91Ó°ÊÓ

Conditional probability offers insight into how the probability of an event changes when we already know the outcome of another related event. In the context of the problem, we aim to find the conditional probability of selecting a number \(X\) given that another number \(Y\) has been chosen already. This is represented mathematically as \(P(X = x \mid Y = i)\). The expression can be read as "the probability of \(X = x\) given \(Y = i\)."

To determine this probability, we use the joint probability \(P(X = x, Y = y)\) and the marginal probability \(P(Y = y)\). The formula for conditional probability is:

• \(P(X = x \mid Y = i) = \frac{P(X = x, Y = i)}{P(Y = i)}\)

Here, \(P(X = x, Y = i)\) is the probability of the joint occurrence of \(X = x\) and \(Y = i\), while \(P(Y = i)\) is the probability that \(Y\) takes value \(i\). Calculating these probabilities helps us figure out the likelihood of different outcomes within certain conditions—a fundamental aspect of answering part b of the exercise.

Random variables \(X\) and \(Y\) are considered independent if the occurrence of \(X\) does not affect the probability of \(Y\), and vice versa. Mathematically, two random variables are independent if the joint probability \(P(X = x, Y = y)\) equals the product of their marginal probabilities \(P(X = x) P(Y = y)\) for all possible values of \(x\) and \(y\).

In our solution, we calculated these probabilities to check for independence but found a contradiction. For instance:

• \(P(X = 2, Y = 1) = \frac{1}{10}\)
• \(P(X = 2) = \frac{1}{5}\)
• \(P(Y = 1) = \frac{1}{3}\)

Multiplying the marginal probabilities: \(\frac{1}{5} \times \frac{1}{3} = \frac{1}{15}\), which does not equal \(\frac{1}{10}\). Because the joint and product probabilities are not equal, \(X\) and \(Y\) are not independent. This means the choice of \(X\) influences the potential choices for \(Y\), showing a dependency between the variables.

The marginal probability distribution provides probabilities of a single random variable by summing or integrating over the probabilities of the other variables. It's essential when analyzing joint probability because it helps us understand the behavior of one variable without considering the influence of another. For the random variable \(X\), the marginal probability function is computed by summing the probabilities of all joint outcomes that include that value of \(X\).

In this exercise, we needed to find \(P(X = x)\), which is simply \(\frac{1}{5}\) for each \(x\) from the set \(\{1, 2, 3, 4, 5\}\) because \(X\) is chosen uniformly. For \(Y\), calculating \(P(Y = y)\) requires us to add the probabilities of all combinations where \(Y\) assumes value \(y\). Each iteration over the example pairs showed:

• \(P(Y = 1) = \frac{1}{3}\)
• \(P(Y = 2) = \frac{13}{60}\)
• \(P(Y = 3)\), etc.

Understanding these distributions helps break down complex joint distributions, making the exploration of conditional probabilities more accessible.