Question

Let \(X\) have the pdf \(f(x)=2 x, 0

Short answer

The probability that \(X\) is at least \(\frac{3}{4}\) given that \(X\) is at least \(\frac{1}{2}\) is the result obtained in Step 4 after calculating the necessary integrals and performing the division.

Step-by-step solution

Understanding the Probability Density Function

The pdf of \(X\), \(f(x)=2x\) for \(0<x<1\), demonstrates the probability distribution of the random variable \(X\). It's crucial to understand the values of the pdf don't give probabilities but density of probabilities. Probabilities can be obtained by integrating the pdf over a range of values. Here, the pdf is 0 for values of \(x\) outside the interval (0,1).

Compute Integral for Denominator of the Conditional Probability Formula

We need to consider the condition that \(X\) is at least \(\frac{1}{2}\). This is represented by the integral \(\int_{1/2}^{1} f(x) dx\), we solve this by integrating our pdf from \(\frac{1}{2}\) to 1.

Compute Integral for Numerator of the Conditional Probability Formula

We need to find out the value when \(X\) is at least \(\frac{3}{4}\). This is represented by the integral \(\int_{3/4}^{1} f(x) dx\), we solve this by integrating our pdf from \(\frac{3}{4}\) to 1.

Compute Conditional Probability

Finally, the conditional probability can be computed by dividing the result from step 3 by the result from step 2. This gives us \(P(X \geq 3/4 | X \geq 1/2)\).