Question

The p.d.f. of a continuous random variable \( X \) is given by \[ f(x) = \begin{cases} \frac{1}{2} & \text{if } 0<x<2, \\ 0 & \text{otherwise}, \end{cases} \] and if \[ a = P(X \leq \frac{1}{2}), \quad b = P(X>\frac{1}{2}), \] then the relation between \( a \) and \( b \) is

• A. \( a - b = 0 \)
• B. \( 2a - b = 0 \)
• C. \( 3a - b = 0 \)
• D. \( a - 2b = 0 \)

Hint:

For probability distributions, the sum of the probabilities for all intervals should be 1.

Answer 1

The Correct Option is A

Step 1: Calculate \( a \) and \( b \).
We are given that \( f(x) = \frac{1}{2} \) for \( 0<x<2 \). The total probability is 1, so \[ P(X \leq \frac{1}{2}) = a = \int_0^{1/2} \frac{1}{2} \, dx = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] Similarly, \[ P(X>\frac{1}{2}) = b = \int_{1/2}^2 \frac{1}{2} \, dx = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \] Step 2: Use the relation \( a + b = 1 \).
Since the total probability is 1, we have \[ a + b = 1 \] Substituting the values of \( a \) and \( b \), \[ \frac{1}{4} + \frac{3}{4} = 1 \] Step 3: Conclusion.
Thus, the relation between \( a \) and \( b \) is \( a - b = 0 \).