Question

If the function \( f \) defined by \[ f(x) = \begin{cases} K(x - x^2), & 0<x<1
0, & \text{otherwise} \end{cases} \] is the p.d.f. of a random variable \( X \), then the value of \( P(X<\tfrac{1}{2}) \) is

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{2} \)
• C. \( \dfrac{1}{3} \)
• D. \( \dfrac{2}{3} \)

Hint:

Always determine the constant in a p.d.f. first using the total probability condition before evaluating probabilities.

Answer 1

The Correct Option is B

Step 1: Use the property of probability density function.
Since \( f(x) \) is a p.d.f., \[ \int_{0}^{1} K(x - x^2)\,dx = 1 \]

Step 2: Find the value of \( K \).
\[ K \int_{0}^{1} (x - x^2)\,dx = K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 \] \[ K \left( \frac{1}{2} - \frac{1}{3} \right) = K \left( \frac{1}{6} \right) = 1 \] \[ K = 6 \]

Step 3: Compute \( P(X<\tfrac{1}{2}) \).
\[ P(X<\tfrac{1}{2}) = \int_{0}^{1/2} 6(x - x^2)\,dx \] \[ = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^{1/2} \] \[ = 6 \left( \frac{1}{8} - \frac{1}{24} \right) \] \[ = 6 \times \frac{1}{12} = \frac{1}{2} \]

Step 4: Conclusion.
The required probability is \[ \boxed{\dfrac{1}{2}} \]