Question

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

Short answer

The probability \(P_{X}\left(D_{2}\right)\) is \(\frac{3}{4}\)

Step-by-step solution

Recognize the Total Probability Space

The random variable \(X\) can take on any value between 0 and 1, exclusive, as defined by the set \(\mathcal{D}\). But given the sets \(D_{1}\) and \(D_{2}\), these two sets cover all possible values that \(X\) can take on, meaning \(D_{1} \cup D_{2} = \mathcal{D}\). Therefore, the total probability should be 1, i.e., \(P_{X}\left(D_{1}\right) + P_{X}\left(D_{2}\right) = 1\)

Solve for the Unknown Probability

We are given \(P_{X}\left(D_{1}\right)=\frac{1}{4}\) and we are asked to find \(P_{X}\left(D_{2}\right)\). Substitute \(P_{X}\left(D_{1}\right)\) into the total probability equation from step 1 and solve for \(P_{X}\left(D_{2}\right)\), therefore, \(P_{X}\left(D_{2}\right) = 1 - \frac{1}{4} = \frac{3}{4}\).