Question

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

Short answer

The probability \(P_{X}(C_{2})\) is less than or equal to \(\frac{5}{8}\).

Step-by-step solution

Understanding the Problem

We should first understand that the total probability of the possible values of \(X\) is 1. This means that for all \(x\) in \(\mathcal{C} = \{x: 0<x<10\}\), when we add up the probability of each individual \(x\), we would get 1. \(C_{1}\) and \(C_{2}\) are two disjoint subsets of \(\mathcal{C}\) which means they don't share any common elements. This would be important in latter steps.

Writing Down the Total Probability

We can write down the total probability as \(P_{X}(\mathcal{C}) = P_{X}(C_{1}) + P_{X}(C_{2}) + P_{X}(C_{3}) = 1\). Here, \(C_{3}\) represents the remaining portion of \(\mathcal{C}\) not covered by \(C_{1}\) and \(C_{2}\). It is represented by two intervals: \(0 < x \leq 1\) and \(5 \leq x <10\). We are given that \(P_{X}(C_{1}) = \frac{3}{8}\). Let's plug this in the equation.

Substituting Known Values And Solving for \(P_{X}(C_2)\)

Replacing the known value in the equation from step 2 gives: \(1 = \frac{3}{8} + P_{X}(C_2) + P_{X}(C_{3})\). We want to find an upper limit for \(P_{X}(C_{2})\), so lets set \(P_{X}(C_{3})\) to its smallest possible value, which is 0 (as probabilities cannot be negative). Then, we get \(1 = \frac{3}{8} + P_{X}(C_2)\), solving for \(P_{X}(C_{2})\) gives \(P_{X}(C_{2}) \leq \frac{5}{8}\). This shows the required inequality.