Question

Consider the probability space \( (\Omega, \mathcal{G}, P) \), where \( \Omega = \{1, 2, 3, 4\} \), \[ \mathcal{G} = \{\emptyset, \Omega, \{4\}, \{2, 3\}, \{1, 4\}, \{1, 2, 3\}, \{2, 3, 4\}\}, \] and \( P(\{1\}) = \frac{1}{4} \). Let \( X \) be the random variable defined on the above probability space as \[ X(1) = 1, X(2) = X(3) = 2, X(4) = 3. \] If \( P(X \leq 2) = \frac{3}{4} \), then \( P(\{1, 4\}) \) (rounded off to two decimal places) equals ................

Hint:

- In a probability space, the total probability is 1. Use this property to calculate missing probabilities.
- The probability of events can be found by adding the probabilities of the individual elements that make up the event.

Answer 1

We are given the probability space \( (\Omega, \mathcal{G}, P) \) and the random variable \( X \). We need to calculate \( P(\{1, 4\}) \).
First, calculate \( P(X \leq 2) \). The values of \( X \) are:
- \( X(1) = 1 \) (for \( \{1\} \)),
- \( X(2) = 2 \) (for \( \{2\} \)),
- \( X(3) = 2 \) (for \( \{3\} \)),
- \( X(4) = 3 \) (for \( \{4\} \)).
The event \( X \leq 2 \) corresponds to \( \{1, 2, 3\} \), so: \[ P(X \leq 2) = P(\{1, 2, 3\}) = P(\{1\}) + P(\{2\}) + P(\{3\}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \] Now, we calculate \( P(\{1, 4\}) \). Since \( \{1, 4\} \) corresponds to the events \( X(1) \) and \( X(4) \), we have: \[ P(\{1, 4\}) = P(\{1\}) + P(\{4\}) = \frac{1}{4} + P(\{4\}). \] To determine \( P(\{4\}) \), we use the fact that \( P(\Omega) = 1 \), and \( P(\{1\}) + P(\{2\}) + P(\{3\}) + P(\{4\}) = 1 \): \[ \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + P(\{4\}) = 1 ⇒ P(\{4\}) = \frac{1}{4}. \] Thus: \[ P(\{1, 4\}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. \] Final Answer: The value of \( P(\{1, 4\}) \) is \( \boxed{0.50} \).