Problem 28 - Calculus for Biology and Medicine

Question

Let \(X\) and \(Y\) be two random variables with the following joint distribution: $$\begin{array}{ccc} \hline & X=0 & X=1 \\ \hline \boldsymbol{Y}=\mathbf{0} & 0.2 & 0.0 \\ \boldsymbol{Y}=\mathbf{1} & 0.3 & 0.5 \\ \hline \end{array}$$ (a) Find \(P(X=0, Y=1)\). (b) Find \(P(X=0)\). (c) Find \(P(Y=1)\). (d) Find \(P(X=0 \mid Y=0)\).

Step-by-Step Solution

Answer

(a) 0.3, (b) 0.5, (c) 0.8, (d) 1.0.

Step 1: Identify Known Probabilities

From the table given, identify the joint and marginal probabilities. The probability provided directly in the table that matches our need for part (a) is \( P(X=0, Y=1) = 0.3 \).

Step 2: Calculate Marginal Probability of X=0

To find \( P(X=0) \), sum the probabilities for all outcomes where \( X=0 \). From the table, \( P(X=0) = P(X=0, Y=0) + P(X=0, Y=1) = 0.2 + 0.3 = 0.5 \).

Step 3: Calculate Marginal Probability of Y=1

To find \( P(Y=1) \), sum the probabilities for all outcomes where \( Y=1 \). From the table, \( P(Y=1) = P(X=0, Y=1) + P(X=1, Y=1) = 0.3 + 0.5 = 0.8 \).

Step 4: Calculate Conditional Probability P(X=0 | Y=0)

The conditional probability \( P(X=0 \mid Y=0) \) is found by dividing \( P(X=0, Y=0) \) by \( P(Y=0) \). First, find \( P(Y=0) = P(X=0, Y=0) + P(X=1, Y=0) = 0.2 + 0.0 = 0.2 \). Then, \( P(X=0 \mid Y=0) = \frac{P(X=0, Y=0)}{P(Y=0)} = \frac{0.2}{0.2} = 1 \).

Key Concepts

Joint Distribution

In probability theory, the joint distribution of two random variables provides a comprehensive overview of their possible combinations and their corresponding probabilities. Let's explore this concept using the example of two random variables, \(X\) and \(Y\), with provided probabilities.

• The joint distribution is essentially a map that assigns a probability to each pair of outcomes \((x, y)\) you could possibly see with \(X\) and \(Y\).
• In the given exercise, the joint distribution table explicitly provides probabilities for each combination of \(X\) and \(Y\). For instance, the probability that \(X=0\) and \(Y=1\) occurs simultaneously is \(P(X=0, Y=1) = 0.3\).

This table format offers a quick snapshot of all potential outcomes and their likelihood, crucial for solving problems involving more complex distributions or dependencies between variables.

Marginal Probability

Marginal probability gives us insight into the likelihood of an individual event without considering the outcomes of the other variables involved. Essentially, it is derived from the joint distribution by summing probabilities over the other variable.

• For the exercise, to find \(P(X=0)\), we sum all probabilities where \(X=0\), irrespective of \(Y\). This results in \(P(X=0) = P(X=0, Y=0) + P(X=0, Y=1) = 0.2 + 0.3 = 0.5\).
• Similarly, \(P(Y=1)\) is found by adding all probabilities where \(Y=1\), thus giving us \(P(Y=1) = P(X=0, Y=1) + P(X=1, Y=1) = 0.3 + 0.5 = 0.8\).

The marginal probability provides a foundational understanding of each variable's behavior independently, which is particularly useful when there's no need to consider the joint occurrences.

Conditional Probability

Conditional probability expresses the likelihood of an event occurring given the backdrop of another event that has already taken place. It is crucial for understanding how knowledge about one variable influences the probability of another.

• In the exercise example, to find \(P(X=0 \mid Y=0)\), we determine the probability of \(X=0\) specifically under the condition that \(Y\) is known to be 0.
• First, we calculate \(P(Y=0)\) by summing the relevant joint probabilities: \(P(Y=0) = P(X=0, Y=0) + P(X=1, Y=0) = 0.2 + 0.0 = 0.2\).
• Subsequently, the conditional probability is given by dividing the pertinent joint probability by the marginal probability: \(P(X=0 \mid Y=0) = \frac{P(X=0, Y=0)}{P(Y=0)} = \frac{0.2}{0.2} = 1\).

Understanding conditional probability is key in many fields, as it helps delineate the impact one set of information has in altering the anticipated outcome for another variable.