Problem 92 - Calculus for Biology and Medicine

Question

\(X\) is Poisson distributed with mean 2 , and \(Y\) is Poisson distributed with mean 3 . (a) Find \(P(X+Y=4)\) (b) Given that \(X+Y=1\), find the probability that \(X=1\).

Step-by-Step Solution

Answer

(a) Approximately 0.1755; (b) 0.

Step 1: Define the Distribution of X + Y

Since both \( X \) and \( Y \) are independently Poisson distributed, the sum \( X + Y \) is also Poisson distributed. The mean of \( X + Y \) is the sum of the means of \( X \) and \( Y \), which is \( 2 + 3 = 5 \). So, \( X + Y \) is Poisson distributed with mean 5.

Step 2: Calculate P(X+Y=4)

The probability that a Poisson-distributed random variable equals \( k \) is given by \[P(X+Y = k) = \frac{{\lambda^k e^{-\lambda}}}{{k!}}\]where \( \lambda \) is the mean. Thus, for \( X + Y = 4 \) with mean 5, \[P(X+Y=4) = \frac{{5^4 e^{-5}}}{{4!}} = \frac{{625 e^{-5}}}{{24}}.\] Calculating this gives approximately 0.1755.

Step 3: Express Conditional Probability

We need to find \( P(X = 1 \mid X + Y = 1) \), which can be expressed using the formula for conditional probability: \[P(A \mid B) = \frac{{P(A \cap B)}}{{P(B)}}.\]In this context, \[A = \{X = 1\} \text{ and } B = \{X + Y = 1\}\].

Key Concepts

Understanding Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring, given that another event has already happened. Imagine you're solving a mystery: knowing one piece of the puzzle impacts how likely other pieces fit in.

For instance, in our exercise, we're tasked with finding the probability that event \(X = 1\) happens, given that \(X + Y = 1\). This is symbolized as \(P(X = 1 \mid X + Y = 1)\). Here, the event \(X + Y = 1\) is the given condition, like our clue in a detective story.

Using the formula for conditional probability:

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

• \(A\) is the event that \(X = 1\)
• \(B\) is the event that \(X + Y = 1\)

This formula essentially demands that we find the probability that both \(X = 1\) and \(X + Y = 1\) occur simultaneously (\(P(A \cap B)\)), then divide by the probability of the condition alone happening (\(P(B)\)).

Understanding conditional probability allows us to make more accurate predictions based on our existing knowledge.

Exploring Probability Distributions

Probability distributions represent the chance of different outcomes in a random scenario. Think of them as maps of uncertainty, showing us which results are more or less likely.

The Poisson distribution, in particular, is useful for counting outcomes in a fixed space or time, especially when these events happen independently at a constant rate. Imagine counting the number of cars passing through a toll booth in an hour.

In our exercise, both \(X\) and \(Y\) follow Poisson distributions, with means of 2 and 3, respectively. This means:

• The average number of occurrences for \(X\) is 2.
• The average number of occurrences for \(Y\) is 3.

A remarkable property of independent Poisson variables is that their sum also follows a Poisson distribution. Hence, \(X + Y\) has a mean of 5 (since \(2 + 3 = 5\)), and it's described by a Poisson distribution too.

Understanding these distributions helps in calculating probabilities, such as \(P(X + Y = 4)\), using the formula:

• \(P(X+Y = k) = \frac{{\lambda^k e^{-}}}{{k!}}\)

This formula efficiently tells us the likelihood of the sum of events being exactly 4.

Grasping Independent Random Variables

When we say random variables are independent, we mean that the outcome of one does not affect the outcome of another. They're like two solitary dice rolls: the numbers don't influence each other.

In the problem we're dealing with, \(X\) and \(Y\) are independent random variables, each following its own Poisson distribution. This property makes it simpler to work with their probabilities since one variable doesn’t mess with the other.

In practical terms:

• The probability of \(X\) producing its outcomes stays the same, no matter the outcome of \(Y\).
• Conversely, \(Y\) acts independently from \(X\) too.

Because of their independence, when we find probabilities like \(P(X + Y = k)\), we can simply keep summing their individual means. This summation feature is rooted in their Poisson nature and independence.

By understanding independence, we not only simplify complex problems, but also ensure that our calculated probabilities reliably reflect reality.