Question

Let the subsets \(C_{1}=\left\\{\frac{1}{4}

Short answer

\(P_{X}(C_{1} \cup C_{2}) = \frac{5}{8}\), \(P_{X}(C_{1}^{c}) = \frac{7}{8}\), and \(P_{X}(C_{1}^{c} \cap C_{2}^{c}) = \frac{3}{8}\)

Step-by-step solution

Finding \( P_{X}(C_{1} \cup C_{2}) \)

Using the addition rule of probability, \( P_{X}(C_{1} \cup C_{2}) = P_{X}(C_{1}) + P_{X}(C_{2}) \). Given that \( P_{X}(C_{1}) = \frac{1}{8} \) and \( P_{X}(C_{2}) = \frac{1}{2} \), the sum of these two probabilities gives \( P_{X}(C_{1} \cup C_{2}) = \frac{5}{8} \).

Finding \( P_{X}(C_{1}^{c}) \)

The complement rule states that \( P(A^{c}) = 1 - P(A) \). Hence, \( P_{X}(C_{1}^{c}) = 1 - P_{X}(C_{1}) = 1 - \frac{1}{8} = \frac{7}{8} \).

Finding \( P_{X}(C_{1}^{c} \cap C_{2}^{c}) \)

We know that \( (C_{1}^{c} \cap C_{2}^{c}) = (C_{1} \cup C_{2})^{c} \). Hence, using the rule of complements, \( P_{X}(C_{1}^{c} \cap C_{2}^{c}) = P_{X}((C_{1} \cup C_{2})^{c}) = 1 - P_{X}(C_{1} \cup C_{2}) = 1 - \frac{5}{8} = \frac{3}{8} \).