and Y equaling 3.

By taking these probabilities, which we calculate using their respective Poisson formulas, we can find the conditional probability that tells us how likely it is for X to be 2 given the sum is 3.

joint distribution

In probability, a joint distribution gives us a way to consider two random variables at the same time. For independent Poisson variables, like our variables \( X \) and \( Y \), the joint distribution is crucial in calculating related probabilities.
For this exercise, because \( X \) and \( Y \) are independent, their joint distribution means we can combine them. Specifically, we find the distribution of their sum \( X + Y \). Here, since both variables are Poisson distributed, the result is also a Poisson distribution, but with a mean that equals the sum of the individual means:
\[ X + Y \sim Pois(4 + 2 = 6) \]
Joint distribution for independent variables simplifies to multiplying the individual distributions, thus allowing us to compute the probability of two occurrences happening together. This comes particularly handy in solving a part of our problem related to finding \( P(X = 2 \cap Y = 1) \).

independent variables

Understanding independent variables is essential when dealing with probability distributions. Two random variables are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. In simpler terms, knowing the outcome of one variable gives no information about the outcome of the other.
In the context of our exercise, \( X \) and \( Y \) are said to be independent Poisson variables. What this means is:

• The probability of \( X \) taking on a particular value is unaffected by what value \( Y \) turns out to be.
• Similarly, \( Y \)'s distribution does not influence \( X \)'s outcomes.

In calculations, this independence tells us that we can calculate their probabilities separately and, for their joint events, simply multiply these probabilities. Calculating the joint probability \( P(X = 2 \cap Y = 1) \) as independent occurrences perfectly illustrates this principle - each event's probability is determined individually and multiplied because the events are independent.