Question

Let the space of the random variable \(X\) be \(\mathcal{C}=\\{x: 0

Short answer

\(P_{X}\left(C_{2}\right) = \frac{5}{8}\). So, certainly, \(P_{X}\left(C_{2}\right) \leq \frac{5}{8}\). Thus, the condition is always true.

Step-by-step solution

Understanding the Space of the Random Variable

The space of the random variable \(X\) is \(\mathcal{C}=\{x: 0<x<10\}\). So the random variable \(X\) can take any value between 0 and 10. The set \(C_{1}=\{x: 1<x<5\}\) represents all values such that 1 < x < 5, and \(P_{X}\left(C_{1}\right)=\frac{3}{8}\), which means the probability of the set \(C_{1}\) is \(\frac{3}{8}\).

Deriving \(P_{X}\left(C_{2}\right)\)

The set \(C_{2}\) is defined as \(C_{2}=\{x: 5 \leq x < 10\}\) and it represents all values such that 5 <= x < 10. Since \(C_{1}\) and \(C_{2}\) are disjoint sets (they don't have any common elements) and they together form the whole space of the random variable \(X\), we know that \(P_{X}(\mathcal{C}) = P_{X}(C_{1} \cup C_{2})\). Since \(C_{1}\) and \(C_{2}\) are disjoint, the probability of their union is the sum of their probabilities, so \(P_{X}(\mathcal{C}) = P_{X}(C_{1}) + P_{X}(C_{2})\).

Solving for \(P_{X}\left(C_{2}\right)\)

Knowing that \(P_{X}(\mathcal{C}) = 1\), we substitute the known values: \(1 = \frac{3}{8} + P_{X}(C_{2})\). Solving for \(P_{X}(C_{2})\), we get \(P_{X}(C_{2}) = 1 - \frac{3}{8} = \frac{5}{8}\).